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Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 16 Chemistry in Everyday Life Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 16 Chemistry in Everyday Life

1. Choose correct option

Question A.
Oxidative Rancidity is …………….. reaction
a. addition
b. subtitution
c. Free radical
d. combination
Answer:
c. Free radical

Question B.
Saponification is carried out by ……………..
a. oxidation
b. alkaline hydrolysis
c. polymarisation
d. Free radical formation
Answer:
b. alkaline hydrolysis

Question C.
Aspirin is chemically named as ……………..
a. Salicylic acid
b. acetyl salicylic acid
c. chloroxylenol
d. thymol
Answer:
b. acetyl salicylic acid

Question D.
Find odd one out from the following
a. dettol
b. chloroxylenol
c. paracetamol
d. trichlorophenol
Answer:
c. paracetamol

Question E.
Arsenic based antibiotic is
a. Azodye
b. prontosil
c. salvarsan
d. sulphapyridine
Answer:
c. salvarsan

Question F.
The chemical used to slow down the browning action of cut fruit is
a. SO₃
b. SO₂
c. H₂SO₄
d. Na₂CO₃
Answer:
b. SO₂

Question G.
The chemical is responsible for the rancid flavour of fats is
a. Butyric acid
b. Glycerol
c. Protein
d. Saturated fat
Answer:
a. Butyric acid

Question H.
Health benefits are obtained by consumption of
a. Saturated fats
b. trans fats
c. monounsaturated fats
d. all of these
Answer:
c. monounsaturated fats

2. Explain the following :

Question A.
Cooking makes food easy to digest.
Answer:

• During the cooking process, high polymers of carbohydrates or proteins are hydrolysed to smaller polymeric units.
• The uncooked food mixture is a heterogeneous suspension which becomes a colloidal matter on cooking.
• As a result, the constituent nutrient molecules present in cooked food are smaller in size and hence, easier to digest, than the uncooked food.

Hence, cooking makes food easy to digest.

Question B.
On cutting some fruits and vegetables turn brown.
Answer:
i. Cutting of fruits and vegetables damage the cells, resulting in release of chemicals.
ii. Depending on the pH of fruits/vegetables, polyphenols are released.
iii. Due to the action of an enzyme, these polyphenols react with oxygen present in the air and get oxidised to form quinones.

iv. Quinones further undergo reactions including polymerization, which results in the formation of brown coloured products called as tannins.
Thus, on cutting, some fruits and vegetables turn brown.

Question C.
Vitmin E is added to packed edible oil.
Answer:

• Vitamin E is a very effective natural antioxidant.
• The phenolic – OH group present in the structure of vitamin E is responsible for its antioxidant activity.
• Also, the long chain of saturated carbon atoms makes it fat soluble.

Therefore, when vitamin E is added to packed edible oil, it prevents the oxidative rancidity of the oil.

Question D.
Browning of cut apple can be prolonged by applying lemon juice.
Answer:

• Browning of cut apple is due to the oxidation of polyphenols at a particular pH to quinones, which further undergoes polymerization to form brown coloured tannins.
• This browning reaction can be prolonged or slowed down by using reducing agents or by changing the pH.
• Applying lemon juice (i.e., citric acid) on the cut apple, lowers the pH at the surface of the apple. This prevents the oxidation reaction. Thus, browning of cut apple can be prolonged by applying lemon juice.

Question E.
A diluted solution (4.8 % w/v) of 2,4,6-trichlorophenol is employed as antiseptic.
Answer:

• 2,4,6-Trichlorophenol (TCP) is more potent antiseptic than phenol.
• It has low corrosive effects as compared to phenol, if used in lower concentrations.

Hence, diluted solution (4.8% w/v) of 2,4,6-trichlorophenol is used as antiseptic.

Question F.
Turmeric powder can be used as antiseptic.
Answer:

• Turmeric powder contains an active ingredient called curcumin.
• Curcumin has antiseptic properties; thus, it is used for wound healing or applied on bruise.

Hence, turmeric powder can be used as antiseptic.

3. Identify the functional groups in the following molecule :

Answer:

4. Give two differences between the following

Question A.
Disinfectant and antiseptic
Answer:

Disinfectant	Antiseptic
1. Disinfectants are applied on non-living surfaces like floors, instruments, sanitary ware, etc. to kill wide range of microorganisms.	1. Antiseptics are applied on the surface of living tissues in order to sterilise them.
2. Disinfectants cannot be applied on wounds.	2. Antiseptics can be directly applied on wounds.
3. p-chloro-o-benzyl phenol	3. Iodine, boric acid, iodoform, dettol, etc.

Question B.
Soap and synthetic detergent
Answer:

Soap	Synthetic detergent
1. Soaps can be broadly classified into two types, i.e., toilet soaps (prepared using KOH) and laundry soaps (prepared using NaOH).	1. Synthetic detergents are of three types, i.e., anionic, cationic and nonionic detergents.
2. Soaps cannot be used in hard water.	2. Synthetic detergents can be used in soft water as well as in hard water.

Question C.
Saturated and unsaturated fats
Answer:

Saturated fats	Unsaturated fats
1. In saturated fat, long chains of tetrahedral carbon atoms in the fatty acid get closely packed together.	1. In unsaturated fats, the presence of one or more C = C bond in long chains of fatty acids, prevent molecules from packing closely together.
2. In saturated fats, the van der Waals forces between long saturated chains are strong. Hence, their melting points are higher than unsaturated fats.	2. In unsaturated fats, the van der Waals forces between long unsaturated chains are weak. Hence, their melting points are lower than saturated fats.

Question D.
Rice flour and cooked rice
Answer:

Rice flour	Cooked rice
1. Rice flour can be stored for a long period of time. It has a long shelf life.	1. Cooked rice cannot be stored for a longer period of time. It has very short shelf life.
2. Rice flour is uncooked food and hence, it is difficult to digest.	2. Cooked rice is easier to digest.

5. Match the pairs.

A group	B group
A. Paracetamol	a. Antibiotic
B. Chloramphenicol	b. Synthetic detergent
C. BHT	c. Soap
D. Sodium stearate	d. Antioxidant
	e. Analgesic

Answer:
A – e,
B – a,
C – d,
D – c

6. Name two drugs which reduce body pain.
Answer:
Aspirin and paracetamol are the two drugs that reduce body pain.

7. Explain with examples

Question A.
Antiseptics
Answer:
i. Antiseptics are used to sterilise surfaces of living tissue when the risk of infection is very high, such as during surgery or on wounds.
ii. Commonly used antiseptics include inorganics like iodine and boric acid or organics like iodoform and some phenolic compounds.

e.g.

• Tincture of iodine (2-3% solution of iodine in alcohol-water mixture) and iodoform serve as powerful antiseptics and is used to apply on wounds.
• A dilute aqueous solution of boric acid is a weak antiseptic used for eyes.
• Various phenols are used as antiseptics. A dilute aqueous solution of phenol (carbolic acid) is used as antiseptic; however, phenol is found to be corrosive in nature. Many chloro derivatives of phenols are more potent antiseptics than the phenol itself. They can be used with much lower concentrations, which reduce their corrosive effects.
• Two of the most common phenol derivatives in use are trichlorophenol (TCP) and chloroxylenol (which is an active ingredient of antiseptic dettol).
• Thymol obtained from oil of thyme (a spice plant) has excellent non-toxic antiseptic properties.

Question B.
Disinfectant
Answer:

• Disinfectants are non-selective antimicrobials.
• They kill a wide range of microorganisms including bacteria.
• They are used on non-living surfaces for example, floors, instruments, sanitary ware, etc.
• Various phenols can be used as disinfectants.
  e.g. p-Chloro-o-benzyl phenol is used as a disinfectant in all-purpose cleaners.

Question C.
Cationic detergents
Answer:
Cationic detergents: These are quaternary ammonium salts having one long chain alkyl group.
e.g. Ethyltrimethylammonium bromide: [CH₃(CH₂)₁₅ – N⁺(CH₃)₃]Br^–

Question D.
Anionic detergents
Answer:
Anionic detergents: These are sodium salts of long chain alkyl sulphonic acids or long chain alkyl substituted benzene sulphonic acids.
e.g. Sodium lauryl sulphate: CH₃(CH₂)₁₀CH₃O\(\mathrm{SO}_{3}^{-}\)Na⁺

Question E.
Non-ionic detergents
Answer:
Nonionic detergents: These are ethers of polyethylene glycol with alkyl phenol or esters of polyethylene glycol with long chain fatty acid.
e.g. a. Nonionic detergent containing ether linkage:

b. Nonionic detergent containing ester linkage: CH₃(CH₂)₁₆ – COO(CH₂CH₂O)_nCH₂CH₂OH

8. Explain : mechnism of cleansing Action of soap with flow chart.
Answer:
The following flow chart shows mechanism of cleansing action of soap:

9. What is meant by broad spectrum antibiotic and narrow spectrum antibiotics?
Answer:
Antibiotics which are effective against wide range of bacteria are known as broad spectrum antibiotics, while antibiotics which are effective against one group of bacteria are known as narrow spectrum antibiotics.

10. Answer in one senetence

Question A.
Name the painkiller obtained from acetylation of salicyclic acid.
Answer:
Aspirin is the pain killer obtained from acetylation of salicylic acid.

Question B.
Name the class of drug often called as painkiller.
Answer:
Analgesics are the class of drug often called as painkiller.

Question C.
Who discovered penicillin?
Answer:
Alexander Fleming discovered penicillin.

Question D.
Draw the structure of chloroxylenol and salvarsan.
Answer:
Structure of chloroxylenol:

Structure of salvarsan:

Question E.
Write molecular formula of Butylated hydroxy toulene.
Answer:
Molecular formula of butylated hydroxytoluene is C₁₅H₂₄O.

Question F.
What is the tincture of iodine ?
Answer:
Tincture of iodine is a 2-3% solution of iodine in alcohol-water mixture.

Question G.
Draw the structure of BHT.
Answer:

Question I.
Write a chemical equation for saponification.
Answer:

Question J.
Write the molecular formula and name of

Answer:
Molecular formula: C₉H₈O₄
Name: Aspirin

11. Answer the following

Question A.
Write two examples of the following.
a. Analgesics
c. Antiseptics
d. Antibiotics
e. Disinfectant
Answer:

No.	Drug type	Examples
i.	Analgesics	Aspirin, paracetamol
ii.	Antiseptics	Dettol, thymol
iii.	Antibiotics	Penicillin, sulphapyridine
iv.	Disinfectant	Phenol, p-Chloro-o-benzyl phenol

Question B.
What do you understand by antioxidant ?
Answer:

• An antioxidant is a substance that delays the onset of oxidant or slows down the rate of oxidation of foodstuff.
• It is used to extend the shelf life of food.
• Antioxidants react with oxygen-containing free radicals and thereby prevent oxidative rancidity.
  e.g. Vitamin E is a very effective natural antioxidant.

Activity :

Collect information about different chemical compounds as per their applications in day-to-day life.
Answer:

No.	Chemical compound	Applications
i.	Vinegar(CH3COOH)	Preservation of food, salad dressing, sauces, etc.
ii.	Magnesium hydroxide [Mg(OH)2]	Common component of antacids (used to relieve heartburn, acid indigestion and stomach upset.)
iii.	Baking soda (NaHCO3)	Cooking, antacid, toothpaste, etc.
iv.	Sodium benzoate (C6H5COONa)	Used as food preservative

[Note: Students can use the above information as reference and collect additional information on their own.]

11th Chemistry Digest Chapter 16 Chemistry in Everyday Life Intext Questions and Answers

Can you recall? (Textbook Page No. 261)

Question i.
What are the components of balanced diet?
Answer:
Carbohydrates, proteins, lipids (fats and oil), vitamins, minerals and water are the components of balanced diet.

Question ii.
Why is food cooked? What is the difference in the physical states of uncooked and cooked food?
Answer:

• Food is cooked in order to make it easy to digest.
• Also, the raw or uncooked food may contain harmful microorganisms which may cause illness. Cooking of food at high temperature kills most of these microorganisms.
• Raw/uncooked food materials like dried pulses, vegetables, meat, etc. are hard and thus, not easily chewable while cooked food is soft and tender, therefore, easily chewable.

Question iii.
What are the chemicals that we come across in everyday life?
Answer:
Detergents, shampoos, medicines, various food flavours, food colours, etc. are different types of chemicals that we come across in everyday life.

Just think (Textbook Page No. 261)

Question i.
Why is food stored for a long time?
Answer:
Food (like various cereals, pulses, pickles) is stored for a long time to make it available in all seasons.

Question ii.
What methods are used for preservation of food?
Answer:
Various physical and chemical methods are used for preservations of food.

• Physical methods like, addition of heat, removal of heat, removal of water, irradiation, etc., are used in order to preserve food.
• Chemical methods like, addition of sugar, salt, vinegar, etc. are employed for preservation of food.

Question iii.
What is meant by quality of food?
Answer:
Food quality can be described in terms of parameters such as flavour, smell, texture, colour and microbial spoilage.

Can you recall? (Textbook Page No. 263)

Question i.
How is Vanaspati ghee made?
Answer:
Vanaspati ghee is prepared by hydrogenation of oils. Hydrogen gas is passed through the oils at about 450 K in the presence of nickel catalyst to form solid edible fats like vanaspati ghee.

Question ii.
What are the physical states of peanut oil, butter, animal fat, Vanaspati ghee at room temperature?
Answer:

Example	Physical state
Peanut oil	Liquid
Butter	Semi-solid
Animal fat	Solid/semi-solid
Vanaspati ghee	Solid/semi-solid

Can you tell? (Textbook Page No. 264)

Question 1.
When is an antipyretic drug used?
Answer:
An antipyretic drug is used to reduce fever (that is, it lowers body temperature when a fever is present).

Question 2.
What type of medicine is applied to a bruise?
Answer:
Antiseptic such as tincture of iodine is applied on a bruise in order to prevent the exposed living tissue from getting infected.

Question 3.
What is meant by a broad spectrum antibiotic?
Answer:
Antibiotics which are effective against wide range of bacteria are known as broad spectrum antibiotic.

Question 4.
What is the active principle ingredient of cinnamon bark?
Answer:
Cinnamaldehyde is the principle active ingredient of cinnamon bark.

Can you tell? (Textbook Page No. 268)

Question i.
Can we use the same soap for bathing as well as cleaning utensils or washing clothes? Why?
Answer:
No, we cannot use the same soap for bathing as well as cleaning utensils or washing clothes due to the following reasons:

• Chemical composition of each type of soap or cleansing material is different.
• Nature, acidity, texture, reactivity towards water (i.e., hard water or soft water), reactivity towards microorganisms, stains are different for each type of soap.
• Depending on these qualities, soaps are classified and used accordingly.
  e.g. pH of soaps used for bathing purpose is different than that of the soap which is used for cleaning utensils.

Thus, we cannot use the same soap for bathing as well as cleaning utensils or washing clothes.

Question ii.
How will you differentiate between soaps and synthetic detergent using borewell water?
Answer:
Borewell water is hard water. Soaps and synthetic detergents react differently with hard water.

1. Soap: Soaps are insoluble in hard water. Borewell water (hard water) contains Ca²⁺ and Mg²⁺ ions. Soaps react with these ions to form insoluble magnesium and calcium salts of fatty acids. These salts precipitate out as gummy substance or form scum.
2. Synthetic detergents: Synthetic detergents can be used in hard water as well. They contain molecules (components) which form soluble calcium and magnesium salts.

