Class 12

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Linear Programming Ex 7.2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.2

I) Find the feasible solution of the following inequations graphically.
Question 1.
3x + 2y ≤ 18, 2x + y ≤ 10, x ≥ 0, y ≥ 0
Solution:
First we draw the lines AB and CD whose equations are 3x + 2y = 18 and 2x + y = 10 respectively.


The feasible solution is OCPBO which is shaded in the graph.

Question 2.
2x + 3y ≤ 6, x + y ≥ 2, x ≥ 0, y ≥ 0
Solution:
First we draw the lines AB and CB whose equations are 2x + 3y = 6 and x + y = 2 respectively.


The feasible solution is ∆ABC which is shaded in the graph.

Question 3.
3x + 4y ≥ 12, 4x + 7y ≤ 28, y ≥ 1, x ≥ 0
Solution:
First we draw the lines AB, CD and EF whose equations are 3x + 4 y = 12, 4x + 7y = 28 and y = 1 respectively.


The feasible solution is PQDBP. which is shaded in the graph.

Question 4.
x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB, CD and EF whose equations are x + 4y = 24, 3x + y = 21 and x + y = 9 respectively.


The feasible solution is OCPQBO. which is shaded in the graph.

Question 5.
0 ≤ x ≤ 3, 0 ≤ y ≤ 3, x + y ≤ 5, 2x + y ≥ 4
Solution:
First we draw the lines AB, CD, EF and GH whose equations are x + y = 5, 2x + y = 4, x = 3 and y = 3 respectively.


The feasible solution is CEPQRC. which is shaded in the graph.

Question 6.
x – 2y ≤ 2, x + y ≥ 3, -2x + y ≤ 4, x ≥ 0, y ≥ 0
Solution:
First we draw the lines AB, CD and EF whose equations are x – 2y = 2, x + y = 3 and -2x + y = 4 respectively.


The feasible solution is shaded in the graph.

Question 7.
A company produces two types of articles A and B which requires silver and gold. Each unit of A requires 3 gm of silver and 1 gm of gold, while each unit of B requires 2 gm of silver and 2 gm of gold. The company has 6 gm of silver and 4 gm of gold. Construct the inequations and find the feasible solution graphically.
Solution:
Let the company produces x units of article A and y units of article B.
The given data can be tabulated as:

Inequations are :
x + 2y ≤ 4 and 3x + 2y ≤ 6
x and y are number of items, x ≥ 0, y ≥ 0
First we draw the lines AB and CD whose equations are x + 2y = 4 and 3x + 2y = 6 respectively.


The feasible solution is OCPBO. which is shaded in the graph.

Question 8.
A furniture dealer deals in tables and chairs. He has Rs.1,50,000 to invest and a space to store at most 60 pieces. A table costs him Rs.1500 and a chair Rs.750. Construct the inequations and find the feasible solution.
Question is modified
A furniture dealer deals in tables and chairs. He has ₹ 15,000 to invest and a space to store at most 60 pieces. A table costs him ₹ 150 and a chair ₹ 750. Construct the inequations and find the feasible solution.
Solution:
Let x be the number of tables and y be the number of chairs. Then x ≥ 0, y ≥ 0.
The dealer has a space to store at most 60 pieces.
∴ x + y ≤ 60
Since, the cost of each table is ₹ 150 and that of each chair is ₹ 750, the total cost of x tables and y chairs is 150x + 750y. Since the dealer has ₹ 15,000 to invest, 150x + 750y ≤ 15,000
Hence the system of inequations are
x + y ≤ 60, 150x + 750y ≤ 15000, x ≥ 0, y ≥ 0.
First we draw the lines AB and CD whose equations are x + y = 60 and 150x + 750y = 15,000 respectively.


The feasible solution is OAPDO. which is shaded in the graph.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Linear Programming Ex 7.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1

Question 1.
Solve graphically :
(i) x ≥ 0
Solution:
Consider the line whose equation is x = 0. This represents the Y-axis.
To find the solution set, we have to check any point other than origin.
Let us check the point (1, 1)
When x = 1, x ≥ 0
∴ (1, 1) lies in the required region
Therefore, the solution set is the Y-axis and the right
side of the Y-axis which is shaded in the graph.

(ii) x ≤ 0
Solution:
Consider the line whose equation is x = 0.
This represents the Y-axis.
To find the solution set, we have to check any point other than origin.
Let us check the point (1, 1).
When x = 1, x ≰ 0
∴ (1, 1) does not lie in the required region.
Therefore, the solution set is the Y-axis and the left side of the Y-axis which is shaded in the graph.

(iii) y ≥ 0
Solution:
Consider the line whose equation is y = 0. This represents the X-axis. To find the solution set, we have to check any point other than origin. Let us check the point (1, 1).
When y = 1, y ≥ 0
∴ (1, 1) lies in the required region.
Therefore, the solution set is the X-axis and above the X-axis which is shaded in the graph.

(iv) y ≤ 0
Solution:
(iv) Consider the line whose equation is y = 0. This represents the X-axis.
To find the solution set, we have to check any point other than origin.
Let us check the point (1, 1).
When y = 1, y ≰ 0.
∴ (1, 1) does not lie in the required region.
Therefore, the solution set is the X-axis and below the X-axis which is shaded in the graph.