Thus, soaps will form scum in borewell water but synthetic detergents will not.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3

Question 1.
In each of the following examples verify that the given expression is a solution of the corresponding differential equation.
(i) xy = log y + c; \(\frac{d y}{d x}=\frac{y^{2}}{1-x y}\)
Solution:
xy = log y + c
Differentiating w.r.t. x, we get

Hence, xy = log y + c is a solution of the D.E.
\(\frac{d y}{d x}=\frac{y^{2}}{1-x y^{\prime}}, x y \neq 1\)

(ii) y = (sin^-1x)² + c; (1 – x²) \(\frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}=2\)
Solution:
y = (sin^-1 x)² + c …….(1)
Differentiating w.r.t. x, we get

Differentiating again w.r.t. x, we get

(iii) y = e^-x + Ax + B; \(e^{x} \frac{d^{2} y}{d x^{2}}=1\)
Solution:
y = e^-x + Ax + B
Differentiating w.r.t. x, we get

∴ \(e^{x} \frac{d^{2} y}{d x^{2}}=1\)
Hence, y = e^-x + Ax + B is a solution of the D.E.
\(e^{x} \frac{d^{2} y}{d x^{2}}=1\)

(iv) y = x^m; \(x^{2} \frac{d^{2} y}{d x^{2}}-m x \frac{d y}{d x}+m y=0\)
Solution:
y = x^m
Differentiating twice w.r.t. x, we get

This shows that y = x^m is a solution of the D.E.
\(x^{2} \frac{d^{2} y}{d x^{2}}-m x \frac{d y}{d x}+m y=0\)

(v) y = a + \(\frac{b}{x}\); \(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}=0\)
Solution:
y = a + \(\frac{b}{x}\)
Differentiating w.r.t. x, we get

Differentiating again w.r.t. x, we get

Hence, y = a + \(\frac{b}{x}\) is a solution of the D.E.
\(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}=0\)

(vi) y = e^ax; x \(\frac{d y}{d x}\) = y log y
Solution:
y = e^ax
log y = log e^ax = ax log e
log y = ax …….(1) ……..[∵ log e = 1]
Differentiating w.r.t. x, we get
\(\frac{1}{y} \cdot \frac{d y}{d x}\) = a × 1
∴ \(\frac{d y}{d x}\) = ay
∴ x \(\frac{d y}{d x}\) = (ax)y
∴ x \(\frac{d y}{d x}\) = y log y ………[By (1)]
Hence, y = e^ax is a solution of the D.E.
x \(\frac{d y}{d x}\) = y log y.

Question 2.
Solve the following differential equations.
(i) \(\frac{d y}{d x}=\frac{1+y^{2}}{1+x^{2}}\)
Solution:

(ii) log(\(\frac{d y}{d x}\)) = 2x + 3y
Solution:

(iii) y – x \(\frac{d y}{d x}\) = 0
Solution:
y – x \(\frac{d y}{d x}\) = 0
∴ x \(\frac{d y}{d x}\) = y
∴ \(\frac{1}{x} d x=\frac{1}{y} d y\)
Integrating both sides, we get
\(\int \frac{1}{x} d x=\int \frac{1}{y} d y\)
∴ log |x| = log |y| + log c
∴ log |x| = log |cy|
∴ x = cy
This is the general solution.

(iv) sec²x . tan y dx + sec²y . tan x dy = 0
Solution:
sec²x . tan y dx + sec²y . tan x dy = 0
∴ \(\frac{\sec ^{2} x}{\tan x} d x+\frac{\sec ^{2} y}{\tan y} d y=0\)
Integrating both sides, we get
\(\int \frac{\sec ^{2} x}{\tan x} d x+\int \frac{\sec ^{2} y}{\tan y} d y=c_{1}\)
Each of these integrals is of the type
\(\int \frac{f^{\prime}(x)}{f(x)} d x\) = log |f(x)| + c
∴ the general solution is
∴ log|tan x| + log|tan y | = log c, where c₁ = log c
∴ log |tan x . tan y| = log c
∴ tan x . tan y = c
This is the general solution.

(v) cos x . cos y dy – sin x . sin y dx = 0
Solution:
cos x . cos y dy – sin x . sin y dx = 0
\(\frac{\cos y}{\sin y} d y-\frac{\sin x}{\cos x} d x=0\)
Integrating both sides, we get
∫cot y dy – ∫tan x dx = c₁
∴ log|sin y| – [-log|cos x|] = log c, where c₁ = log c
∴ log |sin y| + log|cos x| = log c
∴ log|sin y . cos x| = log c
∴ sin y . cos x = c
This is the general solution.

(vi) \(\frac{d y}{d x}\) = -k, where k is a constant.
Solution:
\(\frac{d y}{d x}\) = -k
∴ dy = -k dx
Integrating both sides, we get
∫dy = -k∫dx
∴ y = -kx + c
This is the general solution.

(vii) \(\frac{\cos ^{2} y}{x} d y+\frac{\cos ^{2} x}{y} d x=0\)
Solution:
\(\frac{\cos ^{2} y}{x} d y+\frac{\cos ^{2} x}{y} d x=0\)
∴ y cos²y dy + x cos²x dx = 0
∴ \(x\left(\frac{1+\cos 2 x}{2}\right) d x+y\left(1+\frac{\cos 2 y}{2}\right) d y=0\)
∴ x(1 + cos 2x) dx + y(1 + cos 2y) dy = 0
∴ x dx + x cos 2x dx + y dy+ y cos 2y dy = 0
Integrating both sides, we get
∫x dx + ∫y dy + ∫x cos 2x dx + ∫y cos 2y dy = c₁ ……..(1)
Using integration by parts

Multiplying throughout by 4, this becomes
2x² + 2y² + 2x sin 2x + cos 2x + 2y sin 2y + cos 2y = 4c₁
∴ 2(x² + y²) + 2(x sin 2x + y sin 2y) + cos 2y + cos 2x + c = 0, where c = -4c₁
This is the general solution.

(viii) \(y^{3}-\frac{d y}{d x}=x^{2} \frac{d y}{d x}\)
Solution:

(ix) 2e^x+2y dx – 3 dy = 0
Solution:

(x) \(\frac{d y}{d x}\) = e^x+y + x² e^y
Solution:

∴ 3e^x + 3e^-y + x³ = -3c₁
∴ 3e^x + 3e^-y + x³ = c, where c = -3c₁
This is the general solution.

Question 3.
For each of the following differential equations, find the particular solution satisfying the given condition:
(i) 3e^x tan y dx + (1 + e^x) sec²y dy = 0, when x = 0, y = π
Solution:
3e^x tan y dx + (1 + e^x) sec²y dy = 0

(ii) (x – y²x) dx – (y + x²y) dy = 0, when x = 2, y = 0
Solution:
(x – y²x) dx – (y + x²y) dy = 0
∴ x(1 – y²) dx – y(1 + x²) dy = 0

When x = 2, y = 0, we have
(1 + 4)(1 – 0) = c
∴ c = 5
∴ the particular solution is (1 + x²)(1 – y²) = 5.

(iii) y(1 + log x) \(\frac{d x}{d y}\) – x log x = 0, y = e², when x = e
Solution:
y(1 + log x) \(\frac{d x}{d y}\) – x log x = 0

(iv) (e^y + 1) cos x + e^y sin x \(\frac{d y}{d x}\) = 0, when x = \(\frac{\pi}{6}\), y = 0
Solution:
(e^y + 1) cos x + e^y sin x \(\frac{d y}{d x}\) = 0

\(\int \frac{f^{\prime}(x)}{f(x)} d x\) = log|f(x)| + c
∴ from (1), the general solution is
log|sin x| + log|e^y + 1| = log c, where c₁ = log c
∴ log|sin x . (e^y + 1)| = log c
∴ sin x . (e^y + 1) = c
When x = \(\frac{\pi}{4}\), y = 0, we get
\(\left(\sin \frac{\pi}{4}\right)\left(e^{0}+1\right)=c\)
∴ c = \(\frac{1}{\sqrt{2}}\)(1 + 1) = √2
∴ the particular solution is sin x . (e^y + 1) = √2

(v) (x + 1) \(\frac{d y}{d x}\) – 1 = 2e^-y, y = 0, when x = 1
Solution:

This is the general solution.
Now, y = 0, when x = 1
∴ 2 + e⁰ = c(1 + 1)
∴ 3 = 2c
∴ c = \(\frac{3}{2}\)
∴ the particular solution is 2 + e^y = \(\frac{3}{2}\) (x + 1)
∴ 2(2 + e^y) = 3(x + 1).

(vi) cos(\(\frac{d y}{d x}\)) = a, a ∈ R, y (0) = 2
Solution:
cos(\(\frac{d y}{d x}\)) = a
∴ \(\frac{d y}{d x}\) = cos^-1 a
∴ dy = (cos^-1 a) dx
Integrating both sides, we get
∫dy = (cos^-1 a) ∫dx
∴ y = (cos^-1 a) x + c
∴ y = x cos^-1 a + c
This is the general solution.
Now, y(0) = 2, i.e. y = 2,
when x = 0, 2 = 0 + c
∴ c = 2
∴ the particular solution is
∴ y = x cos^-1 a + 2
∴ y – 2 = x cos^-1 a
∴ \(\frac{y-2}{x}\) = cos^-1a
∴ cos(\(\frac{y-2}{x}\)) = a

Question 4.
Reduce each of the following differential equations to the variable separable form and hence solve:
(i) \(\frac{d y}{d x}\) = cos(x + y)
Solution:

(ii) (x – y)² \(\frac{d y}{d x}\) = a²
Solution:

(iii) x + y \(\frac{d y}{d x}\) = sec(x² + y²)
Solution:

Integrating both sides, we get
∫cos u du = 2 ∫dx
∴ sin u = 2x + c
∴ sin(x² + y²) = 2x + c
This is the general solution.

(iv) cos²(x – 2y) = 1 – 2 \(\frac{d y}{d x}\)
Solution:

Integrating both sides, we get
∫dx = ∫sec²u du
∴ x = tan u + c
∴ x = tan(x – 2y) + c
This is the general solution.

(v) (2x – 2y + 3) dx – (x – y + 1) dy = 0, when x = 0, y = 1
Solution:
(2x – 2y + 3) dx – (x – y + 1) dy = 0
∴ (x – y + 1) dy = (2x – 2y + 3) dx
∴ \(\frac{d y}{d x}=\frac{2(x-y)+3}{(x-y)+1}\) ………(1)
Put x – y = u, Then \(1-\frac{d y}{d x}=\frac{d u}{d x}\)

∴ u – log|u + 2| = -x + c
∴ x – y – log|x – y + 2| = -x + c
∴ (2x – y) – log|x – y + 2| = c
This is the general solution.
Now, y = 1, when x = 0.
∴ (0 – 1) – log|0 – 1 + 2| = c
∴ -1 – o = c
∴ c = -1
∴ the particular solution is
(2x – y) – log|x – y + 2| = -1
∴ (2x – y) – log|x – y + 2| + 1 = 0

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2

Question 1.
Obtain the differential equation by eliminating the arbitrary constants from the following equations:
(i) x³ + y³ = 4ax
Solution:
x³ + y³ = 4ax ……..(1)
Differentiating both sides w.r.t. x, we get
3x² + 3y² \(\frac{d y}{d x}\) = 4a × 1
∴ 3x² + 3y² \(\frac{d y}{d x}\) = 4a
Substituting the value of 4a in (1), we get
x³ + y³ = (3x² + 3y² \(\frac{d y}{d x}\)) x
∴ x³ + y³ = 3x³ + 3xy² \(\frac{d y}{d x}\)
∴ 2x³ + 3xy² \(\frac{d y}{d x}\) – y³ = 0
This is the required D.E.

(ii) Ax² + By² = 1
Solution:
Ax² + By² = 1
Differentiating both sides w.r.t. x, we get
A × 2x + B × 2y \(\frac{d y}{d x}\) = 0
∴ Ax + By \(\frac{d y}{d x}\) = 0 ……..(1)
Differentiating again w.r.t. x, we get

Substituting the value of A in (1), we get

This is the required D.E.

Alternative Method:
Ax² + By² = 1 ……..(1)
Differentiating both sides w.r.t. x, we get
A × 2x + B × 2y \(\frac{d y}{d x}\) = 0
∴ Ax + By \(\frac{d y}{d x}\) = 0 ……….(2)
Differentiating again w.r.t. x, we get,

The equations (1), (2) and (3) are consistent in A and B.
∴ determinant of their consistency is zero.

This is the required D.E.

(iii) y = A cos(log x) + B sin(log x)
Solution:
y = A cos(log x) + B sin (log x) ……. (1)
Differentiating w.r.t. x, we get

(iv) y² = (x + c)³
Solution:
y² = (x + c)³
Differentiating w.r.t. x, we get

This is the required D.E.

(v) y = Ae^5x + Be^-5x
Solution:
y = Ae^5x + Be^-5x ……….(1)
Differentiating twice w.r.t. x, we get

This is the required D.E.

(vi) (y – a)² = 4(x – b)
Solution:
(y – a)² = 4(x – b)
Differentiating both sides w.r.t. x, we get
2(y – a) . \(\frac{d}{d x}\)(y – a) = 4 \(\frac{d}{d x}\)(x – b)
∴ 2(y – a) . (\(\frac{d y}{d x}\) – 0) = 4(1 – 0)
∴ 2(y – a) \(\frac{d y}{d x}\) = 4
∴ (y – a) \(\frac{d y}{d x}\) = 2 ……..(1)
Differentiating w.r.t. x, we get

This is the required D.E.

(vii) y = a + \(\frac{a}{x}\)
Solution:
y = a + \(\frac{a}{x}\)
Differentiating w.r.t. x, we get

Substituting the value of a in (1), we get

This is the required D.E.

(viii) y = c₁e^2x + c₂e^5x
Solution:
y = c₁e^2x + c₂e^5x ………(1)
Differentiating twice w.r.t. x, we get
\(\frac{d y}{d x}\) = c₁e^2x × 2 + c₂e^5x × 5

The equations (1), (2) and (3) are consistent in c₁e^2x and c₂e^5x
∴ determinant of their consistency is zero.

This is the required D.E.

Alternative Method:
y = c₁e^2x + c₂e^5x
Dividing both sides by e^5x, we get

This is the required D.E.

(ix) c₁x³ + c₂y² = 5.
Solution:
c₁x³ + c₂y² = 5 ……….(1)
Differentiating w.r.t. x, we get

Differentiating again w.r.t. x, we get

The equations (1), (2) and (3) in c₁, c₂ are consistent.
∴ determinant of their consistency is zero.

This is the required D.E.

(x) y = e^-2x(A cos x + B sin x)
Solution:
y = e^-2x(A cos x + B sin x)
∴ e^2x . y = A cos x + B sin x ………(1)
Differentiating w.r.t. x, we get

Differentiating again w.r.t. x, we get


This is the required D.E.

Question 2.
Form the differential equation of family of lines having intercepts a and b on the coordinate axes respectively.
Solution:
The equation of the line having intercepts a and b on the coordinate axes respectively, is
\(\frac{x}{a}+\frac{y}{b}=1\) ……….(1)
where a and b are arbitrary constants.
[For different values of a and b, we get, different lines. Hence (1) is the equation of family of lines.]
Differentiating (1) w.r.t. x, we get

Differentiating again w.r.t. x, we get \(\frac{d^{2} y}{d x^{2}}=0\)
This is the required D.E.

Question 3.
Find the differential equation all parabolas having length of latus rectum 4a and axis is parallel to the X-axis.
Solution:

Let A(h, k) be the vertex of the parabola whose length of latus rectum is 4a.
Then the equation of the parabola is (y – k)² = 4a (x – h), where h and k are arbitrary constants.
Differentiating w.r.t. x, we get

Differentiating again w.r.t. x, we get

This is the required D.E.

Question 4.
Find the differential equation of the ellipse whose major axis is twice its minor axis.
Solution:
Let 2a and 2b be lengths of major axis and minor axis of the ellipse.
Then 2a = 2(2b)
∴ a = 2b
∴ equation of the ellipse is
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
i.e., \(\frac{x^{2}}{(2 b)^{2}}+\frac{y^{2}}{b^{2}}=1\)
∴ \(\frac{x^{2}}{4 b^{2}}+\frac{y^{2}}{b^{2}}=1\)
∴ x² + 4y² = 4b²
Differentiating w.r.t. x, we get
2x + 4 × 2y \(\frac{d y}{d x}\) = 0
∴ x + 4y \(\frac{d y}{d x}\) = 0
This is the required D.E.

Question 5.
Form the differential equation of family of lines parallel to the line 2x + 3y + 4 = 0.
Solution:
The equation of the line parallel to the line 2x + 3y + 4 = 0 is 2x + 3y + c = 0, where c is an arbitrary constant.
Differentiating w.r.t. x, we get
2 × 1 + 3 \(\frac{d y}{d x}\) + 0 = 0
∴ 3 \(\frac{d y}{d x}\) + 2 = 0
This is the required D.E.

Question 6.
Find the differential equation of all circles having radius 9 and centre at point (h, k).
Solution:
Equation of the circle having radius 9 and centre at point (h, k) is
(x – h)² + (y – k)² = 81 …… (1)
where h and k are arbitrary constant.
Differentiating (1) w.r.t. x, we get

Differentiating again w.r.t. x, we get

From (2), x – h = -(y – k) \(\frac{d y}{d x}\)
Substituting the value of (x – h) in (1), we get


This is the required D.E.

Question 7.
Form the differential equation of all parabolas whose axis is the X-axis.
Solution:

The equation of the parbola whose axis is the X-axis is
y² = 4a(x – h) …… (1)
where a and h are arbitrary constants.
Differentiating (1) w.r.t. x, we get
2y \(\frac{d y}{d x}\) = 4a(1 – 0)
∴ y \(\frac{d y}{d x}\) = 2a
Differentiating again w.r.t. x, we get

This is the required D.E.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.1

1. Determine the order and degree of each of the following differential equations:

Question (i).
\(\frac{d y}{d x^{2}}+X\left(\frac{d y}{d x}\right)+y=2 \sin x\)
Solution:
The given D.E. is \(\frac{d y}{d x^{2}}+X\left(\frac{d y}{d x}\right)+y=2 \sin x\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 1.
∴ the given D.E. is of order 2 and degree 1.

Question (ii).
\(\sqrt[3]{1+\left(\frac{d y}{d x}\right)^{2}}=\frac{d^{2} y}{d x^{2}}\)
Solution:
The given D.E. is \(\sqrt[3]{1+\left(\frac{d y}{d x}\right)^{2}}=\frac{d^{2} y}{d x^{2}}\)
On cubing both sides, we get
\(1+\left(\frac{d y}{d x}\right)^{2}=\left(\frac{d^{2} y}{d x^{2}}\right)^{3}\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 3.
∴ the given D.E. is of order 2 and degree 3.

Question (iii).
\(\frac{d y}{d x}=\frac{2 \sin x+3}{\frac{d y}{d x}}\)
Solution:
The given D.E. is \(\frac{d y}{d x}=\frac{2 \sin x+3}{\frac{d y}{d x}}\)
∴ \(\left(\frac{d y}{d x}\right)^{2}\) = 2 sin x + 3
This D.E. has highest order derivative \(\frac{d y}{d x}\) with power 2.
∴ the given D.E. is of order 1 and degree 2.

Question (iv).
\(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+x=\sqrt{1+\frac{d^{3} y}{d x^{3}}}\)
Solution:
The given D.E. is \(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+x=\sqrt{1+\frac{d^{3} y}{d x^{3}}}\)
On squaring both sides, we get
\(\left(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+x\right)^{2}=1+\frac{d^{3} y}{d x^{3}}\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 1.
∴ the given D.E. has order 3 and degree 1.

Question (v).
\(\frac{d^{2} y}{d t^{2}}+\left(\frac{d y}{d t}\right)^{2}+7 x+5=0\)
Solution:
The given D.E. is \(\frac{d^{2} y}{d t^{2}}+\left(\frac{d y}{d t}\right)^{2}+7 x+5=0\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 1.
∴ the given D.E. has order 2 and degree 1.

Question (vi).
(y”‘)2 + 3y” + 3xy’ + 5y = 0
Solution:
The given D.E. is (y”‘)2 + 3y” + 3xy’ + 5y = 0
This can be written as:
\(\left(\frac{d^{3} y}{d x^{3}}\right)^{2}+3 \frac{d^{2} y}{d x^{2}}+3 x \frac{d y}{d x}+5 y=0\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 2.
∴ The given D.E. has order 3 and degree 2.

Question (vii).
\(\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\cos \left(\frac{d y}{d x}\right)=0\)
Solution:
The given D.E. is \(\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\cos \left(\frac{d y}{d x}\right)=0\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\)
∴ order = 2
Since this D.E. cannot be expressed as a polynomial in differential coefficients, the degree is not defined.

Question (viii).
\(\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{\frac{3}{2}}=8 \frac{d^{2} y}{d x^{2}}\)
Solution:
The given D.E. is \(\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{\frac{3}{2}}=8 \frac{d^{2} y}{d x^{2}}\)
On squaring both sides, we get
\(\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3}=8^{2} \cdot\left(\frac{d^{2} y}{d x^{2}}\right)^{2}\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 2.
∴ the given D.E. has order 2 and degree 2.

Question (ix).
\(\left(\frac{d^{3} y}{d x^{3}}\right)^{\frac{1}{2}} \cdot\left(\frac{d y}{d x}\right)^{\frac{1}{3}}=20\)
Solution:
The given D.E. is \(\left(\frac{d^{3} y}{d x^{3}}\right)^{\frac{1}{2}} \cdot\left(\frac{d y}{d x}\right)^{\frac{1}{3}}=20\)
∴ \(\left(\frac{d^{3} y}{d x^{3}}\right)^{3} \cdot\left(\frac{d y}{d x}\right)^{2}=20^{6}\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 3.
∴ the given D.E. has order 3 and degree 3.

Question (x).
\(x+\frac{d^{2} y}{d x^{2}}=\sqrt{1+\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}\)
Solution:
The given D.E. is \(x+\frac{d^{2} y}{d x^{2}}=\sqrt{1+\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}\)
On squaring both sides, we get

This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 1.
∴ the given D.E. has order 2 and degree 1.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5

I. Choose the correct option from the given alternatives:

Question 1.
The area bounded by the region 1 ≤ x ≤ 5 and 2 ≤ y ≤ 5 is given by
(a) 12 sq units
(b) 8 sq units
(c) 25 sq units
(d) 32 sq units
Answer:
(a) 12 sq units

Question 2.
The area of the region enclosed by the curve y = \(\frac{1}{x}\), and the lines x = e, x = e² is given by
(a) 1 sq unit
(b) \(\frac{1}{2}\) sq units
(c) \(\frac{3}{2}\) sq units
(d) \(\frac{5}{2}\) sq units
Answer:
(a) 1 sq unit

Question 3.
The area bounded by the curve y = x³, the X-axis and the lines x = -2 and x = 1 is
(a) -9 sq units
(b) \(-\frac{15}{4}\) sq units
(c) \(\frac{15}{4}\) sq units
(d) \(\frac{17}{4}\) sq units
Answer:
(c) \(\frac{15}{4}\) sq units

Question 4.
The area enclosed between the parabola y² = 4x and line y = 2x is
(a) \(\frac{2}{3}\) sq units
(b) \(\frac{1}{3}\) sq units
(c) \(\frac{1}{4}\) sq units
(d) \(\frac{3}{4}\) sq units
Answer:
(b) \(\frac{1}{3}\) sq units

Question 5.
The area of the region bounded between the line x = 4 and the parabola y² = 16x is
(a) \(\frac{128}{3}\) sq units
(b) \(\frac{108}{3}\) sq units
(c) \(\frac{118}{3}\) sq units
(d) \(\frac{218}{3}\) sq units
Answer:
(a) \(\frac{128}{3}\) sq units

Question 6.
The area of the region bounded by y = cos x, Y-axis and the lines x = 0, x = 2π is
(a) 1 sq unit
(b) 2 sq units
(c) 3 sq units
(d) 4 sq units
Answer:
(d) 4 sq units

Question 7.
The area bounded by the parabola y² = 8x, the X-axis and the latus rectum is
(a) \(\frac{31}{3}\) sq units
(b) \(\frac{32}{3}\) sq units
(c) \(\frac{32 \sqrt{2}}{3}\) sq units
(d) \(\frac{16}{3}\) sq units
Answer:
(b) \(\frac{32}{3}\) sq units

Question 8.
The area under the curve y = 2√x, enclosed between the lines x = 0 and x = 1 is
(a) 4 sq units
(b) \(\frac{3}{4}\) sq units
(c) \(\frac{2}{3}\) sq units
(d) \(\frac{4}{3}\) sq units
Answer:
(d) \(\frac{4}{3}\) sq units

Question 9.
The area of the circle x² + y² = 25 in first quadrant is
(a) \(\frac{25 \pi}{3}\) sq units
(b) 5π sq units
(c) 5 sq units
(d) 3 sq units
Answer:
(a) \(\frac{25 \pi}{3}\) sq units

Question 10.
The area of the region bounded by the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is
(a) ab sq units
(b) πab sq units
(c) \(\frac{\pi}{a b}\) sq units ab
(d) πa² sq units
Answer:
(b) πab sq units

Question 11.
The area bounded by the parabola y² = x and the line 2y = x is
(a) \(\frac{4}{3}\) sq units
(b) 1 sq unit
(c) \(\frac{2}{3}\) sq unit
(d) \(\frac{1}{3}\) sq unit
Answer:
(a) \(\frac{4}{3}\) sq units

Question 12.
The area enclosed between the curve y = cos 3x, 0 ≤ x ≤ \(\frac{\pi}{6}\) and the X-axis is
(a) \(\frac{1}{2}\) sq unit
(b) 1 sq unit
(c) \(\frac{2}{3}\) sq unit
(d) \(\frac{1}{3}\) sq unit
Answer:
(d) \(\frac{1}{3}\) sq unit

Question 13.
The area bounded by y = √x and line x = 2y + 3, X-axis in first quadrant is
(a) 2√3 sq units
(b) 9 sq units
(c) \(\frac{34}{3}\) sq units
(d) 18 sq units
Answer:
(b) 9 sq units

Question 14.
The area bounded by the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) and the line \(\frac{x}{a}+\frac{y}{b}=1\) is
(a) (πab – 2ab) sq units
(b) \(\frac{\pi a b}{4}-\frac{a b}{2}\) sq units
(c) (πab – ab) sq units
(d) πab sq units
Answer:
(b) \(\frac{\pi a b}{4}-\frac{a b}{2}\) sq units

Question 15.
The area bounded by the parabola y = x² and the line y = x is
(a) \(\frac{1}{2}\) sq unit
(b) \(\frac{1}{3}\) sq unit
(c) \(\frac{1}{6}\) sq unit
(d) \(\frac{1}{12}\) sq unit
Answer:
(c) \(\frac{1}{6}\) sq unit

Question 16.
The area enclosed between the two parabolas y² = 4x and y = x is
(a) \(\frac{8}{3}\) sq units
(b) \(\frac{32}{3}\) sq units
(c) \(\frac{16}{3}\) sq units
(d) \(\frac{4}{3}\) sq units
Answer:
(c) \(\frac{16}{3}\) sq units

Question 17.
The area bounded by the curve y = tan x, X-axis and the line x = \(\frac{\pi}{4}\) is
(a) \(\frac{1}{3}\) log 2 sq units
(b) log 2 sq units
(c) 2 log 2 sq units
(d) 3 log 2 sq units
Answer:
(a) \(\frac{1}{3}\) log 2 sq units

Question 18.
The area of the region bounded by x² = 16y, y = 1, y = 4 and x = 0 in the first quadrant, is
(a) \(\frac{7}{3}\) sq units
(b) \(\frac{8}{3}\) sq units
(c) \(\frac{64}{3}\) sq units
(d) \(\frac{56}{3}\) sq units
Answer:
(d) \(\frac{56}{3}\) sq units

Question 19.
The area of the region included between the parabolas y² = 4ax and x² = 4ay, (a > 0) is given by
(a) \(\frac{16 a^{2}}{3}\) sq units
(b) \(\frac{8 a^{2}}{3}\) sq units
(c) \(\frac{4 a^{2}}{3}\) sq units
(d) \(\frac{32 a^{2}}{3}\) sq units
Answer:
(a) \(\frac{16 a^{2}}{3}\) sq units

Question 20.
The area of the region included between the line x + y = 1 and the circle x² + y² = 1 is
(a) \(\frac{\pi}{2}-1\) sq units
(b) π – 2 sq units
(c) \(\frac{\pi}{4}-\frac{1}{2}\) sq units
(d) π – \(\frac{1}{2}\) sq units
Answer:
(c) \(\frac{\pi}{4}-\frac{1}{2}\) sq units

(II) Solve the following:

Question 1.
Find the area of the region bounded by the following curve, the X-axis and the given lines:
(i) 0 ≤ x ≤ 5, 0 ≤ y ≤ 2
(ii) y = sin x, x = 0, x = π
(iii) y = sin x, x = 0, x = \(\frac{\pi}{3}\)
Solution:
(i) Required area = \(\int_{0}^{5} y d x\), where y = 2
= \(\int_{0}^{5} 2 d x\)
= \([2 x]_{0}^{5}\)
= 2 × 5 – 0
= 10 sq units.

(ii) The curve y = sin x intersects the X-axis at x = 0 and x = π between x = 0 and x = π.

Two bounded regions A₁ and A₂ are obtained. Both the regions have equal areas.
∴ required area = A₁ + A₂ = 2A₁

(iii) Required area = \(\int_{0}^{\pi / 3} y d x\), where y = sin x

Question 2.
Find the area of the circle x² + y² = 9, using integration.
Solution:
By the symmetry of the circle, its area is equal to 4 times the area of the region OABO.
Clearly, for this region, the limits of integration are 0 and 3.