Question 2.
Solve graphically :
(i) x ≥ 0 and y ≥ 0
Solution:
Consider the lines whose equations are x = 0, y = 0.
These represents the equations of Y-axis and X-axis respectively, which divide the plane into four parts.
(i) Since x ≥ 0, y ≥ 0, the solution set is in the first quadrant which is shaded in the graph.

(ii) x ≤ 0 and y ≥ 0
Solution:
Since x ≤ 0, y ≥ 0, the solution set is in the second quadrant which is shaded in the graph.

(iii) x ≤ 0 and y ≤ 0
Solution:
Since x ≤ 0, y ≤ 0, the solution set is in the third quadrant which is shaded in the graph.

(iv) x ≥ 0 and y ≤ 0
Solution:
Since x ≥ 0, y ≤ 0, the solution set is in the fourth ! quadrant which is shaded in the graph.

Question 3.
Solve graphically :
(i) 2x – 3 ≥ 0
Solution:
Consider the line whose equation is 2x – 3 = 0,
i.e. x = \(\frac{3}{2}\)
This represents a line parallel to Y-axis passing through the point (\(\frac{3}{2}\), 0)
Draw the line x =\(\frac{3}{2}\).
To find the solution set, we have to check the position of the origin (0, 0).
When x = 0, 2x – 3 = 2 × 0 – 3 = -3 ≱ 0
∴ the coordinates of the origin does not satisfy the given inequality.
∴ the solution set consists of the line x = \(\frac{3}{2}\) and the non-origin side of the line which is shaded in the graph.

(ii) 2y – 5 ≥ 0
Solution:
Consider the line whose equation is 2y – 5 = 0, i.e. y = \(\frac{5}{2}\)
This represents a line parallel to X-axis passing through the point (0, \(\frac{5}{2}\)).
Draw the line y = \(\frac{5}{2}\).
To find the solution set, we have to check the position of the origin (0, 0).
When y = 0, 2y – 5 = 2 × 0 – 5 = -5 ≱ 0
∴ the coordinates of the origin does not satisfy the given inequality.
∴ the solution set consists of the line y = \(\frac{5}{2}\) and the non-origin side of the line which is shaded in the graph.

(iii) 3x + 4 ≤ 0
Solution:
(iii) Consider the line whose equation is 3x + 4 = 0,
i.e. x = \(-\frac{4}{3}\)
This represents a line parallel to Y-axis passing through the point (\(-\frac{4}{3}\), 0).
Draw the line x = \(-\frac{4}{3}\).
To find the solution set, we have to check the position of the origin (0, 0).
When x = 0, 3x + 4 = 3 × 0 + 4= 4 ≰ 0
∴ the coordinates of the origin does not satisfy the given inequality.
∴ the solution set consists of the line x = \(-\frac{4}{3}\) and the non-origin side of the line which is shaded in the graph.

(iv) 5y + 3 ≤ 0
Solution:
(iv) Consider the line whose equation is 5y + 3 = 0,
i.e. y = \(\frac{-3}{5}\)
This represents a line parallel to X-axis passing through the point (0, \(\frac{-3}{5}\))
Draw the line y = \(\frac{-3}{5}\).
To find the solution set, we have to check the position of the origin (0, 0).
When y = 0, 5y + 3 = 5 × 0 + 3 = 3 ≰ 0
∴ the coordinates of the origin does not satisfy the given inequality.
∴ the solution set consists of the line y = \(\frac{-3}{5}\) and the non-origin side of the line which is shaded in the graph.

Question 4.
Solve graphically :
(i) x + 2y ≤ 6
Solution:
Consider the line whose equation is x + 2y = 6.
To find the points of intersection of this line with the coordinate axes.
Put y = 0, we get x = 6.
∴ A = (6, 0) is a point on the line.
Put x = 0, we get 2y = 6, i.e. y = 3
∴ B = (0, 3) is another point on the line.

Draw the line AB joining these points. This line divide the line into two parts.
1. Origin side 2. Non-origin side
To find the solution set, we have to check the position of the origin (0, 0) with respect to the line.
When x = 0, y = 0, then x + 2y = 0 which is less than 6.
∴ x + 2y ≤ 6 in this case.
Hence, origin lies in the required region. Therefore, the given inequality is the origin side which is
shaded in the graph.
This is the solution set of x + 2y ≤ 6.

(ii) 2x – 5y ≥ 10
Solution:
Consider the line whose equation is 2x – 5y = 10.
To find the points of intersection of this line with the coordinate axes.
Put y = 0, we get 2x = 10, i.e. x = 5.
∴ A = (5, 0) is a point on the line.
Put x = 0, we get -5y = 10, i.e. y = -2
∴ B = (0, -2) is another point on the line.