From the equation of the circle, y² = 9 – x².
In the first quadrant, y > 0
∴ y = \(\sqrt{9-x^{2}}\)
∴ area of the circle = 4 (area of the region OABO)

Question 3.
Find the area of the ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\) using integration.
Solution:

By the symmetry of the ellipse, its area is equal to 4 times the area of the region OABO.
Clearly, for this region, the limits of integration are 0 and 5.
From the equation of the ellipse
\(\frac{y^{2}}{16}=1-\frac{x^{2}}{25}=\frac{25-x^{2}}{25}\)
∴ y² = \(\frac{16}{25}\) (25 – x²)
In the first quadrant y > 0
∴ y = \(\frac{4}{5} \sqrt{25-x^{2}}\)
∴ area of the ellipse = 4(area of the region OABO)

Question 4.
Find the area of the region lying between the parabolas:
(i) y² = 4x and x² = 4y
(ii) 4y² = 9x and 3x² = 16y
(iii) y² = x and x² = y.
Solution:
(i)

For finding the points of intersection of the two parabolas, we equate the values of y² from their equations.
From the equation x² = 4y, y = \(\frac{x^{2}}{4}\)
y = \(\frac{x^{4}}{16}\)
\(\frac{x^{4}}{16}\) = 4x
∴ x⁴ – 64x = 0
∴ x(x³ – 64) = 0
∴ x = 0 or x³ = 64 i.e. x = 0 or x = 4
When x = 0, y = 0
When x = 4, y = \(\frac{4^{2}}{4}\) = 4
∴ the points of intersection are 0(0, 0) and A(4, 4).
Required area = area of the region OBACO = [area of the region ODACO] – [area of the region ODABO]
Now, area of the region ODACO = area under the parabola y² = 4x, i.e. y = 2√x between x = 0 and x = 4

(ii)

For finding the points of intersection of the two parabolas, we equate the values of 4y² from their equations.
From the equation 3x² = 16y, y = \(\frac{3 x^{2}}{16}\)
∴ y = \(\frac{3 x^{4}}{256}\)
∴ \(\frac{3 x^{4}}{256}\) = 9x
∴ 3x⁴ – 2304x = 0
∴ x(x³ – 2304) = 0
∴ x = 0 or x³ = 2304 i.e. x = 0 or x = 4
When x = 0, y = 0
When x = 4, y = \(\frac{4^{2}}{4}\)
∴ the points of intersection are O(0, 0) and A(4, 4).
Required area = area of the region OBACO = [area of the region ODACO] – [area of the region ODABO]
Now, area of the region ODACO = area under the parabola y² = 4x,
i.e. y = 2√x between x = 0 and x = 4

Area of the region ODABO = area under the rabola x² = 4y,
i.e. y = \(\frac{x^{2}}{4}\) between x = 0 and x = 4

(iii)

For finding the points of intersection of the two parabolas, we equate the values of y² from their equations.
From the equation x² = y, y = \(\frac{x^{2}}{y}\)
∴ y = \(\frac{x^{2}}{y}\)
∴ \(\frac{x^{2}}{y}\) = x
∴ x² – y = 0
∴ x(x³ – y) = 0
∴ x = 0 or x³ = y
i.e. x = 0 or x = 4
When x = 0, y = 0
When x = 4, y = \(\frac{4^{2}}{4}\) = 4
∴ the points of intersection are O(0, 0) and A(4, 4).
Required area = area of the region OBACO = [area of the region ODACO] – [area of the region ODABO]
Now, area of the region ODACO = area under the parabola y² = 4x,
i.e. y = 2√x between x = 0 and x = 4

Area ofthe region ODABO = area under the rabola x² = 4y,
i.e. y = \(\frac{x^{2}}{3}\) between x = 0 and x = 4

Question 5.
Find the area of the region in the first quadrant bounded by the circle x² + y² = 4 and the X-axis and the line x = y√3.
Solution:

For finding the points of intersection of the circle and the line, we solve
x² + y² = 4 ………(1)
and x = y√3 ……..(2)
From (2), x² = 3y²
From (1), x² = 4 – y²
3y² = 4 – y²
4y² = 4
y² = 1
y = 1 in the first quadrant.
When y = 1, r = 1 × √3 = √3
∴ the circle and the line intersect at A(√3, 1) in the first quadrant
Required area = area of the region OCAEDO = area of the region OCADO + area of the region DAED
Now, area of the region OCADO = area under the line x = y√3, i.e. y = \(\frac{x}{\sqrt{3}}\) between x = 0
and x = √3

Question 6.
Find the area of the region bounded by the parabola y² = x and the line y = x in the first quadrant.
Solution:
To obtain the points of intersection of the line and the parabola, we equate the values of x from both equations.

∴ y² = y
∴ y² – y = 0
∴ y(y – 1) = 0
∴ y = 0 or y = 1
When y = 0, x = 0
When y = 1, x = 1
∴ the points of intersection are O(0, 0) and A(1, 1).
Required area = area of the region OCABO = area of the region OCADO – area of the region OBADO
Now, area of the region OCADO = area under the parabola y² = x i.e. y = +√x (in the first quadrant) between x = 0 and x = 1

Area of the region OBADO = area under the line y = x between x = 0 and x = 1

Question 7.
Find the area enclosed between the circle x² + y² = 1 and the line x + y = 1, lying in the first quadrant.
Solution:

Required area = area of the region ACBPA = (area of the region OACBO) – (area of the region OADBO)
Now, area of the region OACBO = area under the circle x² + y² = 1 between x = 0 and x = 1

Area of the region OADBO = area under the line x + y = 1 between x = 0 and x = 1

∴ required area = \(\left(\frac{\pi}{4}-\frac{1}{2}\right)\) sq units.

Question 8.
Find the area of the region bounded by the curve (y – 1)² = 4(x + 1) and the line y = (x – 1).
Solution:
The equation of the curve is (y – 1)² = 4(x + 1)
This is a parabola with vertex at A (-1, 1).
To find the points of intersection of the line y = x – 1 and the parabola.
Put y = x – 1 in the equation of the parabola, we get
(x – 1 – 1)² = 4(x + 1)
∴ x² – 4x + 4 = 4x + 4
∴ x² – 8x = 0
∴ x(x – 8) = 0
∴ x = 0, x = 8
When x = 0, y = 0 – 1 = -1
When x = 8, y = 8 – 1 = 7
∴ the points of intersection are B (0, -1) and C (8, 7).

To find the points where the parabola (y – 1)² = 4(x + 1) cuts the Y-axis.
Put x = 0 in the equation of the parabola, we get
(y – 1)² = 4(0 + 1) = 4
∴ y – 1 = ±2
∴ y – 1 = 2 or y – 1 = -2
∴ y = 3 or y = -1
∴ the parabola cuts the Y-axis at the points B(0, -1) and F(0, 3).
To find the point where the line y = x – 1 cuts the X-axis.
Put y = 0 in the equation of the line, we get
x – 1 = 0
∴ x = 1
∴ the line cuts the X-axis at the point G (1, 0).
Required area = area of the region BFAB + area of the region OGDCEFO + area of the region OBGO
Now, area of the region BFAB = area under the parabola (y – 1)² = 4(x + 1), Y-axis from y = -1 to y = 3

Since, the area cannot be negative,
Area of the region BFAB = \(\left|-\frac{8}{3}\right|=\frac{8}{3}\) sq units.
Area of the region OGDCEFO = area of the region OPCEFO – area of the region GPCDG

Since, area cannot be negative,
area of the region = \(\left|-\frac{1}{2}\right|=\frac{1}{2}\) sq units.
∴ required area = \(\frac{8}{3}+\frac{109}{6}+\frac{1}{2}\)
= \(\frac{16+109+3}{6}\)
= \(\frac{128}{6}\)
= \(\frac{64}{3}\) sq units.

Question 9.
Find the area of the region bounded by the straight line 2y = 5x + 7, X-axis and x = 2, x = 5.
Solution:
The equation of the line is
2y = 5x + 7, i.e., y = \(\frac{5}{2} x+\frac{7}{2}\)
Required area = area of the region ABCDA = area under the line y = \(\frac{5}{2} x+\frac{7}{2}\) between x = 2 and x = 5

Question 10.
Find the area of the region bounded by the curve y = 4x², Y-axis and the lines y = 1, y = 4.
Solution:

By symmetry of the parabola, the required area is 2 times the area of the region ABCD.
From the equation of the parabola, x² = \(\frac{y}{4}\)
In the first quadrant, x > 0
∴ x = \(\frac{1}{2} \sqrt{y}\)
∴ required area = \(\int_{1}^{4} x d y\)

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Application of Definite Integration Ex 5.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1

1. Find the area of the region bounded by the following curves, X-axis, and the given lines:

(i) y = 2x, x = 0, x = 5.
Solution:
Required area = \(\int_{0}^{5} y d x\), where y = 2x
= \(\int_{0}^{5} 2x d x\)
= \(\left[\frac{2 x^{2}}{2}\right]_{0}^{5}\)
= 25 – 0
= 25 sq units.

(ii) x = 2y, y = 0, y = 4.
Solution:
Required area = \(\int_{0}^{4} x d y\), where x = 2y
= \(\int_{0}^{4} 2 y d y\)
= \(\left[\frac{2 y^{2}}{2}\right]_{0}^{4}\)
= 16 – 0
= 16 sq units.

(iii) x = 0, x = 5, y = 0, y = 4.
Solution:
Required area = \(\int_{0}^{5} y d x\), where y = 4
= \(\int_{0}^{5} 4 d x\)
= \([4 x]_{0}^{5}\)
= 20 – 0
= 20 sq units.

(iv) y = sin x, x = 0, x = \(\frac{\pi}{2}\)
Solution:
Required area = \(\int_{0}^{\pi / 2} y d x\), where y = sin x
= \(\int_{0}^{\pi / 2} \sin x d x\)
= \([-\cos x]_{0}^{\pi / 2}\)
= -cos \(\frac{\pi}{2}\) + cos 0
= 0 + 1
= 1 sq unit.

(v) xy = 2, x = 1, x = 4.
Solution:
For xy = 2, y = \(\frac{2}{x}\)
Required area = \(\int_{1}^{4} y d x\), where y = \(\frac{2}{x}\)
= \(\int_{1}^{4} \frac{2}{x} d x\)
= \([2 \log |x|]_{1}^{4}\)
= 2 log 4 – 2 log 1
= 2 log 4 – 0
= 2 log 4 sq units.

(vi) y² = x, x = 0, x = 4.
Solution:

The required area consists of two bounded regions A1 and A2 which are equal in areas.
For y² = x, y = √x
Required area = A₁ + A₂ = 2A₁

(vii) y² = 16x, x = 0, x = 4.
Solution:

The required area consists of two bounded regions A1 and A2 which are equal in areas.
For y² = x, y = √x
Required area = A₁ + A₂ = 2A₁

2. Find the area of the region bounded by the parabola:

(i) y² = 16x and its latus rectum.
Solution:
Comparing y² = 16x with y² = 4ax, we get
4a = 16
∴ a = 4
∴ focus is S(a, 0) = (4, 0)

For y² = 16x, y = 4√x
Required area = area of the region OBSAO
= 2 [area of the region OSAO]

(ii) y = 4 – x² and the X-axis.
Solution:
The equation of the parabola is y = 4 – x²
∴ x² = 4 – y
i.e. (x – 0)² = -(y – 4)
It has vertex at P(0, 4)
For points of intersection of the parabola with X-axis,
we put y = 0 in its equation.
∴ 0 = 4 – x²
∴ x² = 4
∴ x = ± 2
∴ the parabola intersect the X-axis at A(-2, 0) and B(2, 0)

Required area = area of the region APBOA
= 2[area of the region OPBO]

3. Find the area of the region included between:

(i) y² = 2x and y = 2x.
Solution:
The vertex of the parabola y² = 2x is at the origin O = (0, 0).

To find the points of intersection of the line and the parabola, equaling the values of 2x from both the equations we get,
y² = y
∴ y² – y = 0
∴ y = 0 or y = 1
When y = 0, x = \(\frac{0}{2}\) = 0
When y = 1, x = \(\frac{1}{2}\)
∴ the points of intersection are 0(0, 0) and B(\(\frac{1}{2}\), 1)
Required area = area of the region OABCO = area of the region OABDO – area of the region OCBDO
Now, area of the region OABDO = area under the parabola y² = 2x between x = 0 and x = \(\frac{1}{2}\)

Area of the region OCBDO = area under the line y = 2x between x = 0 and x = \(\frac{1}{2}\)

(ii) y² = 4x and y = x.
Solution:
The vertex of the parabola y² = 4x is at the origin O = (0, 0).

To find the points of intersection of the line and the parabola, equaling the values of 4x from both the equations we get,
∴ y² = y
∴ y² – y = 0
∴ y(y – 1) = 0
∴ y = 0 or y = 1
When y = 0, x = \(\frac{0}{2}\) = 0
When y = 1, x = \(\frac{1}{2}\)
∴ the points of intersection are O(0, 0) and B(\(\frac{1}{2}\), 1)
Required area = area of the region OABCO = area of the region OABDO – area of the region OCBDO
Now, area of the region OABDO = area under the parabola y² = 4x between x = 0 and x = \(\frac{1}{2}\)

Area of the region OCBDO = area under the line y = 2x between x = 0 and x = \(\frac{1}{2}\)

(iii) y = x² and the line y = 4x.
Solution:
The vertex of the parabola y = x² is at the origin 0(0, 0)
To find the points of the intersection of a line and the parabola.

Equating the values of y from the two equations, we get
x² = 4x
∴ x² – 4x = 0
∴ x(x – 4) = 0
∴ x = 0, x = 4
When x = 0, y = 4(0) = 0
When x = 4, y = 4(4) = 16
∴ the points of intersection are 0(0, 0) and B(4, 16)
Required area = area of the region OABCO = (area of the region ODBCO) – (area of the region ODBAO)
Now, area of the region ODBCO = area under the line y = 4x between x = 0 and x = 4
= \(\int_{0}^{4} y d x\), where y = 4x
= \(\int_{0}^{4} 4 x d x\)
= 4\(\int_{0}^{4} x d x\)
= 4\([latex]\int_{0}^{4} x d x\)[/latex]
= 2(16 – 0)
= 32
Area of the region ODBAO = area under the parabola y = x² between x = 0 and x = 4
= \(\int_{0}^{4} y d x\), where y = x2
= \(\int_{0}^{4} x^{2} d x\)
= \(\left[\frac{x^{3}}{3}\right]_{0}^{4}\)
= \(\frac{1}{3}\) (64 – 0)
= \(\frac{64}{3}\)
∴ required area = 32 – \(\frac{64}{3}\) = \(\frac{32}{3}\) sq units.

(iv) y² = 4ax and y = x.
Solution:
The vertex of the parabola y² = 4ax is at the origin O = (0, 0).

To find the points of intersection of the line and the parabola, equaling the values of 4ax from both the equations we get,
∴ y² = y
∴ y² – y = 0
∴ y(y – 1) = 0
∴ y = 0 or y = 1
When y = 0, x = \(\frac{0}{2}\) = 0
When y = 1, x = \(\frac{1}{2}\)
∴ the points of intersection are O(0, 0) and B(\(\frac{1}{2}\), 1)
Required area = area of the region OABCO = area of the region OABDO – area of the region OCBDO
Now, area of the region OABDO
= area under the parabola y² = 4ax between x = 0 and x = \(\frac{1}{2}\)

Area of the region OCBDO
= area under the line y
= 4ax between x = 0 and x = \(\frac{1}{4 a x}\)

(v) y = x² + 3 and y = x + 3.
Solution:
The given parabola is y = x² + 3, i.e. (x – 0)² = y – 3
∴ its vertex is P(0, 3).

To find the points of intersection of the line and the parabola.
Equating the values of y from both the equations, we get
x² + 3 = x + 3
∴ x² – x = 0
∴ x(x – 1) = 0
∴ x = 0 or x = 1
When x = 0, y = 0 + 3 = 3
When x = 1, y = 1 + 3 = 4
∴ the points of intersection are P(0, 3) and B(1, 4)
Required area = area of the region PABCP = area of the region OPABDO – area of the region OPCBDO
Now, area of the region OPABDO
= area under the line y = x + 3 between x = 0 and x = 1

Area of the region OPCBDO = area under the parabola y = x² + 3 between x = 0 and x = 1

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Line and Plane Ex 6.2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Line and Plane Ex 6.2

Question 1.
Find the length of the perpendicular from (2, -3, 1) to the line \(\frac{x+1}{2}=\frac{y-3}{3}=\frac{z+1}{-1}\)
Solution:
Let PM be the perpendicular drawn from the point P (2, -3, 1) to the line \(\frac{x+1}{2}=\frac{y-3}{3}=\frac{z+1}{-1}\) = λ …(Say)
The coordinates of any point on the line are given by x = -1 + 2λ, y = 3 + 3λ, z = -1 – λ
Let the coordinates of M be
(-1 + 2λ, 3 + 3λ, -1 – λ) … (1)
The direction ratios of PM are
-1 + 2λ – 2, 3 + 3λ + 3, -1 – λ – 1
i.e. 2λ – 3, 3λ + 6, -λ – 2
The direction ratios of the given line are 2, 3, -1.
Since PM is perpendicular to the given line, we get
2(2λ – 3) + 3(3λ + 6) – 1(-λ – 2) = 0
∴ 4λ – 6 + 9λ + 18 + λ + 2 = 0
∴ 14λ + 14 = 0 ∴ λ = -1.
Put λ = -1 in (1), the coordinats of M are
(-1 – 2, 3 – 3, -1 + 1) i.e. (-3, 0,0).
∴ length of perpendicular from P to the given line

Alternative Method:
We know that the perpendicular distance from the point P\(|\bar{\alpha}|\) to the line \(\bar{r}=\bar{a}+\lambda \vec{b}\) is given by

Substituting these values in (1), we get
length of perpendicular from P to given line

Question 2.
Find the co-ordinates of the foot of the perpendicular drawn from the point \(2 \hat{i}-\hat{j}+5 \hat{k}\) to the line \(\bar{r}=(11 \hat{i}-2 \hat{j}-8 \hat{k})+\lambda(10 \hat{i}-4 \hat{j}-11 \hat{k})\). Also find the length of the perpendicular.
Solution:
Let M be the foot of perpendicular drawn from the point P (\(2 \hat{i}-\hat{j}+5 \hat{k}\)) on the line
\(\bar{r}=(11 \hat{i}-2 \hat{j}-8 \hat{k})+\lambda(10 \hat{i}-4 \hat{j}-11 \hat{k})\).
Let the position vector of the point M be

Then \(\overline{\mathrm{PM}}\) = Position vector of M – Position vector of P
= [(11 + 10λ)\(\hat{i}\) + (-2 – 4λ)\(\hat{j}\) + -8 – 11λ) \(\hat{k}\)] – (2\(\hat{i}\) – \(\hat{j}\) + 5\(\hat{k}\))
= (9 + 10λ)\(\hat{i}\) + (-1 – 4λ)\(\hat{j}\) + (-13 – 11λ)\(\hat{k}\)
Since PM is perpendicular to the given line which is parallel to \(\bar{b}=10 \hat{i}-4 \hat{j}-11 \hat{k}\),
\(\overline{\mathrm{PM}}\) ⊥^r\(\bar{b}\) ∴ \(\overline{\mathrm{PM}} \cdot \bar{b}\) = 0
∴ [(9 + 10λ)\(\hat{i}\) + ( – 1 – 4λ)\(\hat{j}\) + (-13 – 11λ)\(\hat{k}\)]-(10\(\hat{i}\) – 4\(\hat{j}\) – 11\(\hat{k}\)) = 0
∴ 10(9 +10λ) – 4( -1 – 4λ) – 11( -13 – 11λ) = 0
∴ 90 + 100λ + 4 + 16λ + 143 +121λ = 0
∴ 237λ + 237 = 0
∴ λ = -1
Putting this value of λ, we get the position vector of M as \(\hat{i}+2 \hat{j}+3 \hat{k}\).
∴ coordinates of the foot of perpendicular M are (1, 2, 3).

Hence, the coordinates of the foot of perpendicular are (1,2, 3) and length of perpendicular = \(\sqrt {14}\) units.

Question 3.
Find the shortest distance between the lines \(\bar{r}=(4 \hat{i}-\hat{j})+\lambda(\hat{i}+2 \hat{j}-3 \hat{k})\) and \(\bar{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(\hat{i}+4 \hat{j}-5 \hat{k})\)
Solution:
We know that the shortest distance between the skew lines \(\bar{r}=\overline{a_{1}}+\lambda \overline{b_{1}}\) and \(\bar{r}=\overline{a_{2}}+\mu \overline{b_{2}}\) is given by

Question 4.
Find the shortest distance between the lines \(\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\) and \(\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\)
Solution:
The shortest distance between the lines

= 4(-6 + 2) – 6(7 – 1) + 8(-14 + 6)
= -16 – 36 – 64 = -116
and (m₁n₂ – m₂n₁)² + (l₂n₁ – l₁n₂)² + (l₁m₂ – l₂m₁)²
= (-6 + 2)² + (1 – 7)² + (-14 + 6)²
= 16 + 36 + 64 = 116
Hence, the required shortest distance between the given lines = \(\left|\frac{-116}{\sqrt{116}}\right|\) = \(\sqrt{116}\) = \(2 \sqrt{29}\) units

Question 5.
Find the perpendicular distance of the point (1, 0, 0) from the line \(\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}\) Also find the co-ordinates of the foot of the perpendicular.
Solution:
Let PM be the perpendicular drawn from the point (1, 0, 0) to the line \(\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}\) = λ …(Say)
The coordinates of any point on the line are given by x = -1 + 2λ, y = 3 + 2λ, z = 8 – λ
Let the coordinates of M be
(-1 + 2λ, 3 + 3λ, -1 – λ) …..(1)
The direction ratios of PM are
-1 + 2λ – 2, 3 + 3λ + 3, -1 – λ – 1
i.e. 2λ – 3, 3λ = 6, -λ – 2
The direction ratios of the given line are 2, 3, 8.
Since PM is perpendicular to the given line, we get
2(2λ – 3) + 3(3λ + 6) – 1(-λ – 2) = O
∴ 4λ – 6 + 9λ + 18 + λ + 2 = 0
∴ 14λ + 14 = 0
∴ λ = -1
Put λ in (1), the coordinates of M are
(-1 – 2, 3 – 3, -1 + 1) i.e. (-3, 0, 0).
∴ length of perpendicular from P to the given line
= PM
= \(\sqrt{(-3-2)^{2}+(0+3)^{2}+(0-1)^{2}}\)
= \(\sqrt{(25 + 9 + 1)}\)
= \(\sqrt{35}\)units.
Alternative Method :
We know that the perpendicular distance from the point

Substitutng tese values in (1), w get
length of perpendicular from P to given line
= PM

Question 6.
A(1, 0, 4), B(0, -11, 13), C(2, -3, 1) are three points and D is the foot of the perpendicular from A to BC. Find the co-ordinates of D.
Solution:
Equation of the line passing through the points (x₁, y₁, z₁) and (x₂, y₂, z₂) is

AD is the perpendicular from the point A (1, 0, 4) to the line BC.
The coordinates of any point on the line BC are given by x = 2λ, y = -11 + 8λ, z = 13 – 12λ
Let the coordinates of D be (2λ, -11 + 8λ, 13 – 12λ) … (1)
∴ the direction ratios of AD are
2λ – 1, -1λ + 8λ – 0, 13 – 12λ – 4 i.e.
2λ – 1, -11 + 8λ, 9 – 12λ
The direction ratios of the line BC are 2, 8, -12.
Since AD is perpendicular to BC, we get
2(2λ – 1) + 8(-11 + 8λ) – 12(9 – 12λ) = 0
∴ 42λ – 2 – 88 + 64λ – 108 + 144λ = 0
∴ 212λ – 198 = 0

Question 7.
By computing the shortest distance, determine whether following lines intersect each other.
(i) \(\bar{r}=(\hat{i}-\hat{j})+\lambda(2 \hat{i}+\hat{k})\) and \(\bar{r}=(2 \hat{i}-\hat{j})+\mu(\hat{i}+\hat{j}-\hat{k})\)
Solution:
The shortest distance between the lines


Hence, the given lines do not intersect.

(ii) \(\frac{x-5}{4}=\frac{y-7}{-5}=\frac{z+3}{-5}\) and \(\frac{x-8}{7}=\frac{y-7}{1}=\frac{z-5}{3}\)
Solution:
The shortest distance between the lines

∴ x₁ = -1, y₁ = -1, z₁ = -1, x₂ = 3, y₂ = 5, z₂ = 7,
l₁ = 7, m₁ = -6, n₁ = 1, l₂ = 1, m₂ = -2, n₂ = 1
\(\left|\begin{array}{ccc}
x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\
l_{1} & m_{1} & n_{1} \\
l_{2} & m_{2} & n_{2}
\end{array}\right|\) = \(\left|\begin{array}{ccc}
4 & 6 & 8 \\
4 & -5 & -5 \\
7 & 1 & 3
\end{array}\right|\)
= 4(- 6 + 2) – 6(7 – 1) + 8(-14 + 6)
= -16 – 36 – 64
= -116
and
(m₁n₂ – m₂n₁)² + (l₂n₁ – l₁n₂)² + (l₁m₂ – l₂m₁)²
= (-6 + 2)² + (1 – 7)² + (-14 + 6)²
= 16 + 36 + 64
= 116
Hence, the required shortest distance between the given lines
= \(\left|\frac{-116}{\sqrt{116}}\right|\)
= \(\sqrt{116}\)
=\(2 \sqrt{29}\) units
or
The shortest distance between the lines
= \(\frac{282}{\sqrt{3830}}\)units
Hence, the gives lines do not intersect.

Question 8.
If lines \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\) and \(\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}\) intersect each other then find k.
Solution:
The lines

∴ x₁ = 1, y₁ = -1, z₁ = 1, x₂ = 3, y₂ = k, z₂ = 0,
l₁ = 2, m₁ = 3, n₁ = 4, l₂ = 1, m₂ = 2, n₂ = 1.
Since these lines intersect, we get
\(\left|\begin{array}{ccc}
2 & k+1 & -1 \\
2 & 3 & 4 \\
1 & 2 & 1
\end{array}\right|\) = 0
∴ 2 (3 – 8) – (k + 1)(2 – 4) – 1 (4 – 3) = 0
∴ -10 + 2(k + 1) – 1 = 0
∴ 2(k + 1) = 11
∴ k + 1 = \(\frac{11}{2}\)
∴ k = \(\frac{9}{2}\)

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Line and Plane Ex 6.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Line and Plane Ex 6.1

Question 1.
Find the vector equation of the line passing through the point having position vector \(-2 \hat{i}+\hat{j}+\hat{k}\) and parallel to vector \(4 \hat{i}-\hat{j}+2 \hat{k}\).
Solution:
The vector equation of the line passing through A (\(\bar{a}\)) and parallel to the vector \(\bar{b}\) is \(\bar{r}\) = \(\bar{a}\) + λ\(\bar{b}\), where λ is a scalar.
∴ the vector equation of the line passing through the point having position vector \(-2 \hat{i}+\hat{j}+\hat{k}\) and parallel to the vector \(4 \hat{i}-\hat{j}+2 \hat{k}\) is
\(\bar{r}=(-2 \hat{i}+\hat{j}+\hat{k})+\lambda(4 \hat{i}-\hat{j}+2 \hat{k})\).

Question 2.
Find the vector equation of the line passing through points having position vectors \(3 \hat{i}+4 \hat{j}-7 \hat{k}\) and \(6 \hat{i}-\hat{j}+\hat{k}\).
Solution:
The vector equation of the line passing through the A (\(\bar{a}\)) and B(\(\bar{b}\)) is \(\bar{r}=\bar{a}+\lambda(\bar{b}-\bar{a})\), λ is a scalar
∴ the vector equation of the line passing through the points having position vectors \(3 \hat{i}+4 \hat{j}-7 \hat{k}\) and \(6 \hat{i}-\hat{j}+\hat{k}\) is

Question 3.
Find the vector equation of line passing through the point having position vector \(5 \hat{i}+4 \hat{j}+3 \hat{k}\) and having direction ratios -3, 4, 2.
Solution:
Let A be the point whose position vector is \(\bar{a}=5 \hat{i}+4 \hat{j}+3 \hat{k}\).
Let \(\bar{b}\) be the vector parallel to the line having direction ratios -3, 4, 2
Then, \(\bar{b}\) = \(-3 \hat{i}+4 \hat{j}+2 \hat{k}\)
The vector equation of the line passing through A (\(\bar{a}\)) and parallel to \(\bar{b}\) is \(\bar{r}=\bar{a}+\lambda \bar{b}\), where λ is a scalar.
∴ the required vector equation of the line is
\(\bar{r}=5 \hat{i}+4 \hat{j}+3 \hat{k}+\lambda(-3 \hat{i}+4 \hat{j}+2 \hat{k})\).

Question 4.
Find the vector equation of the line passing through the point having position vector \(\hat{i}+2 \hat{j}+3 \hat{k}\) and perpendicular to vectors \(\hat{i}+\hat{j}+\hat{k}\) and \(2 \hat{i}-\hat{j}+\hat{k}\).
Solution:

Since the line is perpendicular to the vector \(\bar{b}\) and \(\bar{c}\), it is parallel to \(\bar{b} \times \bar{c}\). The vector equation of the line passing through A (\(\bar{a}\)) and parallel to \(\bar{b} \times \bar{c}\) is
\(\bar{r}=\bar{a}+\lambda(\bar{b} \times \bar{c})\), where λ is a scalar.
Here, \(\bar{a}\) = \(\hat{i}+2 \hat{j}+3 \hat{k}\)
Hence, the vector equation of the required line is
\(\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+\hat{j}-3 \hat{k})\)

Question 5.
Find the vector equation of the line passing through the point having position vector \(-\hat{i}-\hat{j}+2 \hat{k}\) and parallel to the line \(\bar{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(3 \hat{i}+2 \hat{j}+\hat{k})\).
Solution:
Let A be point having position vector \(\bar{a}\) = \(-\hat{i}-\hat{j}+2 \hat{k}\)
The required line is parallel to the line

The vector equation of the line passing through A(\(\bar{a}\)) and parallel to \(\bar{b}\) is \(\bar{r}\) = \(\bar{a}\) + λ\(\bar{b}\) where λ is a scalar.
∴ the required vector equation of the line is
\(\overline{\mathrm{r}}=(-\hat{i}-\hat{j}+2 \hat{k})+\lambda(3 \hat{i}+2 \hat{j}+\hat{k})\).

Question 6.
Find the Cartesian equations of the line passing through A(-1, 2, 1) and having direction ratios 2, 3, 1.
Solution:
The cartesian equations of the line passing through (x₁, y₁, z₁) and having direction ratios a, b, c are

∴ the cartesian equations of the line passing through the point (-1, 2, 1) and having direction ratios 2, 3, 1 are

Question 7.
Find the Cartesian equations of the line passing through A(2, 2, 1) and B(1, 3, 0).
Solution:
The cartesian equations of the line passing through the points (x₁, y₁, z₁) and (x₂, y₂, z₂) are

Here, (x₁, y₁, z₁) = (2, 2, 1) and (x₂, y₂, z₂) = (1, 3, 0)
∴ the required cartesian equations are

Question 8.
A(-2, 3, 4), B(1, 1, 2) and C(4, -1, 0) are three points. Find the Cartesian equations of the line AB and show that points A, B, C are collinear.
Solution:
We find the cartesian equations of the line AB. The cartesian equations of the line passing through the points (x₁, y₁, z₁) and (x₂, y₂, z₂) are
\(\frac{x-x_{1}}{x_{2}-x_{1}}\) = \(\frac{y-y_{1}}{y_{2}-y_{1}}\) = \(\frac{z-z_{1}}{z_{2}-z_{1}}\)
Here, (x₁, y₁, z₁) = (-2, 3, 4) and (x₂, y₂, z₂) = (4, -1, 0)
∴ the required cartesian equations of the line AB are


∴ coordinates of C satisfy the equations of the line AB.
∴ C lies on the line passing through A and B.
Hence, A, B, C are collinear.