Draw the line AB joining these points. This line J divide the plane in two parts.
1. Origin side 2. Non-origin side
To find the solution set, we have to check the position of the origin (0, 0) with respect to the line. When x = 0, y = 0, then 2x – 5y = 0 which is neither greater nor equal to 10.
∴ 2x – 5y ≱ 10 in this case.
Hence (0, 0) will not lie in the required region.
Therefore, the given inequality is the non-origin side, which is shaded in the graph.
This is the solution set of 2x – 5y ≥ 10.

(iii) 3x + 2y ≥ 0
Solution:
Consider the line whose equation is 3x + 2y = 0.
The constant term is zero, therefore this line is passing through the origin.
∴ one point on the line is O ≡ (0, 0).
To find the another point, we can give any value of x and get the corresponding value of y.
Put x = 2, we get 6 + 2y = 0 i.e. y = – 3
∴ A = (2, -3) is another point on the line.
Draw the line OA.
To find the solution set, we cannot check (0, 0) as it is already on the line.
We can check any other point which is not on the line.
Let us check the point (1, 1)

When x = 1, y = 1, then 3x + 2y = 3 + 2 = 5 which is greater than zero.
∴ 3x + 2y > 0 in this case.
Hence (1, 1) lies in the required region. Therefore, the required region is the upper side which is shaded in the graph.
This is the solution set of 3x + 2y ≥ 0.

(iv) 5x – 3y ≤ 0
Solution:
Consider the line whose equation is 5x – 3y = 0. The constant term is zero, therefore this line is passing through the origin.
∴ one point on the line is the origin O = (0, 0).
To find the other point, we can give any value of x and get the corresponding value of y.
Put x = 3, we get 15 – 3y = 0, i.e. y = 5
∴ A ≡ (3, 5) is another point on the line.
Draw the line OA.
To find the solution set, we cannot check 0(0, 0), as it is already on the line. We can check any other point which is not on the line.
Let us check the point (1, -1).
When x = 1, y = -1 then 5x – 3y = 5 + 3 = 8
which is neither less nor equal to zero.
∴ 5x – 3y ≰ 0 in this case.
Hence (1, -1) will not lie in the required region. Therefore, the required region is the upper side which is shaded in the graph.

This is the solution set of 5x – 3y ≤ 0.

Question 5.
Solve graphically :
(i) 2x + y ≥ 2 and x – y ≤ 1
Solution:
First we draw the lines AB and AC whose equations are 2x + y = 2 and x – y = 1 respectively.


The solution set of the given system of inequalities is shaded in the graph.

(ii) x – y ≤ 2 and x + 2y ≤ 8
Solution:
First we draw the lines AB and CD whose equations are x – y = 2 and x + 2y = 8 respectively.


The solution set of the given system of inequalities is shaded in the graph.

(iii) x + y ≥ 6 and x + 2y ≤ 10
Solution:
First we draw the lines AB and CD whose equations are x + y = 6 and x + 2y = 10 respectively.


The solution set of the given system of inequalities is shaded in the graph.

(iv) 2x + 3y ≤ 6 and x + 4y ≥ 4
Solution:
First we draw the lines AB and CD whose equations are 2x + 3y = 6 and x + 4y = 4 respectively.

The solution set of the given system of inequalities is shaded in the graph.

(v) 2x + y ≥ 5 and x – y ≤ 1
Solution:
First we draw the lines AB and CD whose equations are 2x + y = 5 and x – y = 1 respectively.


The solution set of the given system of inequations is shaded in the graph.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Line and Plane Miscellaneous Exercise 6B Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Line and Plane Miscellaneous Exercise 6B

Question 1.
If the line \(\frac{x}{3}=\frac{y}{4}\) = z is perpendicular to the line \(\frac{x-1}{k}=\frac{y+2}{3}=\frac{z-3}{k-1}\) then the value of k is:

Solution:
(b) \(-\frac{11}{4}\)

Question 2.
The vector equation of line 2x – 1 = 3y + 2 = z – 2 is

Solution:
(a) \(\bar{r}=\left(\frac{1}{2} \hat{i}-\frac{2}{3} \hat{j}+2 \hat{k}\right)+\lambda(3 \hat{i}+2 \hat{j}+6 \hat{k})\)

Question 3.
The direction ratios of the line which is perpendicular to the two lines \(\frac{x-7}{2}=\frac{y+17}{-3}=\frac{z-6}{1}\) and \(\frac{x+5}{1}=\frac{y+3}{2}=\frac{z-6}{-2}\) are
(A) 4, 5, 7
(B) 4, -5, 7
(C) 4, -5, -7
(D) -4, 5, 8
Solution:
(A) 4, 5, 7

Question 4.
The length of the perpendicular from (1, 6, 3) to the line \(\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}\)
(A) 3
(B) \(\sqrt {11}\)
(C) \(\sqrt {13}\)
(D) 5
Solution:
(C ) \(\sqrt {13}\)

Question 5.
The shortest distance between the lines \(\bar{r}=(\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}-\hat{k})\) and \(\bar{r}=(2 \hat{i}-\hat{j}-\hat{k})+\mu(2 \hat{i}+\hat{j}+2 \hat{k})\) is
Question is modified.
The shortest distance between the lines \(\bar{r}=(\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})\) and \(\bar{r}=(2 \hat{i}-\hat{j}-\hat{k})+\mu(2 \hat{i}+\hat{j}+2 \hat{k})\) is