Question 9.
Show that lines \(\frac{x+1}{-10}=\frac{y+3}{-1}=\frac{z-4}{1}\) and \(\frac{x+10}{-1}=\frac{y+1}{-3}=\frac{z-1}{4}\) intersect each other. Find the co-ordinates of their point of intersection.
Solution:
The equations of the lines are

From (1), x = -1 -10λ, y = -3 – 2, z = 4 + λ
∴ the coordinates of any point on the line (1) are
(-1 – 10λ, – 3 – λ, 4 + λ)
From (2), x = -10 – u, y = -1 – 3u, z = 1 + 4u
∴ the coordinates of any point on the line (2) are
(-10 – u, -1 – 3u, 1 + 4u)
Lines (1) and (2) intersect, if
(- 1 – 10λ, – 3 – λ, 4 + 2) = (- 10 – u, -1 – 3u, 1 + 4u)
∴ the equations -1 – 10λ = -10 – u, -3 – 2= – 1 – 3u
and 4 + λ = 1 + 4u are simultaneously true.
Solving the first two equations, we get, λ = 1 and u = 1. These values of λ and u satisfy the third equation also.
∴ the lines intersect.
Putting λ = 1 in (-1 – 10λ, -3 – 2, 4 + 2) or u = 1 in (-10 – u, -1 – 3u, 1 + 4u), we get
the point of intersection (-11, -4, 5).

Question 10.
A line passes through (3, -1, 2) and is perpendicular to lines \(\bar{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(2 \hat{i}-2 \hat{j}+\hat{k})\) and \(\bar{r}=(2 \hat{i}+\hat{j}-3 \hat{k})+\mu(\hat{i}-2 \hat{j}+2 \hat{k})\). Find its equation.
Solution:

The vector perpendicular to the vectors \(\bar{b}\) and \(\bar{c}\) is given by

Since the required line is perpendicular to the given lines, it is perpendicular to both \(\bar{b}\) and \(\bar{c}\).
∴ it is parallel to \(\bar{b} \times \bar{c}\)
The equation of the line passing through A(\(\bar{a}\)) and parallel to \(\bar{b} \times \bar{c}\) is
\(\bar{r}=\bar{a}+\lambda(\bar{b} \times \bar{c})\), where λ is a scalar.
Here, \(\bar{a}\) = \(3 \hat{i}-\hat{j}+2 \hat{k}\)
∴ the equation of the required line is

Question 11.
Show that the line \(\frac{x-2}{1}=\frac{y-4}{2}=\frac{z+4}{-2}\) passes through the origin.
Solution:
The equation of the line is
\(\frac{x-2}{1}=\frac{y-4}{2}=\frac{z+4}{-2}\)
The coordinates of the origin O are (0, 0, 0)

∴ coordinates of the origin O satisfy the equation of the line.
Hence, the line passes through the origin.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Vectors Miscellaneous Exercise 5 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 5 Vectors Miscellaneous Exercise 5

I) Select the correct option from the given alternatives :
Question 1.
If |\(\bar{a}\)| = 2, |\(\bar{b}\)| = 3 |\(\bar{c}\)| = 4 then [\(\bar{a}\) + \(\bar{b}\) \(\bar{b}\) + \(\bar{c}\) \(\bar{c}\) – \(\bar{a}\)] is equal to
(A) 24
(B) -24
(C) 0
(D) 48
Solution:
(C) 0

Question 2.
If |\(\bar{a}\)| = 3, |\(\bar{b}\)| = 4, then the value of λ for which \(\bar{a}\) + λ\(\bar{b}\) is perpendicular to \(\bar{a}\) – λ\(\bar{b}\), is

Solution:
(b) \(\frac{3}{4}\)

Question 3.
If sum of two unit vectors is itself a unit vector, then the magnitude of their difference is
(A) \(\sqrt {2}\)
(B) \(\sqrt {3}\)
(C) 1
(D) 2
Solution:
(B) \(\sqrt {3}\)

Question 4.
If |\(\bar{a}\)| = 3, |\(\bar{b}\)| = 5, |\(\bar{c}\)| = 7 and \(\bar{a}\) + \(\bar{b}\) + \(\bar{c}\) = 0, then the angle between \(\bar{a}\) and \(\bar{b}\) is

Solution:
(b) \(\frac{\pi}{3}\)

Question 5.
The volume of tetrahedron whose vertices are (1, -6, 10), (-1, -3, 7), (5, -1, λ) and (7, -4, 7) is 11 cu. units then the value of λ is
(A) 7
(B) \(\frac{\pi}{3}\)
(C) 1
(D) 5
Solution:
(A) 7

Question 6.
If α, β, γ are direction angles of a line and α = 60º, β = 45º, the γ =
(A) 30º or 90º
(B) 45º or 60º
(C) 90º or 30º
(D) 60º or 120º
Solution:
(D) 60º or 120º

Question 7.
The distance of the point (3, 4, 5) from Y- axis is
(A) 3
(B) 5
(C) \(\sqrt {34}\)
(D) \(\sqrt {41}\)
Solution:
(C) \(\sqrt {34}\)

Question 8.
The line joining the points (-2, 1, -8) and (a, b, c) is parallel to the line whose direction ratios are 6, 2, 3. The value of a, b, c are
(A) 4, 3, -5
(B) 1, 2, \(\frac{-13}{2}\)
(C) 10, 5, -2
(D) 3, 5, 11
Solution:
(A) 4, 3, -5

Question 9.
If cos α, cos β, cos γ are the direction cosines of a line then the value of sin² α + sin²β + sin²γ is
(A) 1
(B) 2
(C) 3
(D) 4
Solution:
(B) 2

Question 10.
If l, m, n are direction cosines of a line then \(\hat{l}+m \hat{j}+n \hat{k}\) is
(A) null vector
(B) the unit vector along the line
(C) any vector along the line
(D) a vector perpendicular to the line
Solution:
(B) the unit vector along the line

Question 11.
If |\(\bar{a}\)| = 3 and –1 ≤ k ≤ 2, then |k\(\bar{a}\)| lies in the interval
(A) [0, 6]
(B) [-3, 6]
(C) [3, 6]
(D) [1, 2]
Solution:
(A) [0, 6]

Question 12.
Let α, β, γ be distinct real numbers. The points with position vectors \(\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}\), \(\beta \hat{i}+\gamma \hat{j}+\alpha \hat{k}\), \(\gamma \hat{i}+\alpha \hat{j}+\beta \hat{k}\)
(A) are collinear
(B) form an equilateral triangle
(C) form a scalene triangle
(D) form a right angled triangle
Solution:
(B) form an equilateral triangle

Question 13.
Let \(\bar{p}\) and \(\bar{q}\) be the position vectors of P and Q respectively, with respect to O and |\(\bar{p}\)| = p, |\(\bar{q}\)| = q. The points R and S divide PQ internally and externally in the ratio 2 : 3 respectively. If OR and OS are perpendicular then.
(A) 9p² = 4q²
(B) 4p² = 9q²
(C) 9p = 4q
(D) 4p = 9q
Solution:
(A) 9p² = 4q²

Question 14.
The 2 vectors \(\hat{j}+\hat{k}\) and \(3 \hat{i}-\hat{j}+4 \hat{k}\) represents the two sides AB and AC, respectively of a ∆ABC. The length of the median through A is
(A) \(\frac{\sqrt{34}}{2}\)
(B) \(\frac{\sqrt{48}}{2}\)
(C) \(\sqrt {18}\)
(D) None of these
Solution:
(A) \(\frac{\sqrt{34}}{2}\)

Question 15.
If \(\bar{a}\) and \(\bar{b}\) are unit vectors, then what is the angle between \(\bar{a}\) and \(\bar{b}\) for \(\sqrt{3} \bar{a}\) – \(\bar{b}\) to be a unit vector ?
(A) 30º
(B) 45º
(C) 60º
(D) 90º
Solution:
(A) 30º

Question 16.
If θ be the angle between any two vectors \(\bar{a}\) and \(\bar{b}\), then \(|\vec{a} \cdot \vec{b}|\) = \(|\vec{a} \times \vec{b}|\), when θ is equal to
(A) 0
(B) \(\frac{\pi}{4}\)
(C) \(\frac{\pi}{2}\)
(D) π
Solution:
(B) \(\frac{\pi}{4}\)

Question 17.
The value of \(\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{i} \times \hat{k})+\hat{k} \cdot(\hat{i} \times \hat{j})\)
(A) 0
(B) -1
(C) 1
(D) 3
Solution:
(C) 1

Question 18.
Let a, b, c be distinct non-negative numbers. If the vectors \(\mathrm{a} \hat{i}+\mathrm{a} \hat{j}+\mathrm{c} \hat{k}\), \(\hat{i}+\hat{k}\) and \(\mathbf{c} \hat{i}+\mathrm{c} \hat{j}+\mathrm{b} \hat{k}\) lie in a plane, then c is
(A) The arithmetic mean of a and b
(B) The geometric mean of a and b
(C) The harmonic man of a and b
(D) 0
Solution:
(B) The geometric mean of a and b

Question 19.
Let \(\bar{a}\) = \(\hat{i} \hat{j}\), \(\bar{b}\) = \(\hat{j} \hat{k}\), \(\bar{c}\) = \(\hat{k} \hat{i}\). If \(\bar{d}\) is a unit vector such that \(\bar{a} . \bar{d}=0=\left[\begin{array}{lll}
\bar{b} & \bar{c} & \bar{d}
\end{array}\right]\), then \(\bar{d}\) equals.

Solution:
(a) \(\pm \frac{\hat{i}+\hat{j}-2 \hat{k}}{\sqrt{6}}\)

Question 20.
If \(\bar{a}\) \(\bar{b}\) \(\bar{c}\) are non coplanar unit vectors such that \(\bar{a} \times(\bar{b} \times \bar{c})\) =\(\frac{(\bar{b}+\bar{c})}{\sqrt{2}}\) then the angle between \(\bar{a}\) and \(\bar{b}\) is
(A) \(\frac{3 \pi}{4}\)
(B) \(\frac{\pi}{4}\)
(C) \(\frac{\pi}{2}\)
(D) π
Solution:
(A) \(\frac{3 \pi}{4}\)

II Answer the following :
1) ABCD is a trapezium with AB parallel to DC and DC = 3AB. M is the mid-point of DC,
\(\overline{A B}\) = \(\bar{p}\) and \(\overline{B C}\) = \(\bar{q}\). Find in terms of \(\bar{p}\) and \(\bar{q}\).
(i) \(\overline{A M}\)
Solution:

(ii) \(\overline{B D}\)
Solution:

(iii) \(\overline{M B}\)
Solution:

(iv) \(\overline{D A}\)
Solution:

Question 2.
The points A, B and C have position vectors \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) respectively. The point P is midpoint of AB. Find in terms of \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) the vector \(\overline{P C}\)
Solution:
P is the mid-point of AB.
∴ \(\bar{p}\) = \(=\frac{\bar{a}+\bar{b}}{2}\), where \(\bar{p}\) is the position vector of P.

Question 3.
In a pentagon ABCDE
Show that \(\overline{A B}\) + \(\overline{A E}\) + \(\overline{B C}\) + \(\overline{D C}\) + \(\overline{E D}\) = 2\(\overline{A C}\)
Solution:

Question 4.
If in parallelogram ABCD, diagonal vectors are \(\overline{A C}\) = \(2 \hat{i}+3 \hat{j}+4 \hat{k}\) and \(\overline{B D}\) = \(-6 \hat{i}+7 \hat{j}-2 \hat{k}\), then find the adjacent side vectors \(\overline{A B}\) and \(\overline{A D}\)
Solution:
ABCD is a parallelogram

Question 5.
If two sides of a triangle are \(\hat{i}+2 \hat{j}\) and \(\hat{i}+\hat{k}\), then find the length of the third side.
Solution:

Let ABC be a triangle with \(\overline{A B}\) = \(\hat{i}+2 \hat{j}\), \(\overline{B C}\) = \(\hat{i}+\hat{k}\).
By triangle law of vectors

Hence, the length of third side is 3 units.

Question 6.
If |\(\bar{a}\)| = |\(\bar{b}\) | = 1 \(\bar{a}\).\(\bar{b}\) = 0 and \(\bar{a}\) + \(\bar{b}\) + \(\bar{c}\) = 0 then find |\(\bar{c}\)|
Solution:
\(\bar{a}\) + \(\bar{b}\) + \(\bar{c}\) = 0
∴ –\(\bar{c}\) = \(\bar{a}\) + \(\bar{b}\)
Taking dot product of both sides with itself, we get

Question 7.
Find the lengths of the sides of the triangle and also determine the type of a triangle.
(i) A(2, -1, 0), B(4, 1, 1,), C(4, -5, 4)
Solution:
The position vectors \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) of the points A, B, C are


∴ ∆ ABC is right angled at A.

(ii) L(3, -2, -3), M(7, 0, 1), N (1, 2, 1)
Solution:
The position vectors bar \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) of the points L M, N are

l(LM) = 6, l(MN) = 2\(\sqrt {10}\) , l(NL) = 6
∆LMN is sosceles

Question 8.
Find the component form of if a if
(i) It lies in YZ plane and makes 60º with positive Y-axis and |\(\bar{a}\)| = 4
Solution:
Let α, β, γ be the direction angles of \(\bar{a}\)
Since \(\bar{a}\) lies in YZ-plane, it is perpendicular to X-axis
∴ α = 90°
It is given that β= 60°
∵ cos²α + cos²β + cos²γ = 1
∴ cos²90° + cos²60° + cos²γ = 1
∴ 0 + \(\left(\frac{1}{2}\right)^{2}\) + cos²γ = 1
∴ cos²γ = 1 – \(\frac{1}{4}=\frac{3}{4}\)
∴ cos γ = \(\pm \frac{\sqrt{3}}{2}\)
Unit vector along a is given by

(ii) It lies in XZ plane and makes 45º with positive Z-axis and |\(\bar{a}\)| = 10
Solution:

Question 9.
Two sides of a parallelogram are \(3 \hat{i}+4 \hat{j}-5 \hat{k}\) and \(-2 \hat{j}+7 \hat{k}\). Find the unit vectors parallel to the diagonals.
Solution:

Question 10.
If D, E, F are the mid-points of the sides BC, CA, AB of a triangle ABC , prove that \(\overline{A D}\) + \(\overline{B E}\) + \(\overline{C F}\) = 0

Solution:
Let \(\bar{a}\), \(\bar{b}\), \(\bar{c}\), \(\bar{d}\), \(\bar{e}\), \(\bar{f}\) be the position vectors of the points A, B, C, D, E, F respectively.
Since D, E, F are the midpoints of BC, CA, AB respec-tively, by the midpoint formula

Question 11.
Find the unit vectors that are parallel to the tangent line to the parabola y = x² at the point (2, 4)
Solution:
Differentiating y = x² w.r.t. x, we get \(\) = 2x
Slope of tangent at P(2, 4) = \(\left(\frac{d y}{d x}\right)_{\text {at } \mathrm{P}(2, 4)}\) = 2 × 2 = 4
∴ the equation of tangent at P is
y – 4 = 4(-2)
∴ y = 4x – 4
∴ y = 4x is equation of line parallel to the tangent at P and passing through the origin O.
4x = y, z = 0 ∴ \(\frac{x}{1}=\frac{y}{4}\), z = 0
∴ the direction ratios of this line are 1, 4, 0
∴ its direction cosines are

Question 12.
Express the vector \(\hat{i}+4 \hat{j}-4 \hat{k}\) as a linear combination of the vectors \(2 \hat{i}-\hat{j}+3 \hat{k}\), \(\hat{i}-2 \hat{j}+4 \hat{k}\) and \(-\hat{i}+3 \hat{j}-5 \hat{k}\)
Solution:

By equality of vectors,
2x + 2y – z = 1
-x – 2y + 3z = 4
3x + 4y – 5z = -4
We have to solve these equations by using Cramer’s Rule.
D = \(\left|\begin{array}{ccc}
2 & 2 & -1 \\
-1 & -2 & 3 \\
3 & 4 & -5
\end{array}\right|\)
= 2(10 – 12) – 2(5 – 9) – 1(-4 + 6)
= -4 + 8 – 2
= 2 ≠ 0
D_x = \(\left|\begin{array}{ccc}
1 & 2 & -1 \\
4 & -2 & 3 \\
-4 & 4 & -5
\end{array}\right|\)
= 1(10 – 12) – 2(-20 + 12) – 1 (16 – 8)
= -2 + 16 – 8
= 6
D_y = \(\left|\begin{array}{ccc}
2 & 1 & -1 \\
-1 & 4 & 3 \\
3 & -4 & -5
\end{array}\right|\)
= 2(-20 + 12) – 1(5 – 9) – 1(4 – 12)
= -16 – 4 – 8
= -28
D_z = \(\left|\begin{array}{ccc}
2 & 2 & 1 \\
-1 & -2 & 4 \\
3 & 4 & -4
\end{array}\right|\)
= 2(8 – 16) – 2(4 – 12) + 1(-4 + 6)
= -16 – 16 + 2
= -30

Question 13.
If \(\overline{O A}\) = \(\bar{a}\) and \(\overline{O B}\) = \(\bar{b}\) then show that the vector along the angle bisector of angle AOB is
given by \(\bar{d}\) = λ\(\left(\frac{\bar{a}}{|\bar{b}|}+\frac{\bar{b}}{|\bar{b}|}\right)\)
Question is modified
If \(\overline{O A}\) = \(\bar{a}\) and \(\overline{O B}\) = \(\bar{b}\) then show that the vector along the angle bisector of ∠AOB is
given by \(\bar{d}\) = λ\(\left(\frac{\bar{a}}{|\bar{a}|}+\frac{\bar{b}}{|\bar{b}|}\right)\)
Solution:

Choose any point P on the angle bisector of ∠AOB. Draw PM parallel to OB.
∴ ∠OPM = ∠POM
= ∠POB
Hence, OM = MP
∴ OM and MP is the same scalar multiple of unit vectors \(\hat{a}\) and \(\hat{b}\) along these directions,

Question 14.
The position vectors f three consecutive vertices of a parallelogram are \(\hat{i}+\hat{j}+\hat{k}\), \(\hat{i}+3 \hat{j}+5 \hat{k}\) and \(7 \hat{i}+9 \hat{j}+11 \hat{k}\) Find the position vector of the fourth vertex.
Solution:
Let ABCD be a parallelogram.
Let \(\bar{a}\), \(\bar{b}\), \(\bar{c}\), \(\bar{d}\) be the position vectors of the vertices
A, B, C, D of the parallelogram,

Hence, the position vector of the fourth vertex is 7(\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)).

Question 15.
A point P with position vector \(\frac{-14 \hat{i}+39 \hat{j}+28 \hat{k}}{5}\) divides the line joining A(-1, 6, 5) and B in the ratio 3 : 2 then find the point B.
Solution:
Let A, B and P have position vectors \(\bar{a}\), \(\bar{b}\) and \(\bar{p}\) respectively.

∴ coordinates of B are (-4, 9, 6).

Question 16.
Prove that the sum of the three vectors determined by the medians of a triangle directed from the vertices is zero.
Solution:
Let \(\overrightarrow{\mathrm{a}}\), \(\overrightarrow{\mathrm{b}}\) and \(\overrightarrow{\mathrm{c}}\) are the position vectors of the vertices A, B and C respectively.
Then we know that the position vector of the centroid O of the triangle is \(\frac{\vec{a}+\vec{b}+\vec{c}}{3}\)
Therefore sum of the three vectors \(\overrightarrow{\mathrm{OA}}\), \(\overrightarrow{\mathrm{OB}}\) and \(\overrightarrow{\mathrm{OC}}\), is

Hence, Sum os the three vectors determined by the medians of a triangle directed from the vertices is zero.

Question 17.
ABCD is a parallelogram E, F are the mid points of BC and CD respectively. AE, AF meet the diagonal BD at Q and P respectively. Show that P and Q trisect DB.
Solution:


LHS is the position vector of the point on AE and RHS is the position vector of the point on DB. But AE and DB meet at Q.

LHS is the position vector of the point on AF and RHS is the position vector of the point on DB.
But AF and DB meet at P.
∴ \(\bar{p}=\frac{\bar{b}+2 \bar{d}}{1+2}\)
∴ P divides DB in the ratio 1 : 2 … (5)
From (4) and (5), if follows that P and Q trisect DB.

Question 18.
If aBC is a triangle whose orthocenter is P and the circumcenter is Q, then prove that \(\overline{P A}\) + \(\overline{P C}\) + \(\overline{P B}\) = 2 \(\overline{P Q}\)
Solution:
Let G be the centroid of the ∆ ABC.
Let A, B, C, G, Q have position vectors \(\bar{a}\), \(\bar{b}\), \(\bar{c}\), \(\bar{g}\), \(\bar{q}\) w.r.t. P. We know that Q, G, P are collinear and G divides segment QP internally in the ratio 1 : 2.

Question 19.
If P is orthocenter, Q is circumcenter and G is centroid of a triangle ABC, then prove that \(\overline{Q P}\) = 3\(\overline{Q G}\)
Solution:
Let \(\bar{p}\) and \(\bar{g}\) be the position vectors of P and G w.r.t. the circumcentre Q.
i.e. \(\overline{\mathrm{QP}}\) = p and \(\overline{\mathrm{QG}}\) = g.
We know that Q, G, P are collinear and G divides segment QP internally in the ratio 1 : 2
∴ by section formula for internal division,

Question 20.
In a triangle OAB, E is the midpoint of BO and D is a point on AB such that AD: DB = 2:1. If OD and AE intersect at P, determine the ratio OP:PD using vector methods.
Solution:

Let A, B, D, E, P have position vectors \(\bar{a}\), \(\bar{b}\), \(\bar{d}\), \(\bar{e}\), \(\bar{p}\) respectively w.r.t. O.
∵ AD : DB = 2 : 1.
∴ D divides AB internally in the ratio 2 : 1.
Using section formula for internal division, we get

LHS is the position vector of the point which divides OD internally in the ratio 3 : 2.
RHS is the position vector of the point which divides AE internally in the ratio 4 : 1.
But OD and AE intersect at P
∴ P divides OD internally in the ratio 3 : 2.
Hence, OP : PD = 3 : 2.

Question 21.
Dot-product of a vector with vectors \(3 \hat{i}-5 \hat{k}, 2 \hat{i}+7 \hat{j}\) and \(\hat{i}+\hat{j}+\hat{k}\) are respectively -1, 6 and 5. Find the vector.
Solution:

∴ 3x – 5z= -1 … (1)
∴ 2x + 7y = 6 … (2)
∴ x + y + z = 5 … (3)
From (3), z = 5 – x – y
Substituting this value of z in (1), we get
∴ 3x – 5(5 – x – y)= -1
∴ 8x + 5y = 24 … (4)
Multiplying (2) by 4 and subtracting from (4), we get
8x + 5y – 4(2x + 7y) = 24 – 6 × 4
∴ -23y = 0 ∴ y = 0
Substituting y = 0 in (2), we get
∴ 2x = 6 ∴ x = 3
Substituting x = 3 in (1), we get
∴ 3(3) – 5z = -1
∴ 5z = -10 ∴ z = 2
∴ \(\bar{r}=3 \hat{i}+0 \cdot \hat{j}+2 \hat{k}=3 \hat{i}+2 \hat{k}\)
Hence, the required vector is \(3 \hat{i}+2 \hat{k}\)

Question 22.
If \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) are unit vectors such that \(\bar{a}\) + \(\bar{b}\) + \(\bar{c}\) = 0, then find the value of \(\bar{a}\).\(\bar{b}\) + \(\bar{b}\).\(\bar{c}\) + \(\bar{c}\).\(\bar{a}\)
Solution:
\(\bar{a}\), \(\bar{b}\), \(\bar{c}\) are unit vectors

Adding (2), (3), (4) and using the fact that scalar product commutative, we get

Question 23.
If a parallelogram is constructed on the vectors \(\bar{a}=3 \bar{p}-\bar{q}\), \(\bar{b}=\bar{p}+3 \bar{q}\) and \(|\bar{p}|=|\bar{q}|=2\) and angle between \(\bar{p}\) and \(\bar{q}\) is\(\frac{\pi}{3}\) show that the ratio of the lengths of the sides is \(\sqrt {7}\) : \(\sqrt {13}\)
Solution:


Hence, the ratio of the lengths of the sides is \(\sqrt {7}\) : \(\sqrt {13}\).

Question 24.
Express the vector \(\bar{a}=5 \hat{i}-2 \hat{j}+5 \hat{k}\) as a sum of two vectors such that one is parallel to the vector \(\bar{b}=3 \hat{i}+\hat{k}\) and other is perpendicular to \(\bar{b}\).
Solution:

By equality of vectors
3m + x = 5 … (1)
y = -2
and m – 3x = 5
From (1) and (2)
3m + x = m – 3x
∴ 2m = -4x m ∴ m = -2x
Substituting m = -2x in (1), we get
∴ -6x + x = 5 ∴ -5x = 5 ∴ x = -1
∴ m = -2x = 2

Question 25.
Find two unit vectors each of which makes equal angles with \(\bar{u}\), \(\bar{v}\) and \(\bar{w}\). \(\bar{u}=2 \hat{i}+\hat{j}-2 \hat{k}\), \(\bar{v}=\hat{i}+2 \hat{j}-2 \hat{k}\) and \(\bar{W}=2 \hat{i}-2 \hat{j}+\hat{k}\)
Solution:

Question 26.
Find the acute angles between the curves at their points of intersection. y = x², y = x³
Solution:
The angle between the curves is same as the angle between their tangents at the points of intersection. We find the points of intersection of y = x² … (1)
and y = x³ … (2)
From (1) and (2)
x³ = x²
∴ x³ – x² = 0
∴ x²(x – 1) = 0
∴ x = 0 or x = 1
When x = 0, y = 0.
When x = 1, y = 1.

∴ equation of tangent to y = x³ at P is y = 0.
∴ the tangents to both curves at (0, 0) are y = 0
∴ angle between them is 0.
Angle at P = (1, 1)
Slope of tangent to y = x² at P

∴ equation of tangent to y = x³ at P is y – 1 = 3(x – 1) y = 3x – 2
We have to find angle between y = 2x – 1 and y = 3x – 2
Lines through origin parallel to these tagents are y = 2x and y = 3x
∴ \(\frac{x}{1}=\frac{y}{2}\) and \(\frac{x}{1}=\frac{y}{3}\)
These lines lie in XY-plane.
∴ the direction ratios of these lines are 1, 2, 0 and 1, 3, 0.
The angle θ between them is given by

Question 27.
Find the direction cosines and direction angles of the vector.
(i) \(2 \hat{i}+\hat{j}+2 \hat{k}\)
Solution:
Let \(\bar{a}\) = \(2 \hat{i}+\hat{j}+2 \hat{k}\)

(ii) \((1 / 2) \hat{i}+\hat{j}+\hat{k}\)
Solution:

Question 28.
Let \(\bar{b}\) = \(4 \hat{i}+3 \hat{j}\) and \(\bar{c}\) be two vectors perpendicular to each other in the XY-plane. Find vectors in the same plane having projection 1 and 2 along \(\bar{b}\) and \(\bar{c}\), respectively, are given y.
Solution:
\(\bar{b}\) = \(4 \hat{i}+3 \hat{j}\)

Question 29.
Show that no line in space can make angle \(\frac{\pi}{6}\) and \(\frac{\pi}{4}\) with X- axis and Y-axis.
Solution:
Let, if possible, a line in space make angles \(\frac{\pi}{6}\) and \(\frac{\pi}{4}\) with X-axis and Y-axis.

∴ cos²γ = 1 – \(\frac{3}{4}-\frac{1}{2}=-\frac{1}{4}\)
This is not possible, because cos γ is real
∴ cos²γ cannot be negative.
Hence, there is no line in space which makes angles \(\frac{\pi}{6}\) and \(\frac{\pi}{4}\) with X-axis and Y-axis.

Question 30.
Find the angle between the lines whose direction cosines are given by the equation 6mn – 2nl + 5lm = 0, 3l + m + 5n = 0
Solution:
Given 6mn – 2nl + 5lm = o
3l + m +5n = 0.
From (2), m = 3l – 5n
Putting the value of m in equation (1), we get,
⇒ 6n(-3l – 5n) – 2nl + 5l(-3l – 5n) = 0
⇒ -18nl- 30n – 2nl- 15l2 – 25nl = 0
⇒ – 30n² – 45nl – 15l² = 0
⇒ 2n² + 3nl + l² = 0
⇒ 2n² + 2nl + nl + l² = 0
⇒ (2n + l) (n + l) = 0
∴ 2n + l = 0 OR n + l = 0
∴ l = -2n OR l = -n
∴ l = -2n
From (2), 3l + m + 5n = 0
∴ -6n + m + 5n = 0
∴ m = n
i.e. (-2n, n, n) = (-2, 1, 1)
∴ l = -n
∴ -3n + m + 5n = 0
∴ m = -2n
i.e. (-n, -2n, n) = (1, 2, -1)
(a₁, b₁, c₁) = (-2, 1, 1) and (a₂, b₃, c₃) = (1, 2, -1)

Question 31.
If Q is the foot of the perpendicular from P(2, 4, 3) on the line joining the points A(1, 2, 4) and B(3, 4, 5), find coordinates of Q.
Solution:
Let PQ be the perpendicular drawn from point P(2, 4, 3) to the line joining the points A(1, 2, 4) and B (3, 4, 5).
Let Q divides AB internally in the ratio λ : 1

Now, direction ratios of AB are, 3 – 1, 4 – 2, 5 – 4 i.e., 2, 2, 1.

Coordinates of Q are,

Question 32.
Show that the area of a triangle ABC, the position vectors of whose vertices are a, b and c is \(\frac{1}{2}[\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}]\)
Question is modified.
Show that the area of a triangle ABC, the position vectors of whose vertices are \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) is \(\frac{1}{2}[\bar{a} \times \bar{b}+\bar{b} \times \bar{c}+\bar{c} \times \bar{a}]\).
Solution:

Consider the triangle ABC.
Complete the parallelogram ABDC.
Vector area of ∆ABC

Question 33.
Find a unit vector perpendicular to the plane containing the point (a, 0, 0), (0, b, 0), and (0, 0, c). What is the area of the triangle with these vertices?
Solution:

Question 34.
State whether each expression is meaningful. If not, explain why ? If so, state whether it is a vector or a scalar.
(a) \(\bar{a} \cdot(\bar{b} \times \bar{c})\)
Solution:
This is the scalar product of two vectors. Therefore, this expression is meaningful and it is a scalar.

(b) \(\bar{a} \times(\bar{b} \cdot \bar{c})\)
Solution:
This expression is meaningless because \(\bar{a}\) is a vector, \(\bar{b} \cdot \bar{c}\) is a scalar and vector product of vector and scalar is not defined.

(c) \(\bar{a} \times(\bar{b} \times \bar{c})\)
Solution:
This is vector product of two vectors. Therefore, this expression is meaningful and it is a vector.

(d) \(\bar{a} \cdot(\bar{b} \cdot \bar{c})\)
Solution:
This is meaningless because \(\bar{a}\) is a vector, \(\bar{b} \cdot \bar{c}\) is a scalar and scalar product of vector and scalar is not defined.

(e) \((\bar{a} \cdot \bar{b}) \times(\bar{c} \cdot \bar{d})\)
Solution:
This is meaningless because \(\bar{a} \cdot \bar{b}, \bar{c} \cdot \bar{d}\) are scalars and cross product of two scalars is not defined.

(f) \((\bar{a} \times \bar{b}) \cdot(\bar{c} \times \bar{d})\)
Solution:
This is scalar product of two vectors. Therefore, this expression is meaningful and it is a scalar.