Solution:
(c) \(\frac{3}{\sqrt{2}}\)

Question 6.
The lines \(\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}\) and \(\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}\). and coplanar if
(A) k = 1 or -1
(B) k = 0 or -3
(C) k = + 3
(D) k = 0 or -1
Solution:
(B ) k = 0 or -3

Question 7.
The lines \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) and \(\frac{x-1}{-2}=\frac{y-2}{-4}=\frac{z-3}{6}\) and are
(A) perpendicular
(B) inrersecting
(C) skew
(D) coincident
Solution:
(B) inrersecting

Question 8.
Equation of X-axis is
(A) x = y = z
(B) y = z
(C) y = 0, z = 0
(D) x = 0, y = 0
Solution:
(C) y = 0, z = 0

Question 9.
The angle between the lines 2x = 3y = -z and 6x = -y = -4z is
(A ) 45º
(B ) 30º
(C ) 0º
(D ) 90º
Solution:
(D ) 90º

Question 10.
The direction ratios of the line 3x + 1 = 6y – 2 = 1 – z are
(A ) 2, 1, 6
(B ) 2, 1, -6
(C ) 2, -1, 6
(D ) -2, 1, 6
Solution:
(B ) 2, 1, -6

Question 11.
The perpendicular distance of the plane 2x + 3y – z = k from the origin is \(\sqrt {14}\) units, the value
of k is
(A ) 14
(B ) 196
(C ) \(2\sqrt {14}\)
(D ) \(\frac{\sqrt{14}}{2}\)
Solution:
(A ) 14

Question 12.
The angle between the planes and \(\bar{r} \cdot(\bar{i}-2 \bar{j}+3 \bar{k})+4=0\) and \(\bar{r} \cdot(2 \bar{i}+\bar{j}-3 \bar{k})+7=0\) is

Solution:
(d) cos^-1\(\left(\frac{9}{14}\right)\)

Question 13.
If the planes \(\bar{r} \cdot(2 \bar{i}-\lambda \bar{j}+\bar{k})=3\) and \(\bar{r} \cdot(4 \bar{i}-\bar{j}+\mu \bar{k})=5\) are parallel, then the values of λ and μ are respectively.

Solution:
(d) \(\frac{1}{2}\), 2

Question 14.
The equation of the plane passing through (2, -1, 3) and making equal intercepts on the coordinate axes is
(A ) x + y + z =1
(B ) x + y + z = 2
(C ) x + y + z = 3
(D ) x + y + z = 4
Solution:
(D ) x + y + z = 4

Question 15.
Measure of angle between the planes 5x – 2y + 3z – 7 = 0 and 15x – 6y + 9z + 5 = 0 is
(A ) 0º
(B ) 30º
(C ) 45º
(D ) 90º
Solution:
(A ) 0º

Question 16.
The direction cosines of the normal to the plane 2x – y + 2z = 3 are

Solution:
(a) \(\frac{2}{3}, \frac{-1}{3}, \frac{2}{3}\)

Question 17.
The equation of the plane passing through the points (1, -1, 1), (3, 2, 4) and parallel to Y-axis is :
(A ) 3x + 2z – 1 = 0
(B ) 3x – 2z = 1
(C ) 3x + 2z + 1 = 0
(D ) 3x + 2z = 2
Solution:
(B ) 3x – 2z = 1

Question 18.
The equation of the plane in which the line \(\frac{x-5}{4}=\frac{y-7}{4}=\frac{z+3}{-5}\) and \(\frac{x-8}{7}=\frac{y-4}{1}=\frac{z+5}{3}\) lie, is
(A ) 17x – 47y – 24z + 172 = 0
(B ) 17x + 47y – 24z + 172 = 0
(C ) 17x + 47y + 24z +172 = 0
(D ) 17x – 47y + 24z + 172 = 0
Solution:
(A ) 17x – 47y – 24z + 172 = 0

Question 19.
If the line \(\frac{x+1}{2}=\frac{y-m}{3}=\frac{z-4}{6}\) lies in the plane 3x – 14y + 6z + 49 = 0, then the value of m is:
(A ) 5
(B ) 3
(C ) 2
(D ) -5
Solution:
(A ) 5

Question 20.
The foot of perpendicular drawn from the point (0,0,0) to the plane is (4, -2, -5) then the equation of the plane is
(A ) 4x + y + 5z = 14
(B ) 4x – 2y – 5z = 45
(C ) x – 2y – 5z = 10
(D ) 4x + y + 6z = 11
Solution:
(B ) 4x – 2y – 5z = 45

II. Solve the following :
Question 1.
Find the vector equation of the plane which is at a distance of 5 unit from the origin and which is normal to the vector \(2 \hat{i}+\hat{j}+2 \hat{k}\)
Solution:
If \(\hat{n}\) is a unit vector along the normal and p i the length of the perpendicular from origin to the plane, then the vector equation of the plane \(\bar{r} \cdot \hat{n}\) = p
Here, \(\overline{\mathrm{n}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}\) and p = 5

Question 2.
Find the perpendicular distance of the origin from the plane 6x + 2y + 3z – 7 = 0
Solution:
The distance of the point (x₁, y₁, z₁) from the plane ax + by + cz + d is \(\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|\)
∴ the distance of the point (1, 1, -1) from the plane 6x + 2y + 3z – 7 = 0 is

= 1units.

Question 3.
Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x + 3y + 6z = 49.
Solution:
The equation of the plane is 2x + 3y + 6z = 49
Dividing each term by
\(\sqrt{2^{2}+3^{2}+(-6)^{2}}\)
= \(\sqrt{49}\)
= 7
we get
\(\frac{2}{7}\)x + \(\frac{3}{7}\)y – \(\frac{6}{7}\)z = \(\frac{49}{7}\) = 7
This is the normal form of the equation of plane.
∴ the direction cosines of the perpendicular drawn from the origin to the plane are
l = \(\frac{2}{7}\), m = \(\frac{3}{7}\), n = \(\frac{6}{7}\)
and length of perpendicular from origin to the plane is p = 7.
the coordinates of the foot of the perpendicular from the origin to the plane are
(lp, ∓, np)i.e.(2, 3, 6)

Question 4.
Reduce the equation \(\bar{r} \cdot(\hat{i}+8 \hat{j}+24 \hat{k})=13\) to normal form and hence find
(i) the length of the perpendicular from the origin to the plane
(ii) direction cosines of the normal.
Solution:
The normal form of equation of a plane is \(\bar{r} \cdot \hat{n}\) = p where \(\hat{n}\) is unit vector along the normal and p is the length of perpendicular drawn from origin to the plane.
Given pane is \(\text { r. }(6 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}+24 \hat{\mathrm{k}})=13\) …(1)

This is the normal form of the equation of plane.
Comparing with \(\bar{r} \cdot \hat{n}\) = p,
(i) the length of the perpendicular from the origin to plane is \(\frac{1}{2}\).
(ii) direction cosines of the normal are \(\frac{3}{13}, \frac{4}{13}, \frac{12}{13}\)

Question 5.
Find the vector equation of the plane passing through the points A(1, -2, 1), B (2, -1, -3) and C (0, 1, 5).
Solution:
The vector equation of the plane passing through three non-collinear points A(\(\bar{a}\)), B(\(\bar{b}\)) and C(\(\bar{c}\)) is \(\bar{r} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}})=\bar{a} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}})\) … (1)

Question 6.
Find the Cartesian equation of the plane passing through A(1, -2, 3) and the direction ratios of whose normal are 0, 2, 0.
Solution:
The Cartesian equation of the plane passing through (x₁, y₁, z₁), the direction ratios of whose normal are a, b, c, is
a(x – x₁) + b(y – y₁) + c(z – z₁) = 0
∴ the cartesian equation of the required plane is
o(x + 1) + 2(y + 2) + 5(z – 3) = 0
i.e. 0 + 2y – 4 + 10z – 15 = 0
i.e. y + 2 = 0.

Question 7.
Find the Cartesian equation of the plane passing through A(7, 8, 6) and parallel to the plane \(\bar{r} \cdot(6 \hat{i}+8 \hat{j}+7 \hat{k})=0\)
Solution:
The cartesian equation of the plane \(\bar{r} \cdot(6 \hat{i}+8 \hat{j}+7 \hat{k})=0\) is 6x + 8y + 7z = 0 The required plane is parallel to it
∴ its cartesian equation is
6x + 8y + 7z = p …(1)
A (7, 8, 6) lies on it and hence satisfies its equation
∴ (6)(7) + (8)(8) + (7)(6) = p
i.e., p = 42 + 64 + 42 = 148.
∴ from (1), the cartesian equation of the required plane is 6x + 8y + 7z = 148.

Question 8.
The foot of the perpendicular drawn from the origin to a plane is M(1, 2,0). Find the vector equation of the plane.
Solution:
The vector equation of the plane passing through A(\(\bar{a}\)) and perpendicular to \(\bar{n}\) is \(\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n}\).
M(1, 2, 0) is the foot of the perpendicular drawn from origin to the plane. Then the plane is passing through M and is
perpendicular to OM.
If \(\bar{m}\) is the position vector of M, then \(\bar{m}\) = \(\hat{\mathrm{i}}\).
Normal to the plane is
\(\bar{n}\) = \(\overline{\mathrm{OM}}\) = \(\hat{\mathrm{i}}\)
\(\overline{\mathrm{m}} \cdot \overline{\mathrm{n}}\) = \(\hat{\mathrm{i}}, \hat{i}\) = 5
∴ the vector equation of the required plane is
\(\bar{r} \cdot(\hat{i}+2 \hat{j})\) = 5

Question 9.
A plane makes non zero intercepts a, b, c on the co-ordinates axes. Show that the vector equation of the plane is \(\bar{r} \cdot(b c \hat{i}+c a \hat{j}+a b \hat{k})\) = abc
Solution:
The vector equation of the plane passing through A(\(\bar{a}\)), B(\(\bar{b}\)).. C(\(\bar{c}\)), where A, B, C are non collinear is
\(\overline{\mathrm{r}} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}})=\overline{\mathrm{a}} \cdot(\overline{\mathrm{AB}} \times \overline{\mathrm{AC}})\) …(1)
The required plane makes intercepts 1, 1, 1 on the coordinate axes.
∴ it passes through the three non collinear points A = (1, 0, 0), B = (0, 1, 0), C = (0, , 1)

Question 10.
Find the vector equation of the plane passing through the pointA(-2, 3, 5) and parallel to vectors \(4 \hat{i}+3 \hat{k}\) and \(\hat{i}+\hat{j}\)
Solution:
The vector equation of the plane passing through the point A(\(\bar{a}\)) and parallel to the vectors \(\bar{b}\) and \(\bar{c}\) is


= (-2)(-4) + (7)(-1) + (5)(4)
= 8 – 7 + 8
= 35
∴ From (1), the vector equation of the required plane is \(\overline{\mathrm{r}} \cdot(-3 \hat{\mathrm{i}}-3 a t \mathrm{j}+4 \hat{\mathrm{k}})\) = 35.

Question 11.
Find the Cartesian equation of the plane \(\bar{r}=\lambda(\hat{i}+\hat{j}-\hat{k})+\mu(\hat{i}+2 \hat{j}+3 \hat{k})\)
Solution:
The equation \(\bar{r}=\bar{a}+\lambda \bar{b}+\mu \bar{c}\) represents a plane passing through a point having position vector \(\overline{\mathrm{a}}\) and parallel to vectors \(\overline{\mathrm{b}}\) and \(\overline{\mathrm{c}}\).
Here,

Question 12.
Find the vector equations of planes which pass through A(1, 2, 3), B (3, 2, 1) and make equal intercepts on the co-ordinates axes.
Question is modified
Find the cartesian equations of the planes which pass through A(1, 2, 3), B(3, 2, 1) and make equal intercepts on the coordinate axes.
Solution:
Case 1 : Let all the intercepts be 0.
Then the plane passes through the origin.
Then the cartesian equation of the plane is
ax + by + cz = 0 …..(1)
(1, 2, 3) and (3, 2, 1) lie on the plane.
∴ a + 2b + 3c = 0 and 3a + 2b + c = 0

∴ a, b, c are proportional to 1, -2, 1
∴ from (1), the required cartesian equation is x – 2y + z = 0.
Case 2 : Let the plane make non zero intercept p on each axis.
then its equation is \(\frac{x}{p}+\frac{y}{p}+\frac{z}{p}\) = 1
i.e. x + y + z = p …(2)
Since this plane pass through (1, 2, 3) and (3, 2, 1)
∴ 1 + 2 + 3 = p and 3 + 2 + 1 = p
∴ p = 6
∴ from (2), the required cartesian equation is
x + y + z = 6
Hence, the cartesian equations of required planes are x + y + z = 6 and x – 2y + z = 0.

Question 13.
Find the vector equation of the plane which makes equal non-zero intercepts on the co-ordinates axes and passes through (1, 1, 1).
Solution:
Case 1 : Let all the intercepts be 0.
Then the plane passes through the origin.
Then the vector equation of the plane is ax + by + cz …(1)
(1, 1, 1) lie on the plane.
∴ 1a + 1b + 1c = 0

∴ from (1), the required cartesian equation is x – y + z = 0
Case 2 : Let he plane make non zero intercept p on each axis.
then its equation is \(\frac{\hat{\mathrm{i}}}{p}+\frac{\hat{\mathrm{j}}}{p}+\frac{\hat{\mathrm{k}}}{p}=1\) = 1
i.e. \(\hat{i}+\hat{j}+\hat{k}=p\) = p ….(2)
Since this plane pass through (1, 1, 1)
∴ 1 + 1 + 1 = p
∴ p = 3
∴ from (2), the required cartesian equation is \(\hat{i}+\hat{j}+\hat{k}\) = 3
Hence, the cartesian equations of required planes are \(\bar{r} \cdot(\hat{i}+\hat{j}+\hat{k})=3\)

Question 14.
Find the angle between planes \(\bar{r} \cdot(-2 \hat{i}+\hat{j}+2 \hat{k})=17\) and \(\bar{r} \cdot(2 \hat{i}+2 \hat{j}+\hat{k})=71\).
Solution:
The acute angle between the planes

= (1)(2) + (1)(1) + (2)(1)
= 2 + 1 + 2
= 5
Also,

Question 15.
Find the acute angle between the line \(\bar{r}=\lambda(\hat{i}-\hat{j}+\hat{k})\) and the plane \(\bar{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=23\)
Solution:
The acute angle θ between the line \(\overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}\) and the plane \(\overline{\mathrm{r}} \cdot \overline{\mathrm{n}}\) = d is given by

= (2)(2) + (3)(-1) + (-6)(1)
= 4 – 3 – 6
= -5

Question 16.
Show that lines \(\bar{r}=(\hat{i}+4 \hat{j})+\lambda(\hat{i}+2 \hat{j}+3 \hat{k})\) and \(\bar{r}=(3 \hat{j}-\hat{k})+\mu(2 \hat{i}+3 \hat{j}+4 \hat{k})\)
Solution:

Question 17.
Find the distance of the point \(3 \hat{i}+3 \hat{j}+\hat{k}\) from the plane \(\bar{r} \cdot(2 \hat{i}+3 \hat{j}+6 \hat{k})=21\)
Solution:
The distance of the point A(\(\bar{a}\)) from the plane \(\bar{r} \cdot \bar{n}\) = p is given by d = \(\frac{|\bar{a} \cdot \bar{n}-p|}{|\bar{n}|}\) ……(1)

= (3)(2) + (3)(3) + (1)(-6)
= 6 + 9 – 6
= 9

Question 18.
Find the distance of the point (13, 13, -13) from the plane 3x + 4y – 12z = 0.
Solution:

= 19units.

Question 19.
Find the vector equation of the plane passing through the origln and containing the line \(\bar{r}=(\hat{i}+4 \hat{j}+\hat{k})+\lambda(\hat{i}+2 \hat{j}+\hat{k})\).
Solution:
The vector equation of the plane passing through A(\(\bar{a}\)) and perpendicular to the vector \(\bar{n}\) is \(\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n}\) … (1)
We can take \(\bar{a}\) = \(\bar{0}\) since the plane passes through the origin.
The point M with position vector \(\bar{m}\) =\(\hat{i}+4 \hat{j}+\hat{k}\) lies on the line and hence it lies on the plane.
.’. \(\overline{\mathrm{OM}}=\bar{m}=\hat{i}+4 \hat{j}+\hat{k}\) lies on the plane.
The plane contains the given line which is parallel to \(\bar{b}=\hat{i}+2 \hat{j}+\hat{k}\)
Let \(\bar{n}\) be normal to the plane. Then \(\bar{n}\) is perpendicular to \(\overline{\mathrm{OM}}\) as well as \(\bar{b}\)

Question 20.
Find the vector equation of the plane which bisects the segment joining A(2, 3, 6) and B( 4, 3, -2) at right angle.
Solution:
The vector equation of the plane passing through A(\(\bar{a}\)) and perpendicular to the vector \(\bar{n}\) is \(\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n}\) ….(1)
The position vectors \(\bar{a}\) and \(\bar{b}\) of the given points A and B are \(\bar{a}=2 \hat{i}+3 \hat{j}+6 \hat{k}\) and \(\bar{b}=4 \hat{i}+3 \hat{j}-2 \hat{k}\)
If M is the midpoint of segment AB, the position vector \(\bar{m}\) of M is given by

Question 21.
Show thatlines x = y, z = 0 and x + y = 0, z = 0 intersect each other. Find the vector equation of the plane determined by them.
Solution:
Given lines are x = y, z = 0 and x + y = 0, z = 0.
It is clear that (0, 0, 0) satisfies both the equations.
∴ the lines intersect at O whose position vector is \(\overline{0}\)
Since z = 0 for both the lines, both the lines lie in XY- plane.
Hence, we have to find equation of XY-plane.
Z-axis is perpendicular to XY-plane.
∴ normal to XY plane is \(\hat{k}\).
0(\(\overline{0}\)) lies on the plane.
By using \(\bar{r} \cdot \bar{n}=\bar{a} \cdot \bar{n}\), the vector equation of the required plane is \(\bar{r} \cdot \hat{k}=\overline{0} \cdot \bar{k}\)
i.e. \(\bar{r} \cdot \hat{k}=0\).
Hence, the given lines intersect each other and the vector equation of the plane determine by them is \(\bar{r} \cdot \hat{k}=0\).

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Line and Plane Miscellaneous Exercise 6A Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Line and Plane Miscellaneous Exercise 6A

Question 1.
Find the vector equation of the line passing through the point having position vector \(3 \hat{i}+4 \hat{j}-7 \hat{k}\) and parallel to \(6 \hat{i}-\hat{j}+\hat{k}\).
Solution:
The vector equation of the line passing through A(\(\bar{a}\)) and parallel to the vector \(\bar{b}\) is \(\overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}\), where λ is a scalar.
∴ the vector equation of the line passing through the point having position vector
\(3 \hat{i}+4 \hat{j}-7 \hat{k}\) and parallel to the vector \(6 \hat{i}-\hat{j}+\hat{k}\) is
\(\overline{\mathrm{r}}=(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-7 \hat{\mathrm{k}})+\lambda(\hat{6 \hat{\mathrm{i}}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})\).

Question 2.
Find the vector equation of the line which passes through the point (3, 2, 1) and is parallel to the vector \(2 \hat{i}+2 \hat{j}-3 \hat{k}\).
Solution:
The vector equation of the line passing through A(\(\bar{a}\)) and parallel to the vector \(\bar{b}\) is \(\overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}\), where λ is a scalar.
∴ the vector equation of the line passing through the point having position vector \(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\) and parallel to the vector
\(2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \text { is } \overline{\mathrm{r}}=(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})\)

Question 3.
Find the Cartesian equations of the line which passes through the point (-2, 4, -5) and parallel to the line \(\frac{x+2}{3}=\frac{y-3}{5}=\frac{z+5}{6}\)
Solution:
The line \(\frac{x+2}{3}=\frac{y-3}{5}=\frac{z+5}{6}\) has direction ratios 3, 5, 6. The required line has direction ratios 3, 5, 6 as it is parallel to the given line.
It passes through the point (-2, 4, -5).
The cartesian equations of the line passing through (x₁, y₁, z₁) and having direction ratios a, b, c are

Question 4.
Obtain the vector equation of the line \(\frac{x+5}{3}=\frac{y+4}{5}=\frac{z+5}{6}\).
Solution:
The cartesian equations of the line are \(\frac{x+5}{3}=\frac{y+4}{5}=\frac{z+5}{6}\).
This line is passing through the point A(-5, -4, -5) and having direction ratios 3, 5, 6.
Let \(\bar{a}\) be the position vector of the point A w.r.t. the origin and \(\bar{b}\) be the vector parallel to the line.
Then \(\bar{a}=-5 \hat{i}-4 \hat{j}-5 \hat{k}\) and \(\bar{b}=3 \hat{i}+5 \hat{j}+6 \hat{k}\).
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 vector equation of the required line is \(\bar{r}=(-5 \hat{i}-4 \hat{j}-6 \hat{k})+\lambda(3 \hat{i}+5 \hat{j}+6 \hat{k})\)

Question 5.
Find the vector equation of the line which passes through the origin and the point (5, -2, 3).
Solution:
Let \(\bar{b}\) be the position vector of the point B(5, -2, 3).
Then \(\bar{b}=5 \hat{i}-2 \hat{j}+3 \hat{k}\)
Origin has position vector \(\overline{0}=0 \hat{i}+0 \hat{j}+0 \hat{k}\).
The vector equation the line passing through A(\(\bar{a}\)) and B(\(\bar{b}\)) is \(\bar{r}=\bar{a}+\lambda(\bar{b}-\bar{a})\) where λ is a scalar.
∴ the vector equation of the required line is \(\bar{r}=\overline{0}+\lambda(\bar{b}-\overline{0})=\lambda(5 \hat{i}-2 \hat{j}+3 \hat{k})\)

Question 6.
Find the Cartesian equations of the line which passes through points (3, -2, -5) and (3, -2, 6).
Solution:
Let A = (3, -2, -5), B = (3, -2, 6)
The direction ratios of the line AB are
3 – 3, -2 – (-2), 6 – (-5) i.e. 0, 0, 11.
The parametric equations of the line passing through (x₁, y₁, z₁) and having direction ratios a, b, c are
x = x₁ + aλ, y = y₁ + bλ, z = z₁ + cλ
∴ the parametric equattions of the line passing through (3, -2, -5) and having direction ratios are 0, 0, 11 are
x = 3 + (0)λ, y = -2 + 0(λ), z = -5 + 11λ
i.e. x = 3, y = -2, z = 11λ – 5
∴ the cartesian equations of the line are
x = 3, y = -2, z = 11λ – 5, λ is a scalar.

Question 7.
Find the Cartesian equations of the line passing through A(3, 2, 1) and B(1, 3, 1).
Solution:
The direction ratios of the line AB are 3 – 1, 2 – 3, 1 – 1 i.e. 2, -1, 0.
The parametric equations of the line passing through (x₁, y₁, z₁) and having direction ratios a, b, c are
x = x₁ + aλ, y = y₁ + bλ, z = z₁ + cλ
∴ the parametric equattions of the line passing through (3, 2, 1) and having direction ratios 2, -1, 0 are
x = 3 + 2λ, y = 2 – λ, z = 1 + 0(λ)
x – 3 = 2λ, y – 2 = -λ, z = 1
∴ \(\frac{x-3}{2}=\frac{y-2}{-1}\) = λ, z = 1
∴ the cartesian equations of the line are
\(\frac{x-3}{2}=\frac{y-2}{-1}\), z = 1.

Question 8.
Find the Cartesian equations of the line passing through the point A(1, 1, 2) and perpendicular to vectors \(\bar{b}=\hat{i}+2 \hat{j}+\hat{k}\) and \(\bar{c}=3 \hat{i}+2 \hat{j}-\hat{k}\).
Solution:
Let the required line have direction ratios p, q, r. ,
It is perpendicular to the vectors \(\bar{b}=\hat{i}+2 \hat{j}+\hat{k}\) and \(\bar{c}=3 \hat{i}+2 \hat{j}-\hat{k}\).
∴ it is perpendicular to lines whose direction ratios are 1, 2, 1 and 3, 2, -1.
∴ p + 2q + r = 0, 3p + 2q – r = 0

∴ the required line has direction ratios -1, 1, -1.
The cartesian equations of the line passing through (x₁, y₁, z₁) and having direction ratios a, b, c are
\(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\)
∴ the cartesian equations of the line passing through the point (1, 1, 2) and having direction ratios -1, 1, -1 are
\(\frac{x-1}{-1}=\frac{y-1}{1}=\frac{z-2}{-1}\)