(g) \((\bar{a} \cdot \bar{b}) \cdot \bar{c}\)
Solution:
This is meaningless because \(\bar{c}\) is a vector, \(\bar{a} \cdot \bar{b}\) scalar and scalar product of vector and scalar is not defined.

(h) \((\bar{a} \cdot \bar{b}) \bar{c}\)
Solution:
This is a scalar multiplication of a vector. Therefore, this expression is meaningful and it is a vector.

(i) \((|\bar{a}|)(\bar{b} \cdot \bar{c})\)
Solution:
This is the product of two scalars. Therefore, this expression is meaningful and it is a scalar.

(j) \(\bar{a} \cdot(\bar{b}+\bar{c})\)
Solution:
This is the scalar product of two vectors. Therefore, this expression is meaningful and it is a scalar.

(k) \(\bar{a} \cdot \bar{b}+\bar{c}\)
Solution:
This is the sum of scalar and vector which is not defined. Therefore, this expression is meaningless.

(l) \(|\bar{a}| \cdot(\bar{b}+\bar{c})\)
Solution:
This is meaningless because \(\bar{a}\) is a vector, \(\overline{\mathrm{b}}+\overline{\mathrm{c}}\) is a scalar and the scalar product of vector and scalar is not defined.

Question 35.
Show that, for any vectors \(\bar{a}, \bar{b}, \bar{c}\)
\((\bar{a}+\bar{b}+\bar{c}) \times \bar{c}+(\bar{a}+\bar{b}+\bar{c}) \times \bar{b}+(\bar{b}+\bar{c}) \times \bar{a}=2 \bar{a} \times \bar{c}\)
Question is modified.
For any vectors \(\bar{a}, \bar{b}, \bar{c}\) show that
\(\begin{aligned}
&(\bar{a}+\bar{b}+\bar{c}) \times \bar{c}+(\bar{a}+\bar{b}+\bar{c}) \times \bar{b}+(\bar{b}-\bar{c}) \times \bar{a} \\
&=2 \bar{a} \times \bar{c} .
\end{aligned}\)
Solution:

Question 36.
Suppose that \(\bar{a}\) = 0.
(a) If \(\bar{a} \cdot \bar{b}=\bar{a} \cdot \bar{c}\) then is \(\bar{b}=\bar{c}\)?
Solution:

(b) If \(\bar{a} \times \bar{b}=\bar{a} \times \bar{c}\) then is \(\bar{b}=\bar{c}\)?
Solution:

(c) If \(\bar{a} \cdot \bar{b}=\bar{a} \cdot \bar{c}\) and \(\overline{\mathrm{a}} \times \overline{\mathrm{b}}=\overline{\mathrm{a}} \times \overline{\mathrm{c}}\) then is \(\bar{b}=\bar{c}\)?
Solution:

Question 37.
If A(3, 2, -1), B(-2, 2, -3), C(3, 5, -2), D(-2, 5, -4) then
(i) verify that the points are the vertices of a parallelogram and
Solution:

∴ opposite sides AB and DC of ABCD are parallel and equal.
∴ ABCD is a parallelogram.

(ii) find its area.
Solution:

Question 38.
Let A, B, C, D be any four points in space. Prove that \(|\overline{A B} \times \overline{C D}+\overline{B C} \times \overline{A D}+\overline{C A} \times \overline{B D}|\) = 4 (area of ∆ABC)
Solution:
Let A, B, C, D have position vectors \(\bar{a}\), \(\bar{b}\), \(\bar{c}\), \(\bar{d}\) respectively.
Consider \(|\overline{A B} \times \overline{C D}+\overline{B C} \times \overline{A D}+\overline{C A} \times \overline{B D}|\)

Question 39.
Let \(\hat{a}, \hat{b}, \hat{c}\) be unit vectors such that \(\hat{a} \cdot \hat{b}=\hat{a} \cdot \hat{c}=0\) and the angle between \(\hat{b}\) and \(\hat{c}\) be\(\frac{\pi}{6}\).
Prove that \(\hat{a}=\pm 2(\hat{b} \times \hat{c})\)
Solution:
\(\hat{a} \cdot \hat{b}=\hat{a} \cdot \hat{c}=0\)
∴ \(\hat{a}\) is perpendicular to \(\hat{b}\) and \(\hat{c}\) both
∴ \(\hat{a}\) is parallel to \(\hat{b}\) × \(\hat{c}\)
∴ \(\hat{a}\) = m(\(\hat{b}\) × \(\hat{c}\)), m is a scalar.

Question 40.
Find the value of ‘a’ so that the volume of parallelopiped a formed by \(\hat{i}+\hat{j}+\hat{k}+a \hat{k}\) aand \(a j+\hat{k}\) becomes minimum.
Question is modified.
Find the value of ‘a’ so that the volume of parallelopiped formed by \(\hat{i}+a \hat{j}+\hat{k}, \hat{j}+a \hat{k}\) and \(a \hat{i}+\hat{k}\) becomes minimum.
Solution:
Let \(\bar{p}\) = \(\hat{i}+a \hat{j}+\hat{k}\), \(\bar{q}\) = \(\hat{j}+a \hat{k}\), \(\bar{r}\) = \(a \hat{i}+\hat{k}\)
Let V be the volume of the parallelopiped formed by \(\bar{p}, \bar{q}, \bar{r}\).

Question 41.
Find the volume of the parallelepiped spanned by the diagonals of the three faces of a cube of side a that meet at one vertex of the cube.
Solution:

Take origin O as one vertex of the cube and OA, OB and OC as the positive directions of the X-axis, the Y-axis and the Z-axis respectively.
Here, the sides of the cube are
OA = OB = OC = a
∴ the coordinates of all the vertices of the cube will be
O = (0, 0, 0) A = (a, 0, 0)
B = (0, a, 0) C = (0, 0, a)
N = (a, a, 0) L = (0, a, a)
M = (a, 0, a) P = (a, a, a)
ON, OL, OM are the three diagonals which meet at the vertex O

Question 42.
If \(\bar{a}, \bar{b}, \bar{c}\) are three non-coplanar vectors, then show that \(\frac{\bar{a} \cdot(\bar{b} \times \bar{c})}{(\bar{c} \times \bar{a}) \cdot \bar{b}}+\frac{\bar{b} \cdot(\bar{a} \times \bar{c})}{(\bar{c} \times \bar{a}) \cdot \bar{b}}\) = 0
Solution:

Question 43.
Prove that \((\bar{a} \times \bar{b}) \cdot(\bar{c} \times \bar{d})\left|\begin{array}{ll}
\bar{a} \cdot \bar{c} & \bar{b} \cdot \bar{c} \\
\bar{a} \cdot \bar{d} & \bar{b} \cdot \bar{d}
\end{array}\right|\)
Solution:

Question 44.
Find the volume of a parallelopiped whose coterminus edges are represented by the vector \(\hat{j}+\hat{k} \cdot \hat{i}+\hat{k}\) and \(\hat{i}+\hat{j}\). Also find volume of tetrahedron having these coterminous edges.
Solution:
Let \(\bar{a}\) = \(\hat{j}+\hat{k}\), \(\bar{b}\) = \(\hat{i}+\hat{k}\) and \(\bar{c}\) = \(\hat{i}+\hat{j}\) be the co-terminus edges of a parallelopiped.
Then volume of the parallelopiped = \([\bar{a} \bar{b} \bar{c}]\)
= \(\left|\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right|\)
= 0(0 – 1) – 1(0 – 1) + 1(1 – 0)
= 0 + 1 + 1 = 2cu units.
Also, volume of tetrahedron = \(\frac{1}{6}[\bar{a} \bar{b} \bar{c}]\)
= \(\frac{1}{6}(2)=\frac{1}{3}\) cubic units.

Question 45.
Using properties of scalar triple product, prove that \(\left[\begin{array}{llll}
\bar{a}+\bar{b} & \bar{b}+\bar{c} & \bar{c}+\bar{a}
\end{array}\right]=2\left[\begin{array}{lll}
\bar{a} & \bar{b} & \bar{c}
\end{array}\right]\).
Solution:

Question 46.
If four points A(\(\bar{a}\)), B(\(\bar{b}\)), C(\(\bar{c}\)) and D(\(\bar{d}\)) are coplanar then show that \(\left[\begin{array}{lll}
\bar{a} \bar{b} \bar{d}]+\left[\begin{array}{lll}
\bar{b} & \bar{c} & \bar{d}
\end{array}\right]+\left[\begin{array}{lll}
\bar{c} & \bar{a} & \bar{d}
\end{array}\right]=[\overline{\bar{a}} \bar{b} & \bar{c}
\end{array}\right]\)
Solution:

Question 47.
If \(\bar{a}\) \(\bar{b}\) and \(\bar{c}\) are three non coplanar vectors, then \((\bar{a}+\bar{b}+\bar{c}) \cdot[(\bar{a}+\bar{b}) \times(\bar{a}+\bar{c})]=-\left[\begin{array}{lll}
\bar{a} & \bar{b} & \bar{c}
\end{array}\right]\).
Solution:

Question 48.
If in a tetrahedron, edges in each of the two pairs of opposite edges are perpendicular, then show that the edges in the third pair are also perpendicular.
Solution:

Let O-ABC be a tetrahedron. Then o
(OA, BC), (OB, CA) and (OC, AB) are the pair of opposite edges.
Take O as the origin of reference and let \(\bar{a}\) \(\bar{b}\) and
∴ the third pair (OC, AB) is perpendicular.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Vectors Ex 5.5 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 5 Vectors Ex 5.5

Question 1.
Find \(\bar{a}\)∙(\(\bar{b}\) × \(\bar{c}\)), if \(\bar{a}\) = \(3 \hat{i}-\hat{j}+4 \hat{k}\), \(\bar{b}\) = \(2 \hat{i}+3 \hat{j}-\hat{k}\) and \(\bar{c}\) = \(-5 \hat{i}+2 \hat{j}+3 \hat{k}\)
Solution:
\(\bar{a}\)∙(\(\bar{b}\) × \(\bar{c}\)) = \(\left|\begin{array}{rrr}
3 & -1 & 4 \\
2 & 3 & -1 \\
-5 & 2 & 3
\end{array}\right|\)
= 3(9 + 2) + 1 (6 – 5) + 4(4 + 15)
= 33 + 1 + 76
= 110.

Question 2.
If the vectors \(3 \hat{i}+5 \hat{k}, 4 \hat{i}+2 \hat{j}-3 \hat{k}\) and \(3 \hat{i}+\hat{j}+4 \hat{k}\) are to co-terminus edges of the parallelo piped, then find the volume of the parallelopiped.
Solution:

= 3(8 + 3) – 0(16 + 9) + 5(4 – 6)
= 33 – 0 – 10 = 23
∴ volume of the parallelopiped = \([\bar{a} \bar{b} \bar{c}]\)
= 23 cubic units.

Question 3.
If the vectors \(-3 \hat{i}+4 \hat{j}-2 \hat{k}, \hat{i}+2 \hat{k}\) and\(\hat{i}-p \hat{j}\) are coplanar, then find the value of p.
Solution:

∴ -3(0 + 2p) – 4(0 – 2) – 2(-p – 0) = 0
∴ -6p + 8 + 2p = 0
∴ -4p = -8
P = 2.

Question 4.
Prove that :
(i) [latex]\bar{a} \bar{b}+\bar{c} \bar{a}+\bar{b}+\bar{c}[/latex] = 0
Solution:

(ii) (\(\bar{a}\) – \(2 \bar{b}\) – \(\bar{c}\))∙[(\(\bar{a}\) – \(\bar{b}\)) × \(\bar{a}\) – \(\bar{b}\) – \(\bar{c}\)] = 3[\(\bar{a}\) – \(\bar{b}\) – \(\bar{c}\)]
Question is modified.
(\(\bar{a}\) – \(2 \bar{b}\) – \(\bar{c}\)) [(\(\bar{a}\) – \(\bar{b}\)) × \(\bar{a}\) – \(\bar{b}\) – \(\bar{c}\)] = 3[\(\bar{a}\) \(\bar{b}\) \(\bar{c}\)]
Solution:

Question 5.
If \(\bar{c}\) =3\(\bar{a}\) – 2\(\bar{b}\) prove that [\(\bar{a}\) \(\bar{b}\) \(\bar{c}\)] = 0
Solution:
We use the results :\(\bar{b}\) × \(\bar{b}\) = 0 and if in a scalar triple product, two vectors are equal, then the scalar triple product is zero.

Question 6.
If u = \(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{k}\), \(\bar{v}\) = \(3 \hat{\mathbf{i}}+\hat{k}\) and \(\bar{w}\) = \(\hat{\mathrm{j}}-\hat{\mathrm{k}}\) are given vectors, then find
(i) [\(\bar{u}\) + \(\bar{w}\)]∙[(\(\bar{w}\) × \(\bar{r}\)) × (\(\bar{r}\) × \(\bar{w}\))]
Question is modified.
If \(\bar{u}\) = \(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{k}\), \(\bar{r}\) = \(3 \hat{\mathbf{i}}+\hat{k}\) and \(\bar{w}\) = \(\hat{\mathrm{j}}-\hat{\mathrm{k}}\) are given vectors, then find [\(\bar{u}\) + \(\bar{w}\)]∙[(\(\bar{u}\) × \(\bar{r}\)) × (\(\bar{r}\) × \(\bar{w}\))]
Solution:

= 1(6 – 18) + 1 (-6 + 6) + 0
= -12 + 0 + 0 = -12.

Question 7.
Find the volume of a tetrahedron whose vertices are A( -1, 2, 3) B(3, -2, 1), C (2, 1, 3) and D(-1, -2, 4).
Solution:
The position vectors \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) and \(\bar{d }\) of the points A, B, C and D w.r.t. the origin are \(\bar{a}\) = \(-\hat{i}+2 \hat{j}+3 \hat{k}\), \(\bar{b}\) = \(3 \hat{i}-2 \hat{j}+\hat{k}\), \(\bar{c}\) = \(2 \hat{i}+\hat{j}+3 \hat{k}\) and

Question 8.
If \(\bar{a}\) = \(\hat{i}+2 \hat{j}+3\), \(\bar{b}\) = \(3 \hat{i}+2 \hat{j}\) and \(\bar{c}\) = ,\(2 \hat{i}+\hat{j}+3\) then verify that \(\bar{a}\) × (\(\bar{b}\) × \(\bar{c}\)) = (\(\bar{a}\) ⋅ \(\bar{c}\))\(\bar{b}\) – (\(\bar{a}\) ⋅ \(\bar{b}\))\(\bar{c}\)
Solution:


From (1) and (2), we get
\(\bar{a}\) × (\(\bar{b}\) × \(\bar{c}\)) = (\(\bar{a}\) ⋅ \(\bar{c}\))\(\bar{b}\) – (\(\bar{a}\) ⋅ \(\bar{b}\))\(\bar{c}\)

Question 9.
If, \(\bar{a}\) = \(\hat{i}-2 \hat{j}\), \(\bar{b}\) = \(\hat{i}+2 \hat{j}\) and \(\bar{c}\) =\(2 \hat{i}+\hat{j}-2\) then find
(i) \(\bar{a}\) × (\(\bar{b}\) × \(\bar{c}\))
Solution:

(ii) (\(\bar{a}\) × \(\bar{b}\)) × \(\bar{c}\) Are the results same? Justify.
Solution:

\(\bar{a}\) × (\(\bar{b}\) × \(\bar{c}\)) ≠ (\(\bar{a}\) × \(\bar{b}\)) × \(\bar{c}\)

Question 10.
Show that \(\bar{a}\) × (\(\bar{b}\) × \(\bar{c}\)) + \(\bar{b}\) × (\(\bar{c}\) × \(\bar{a}\)) + \(\bar{c}\) × (\(\bar{a}\) × \(\bar{b}\)) = 0
Solution: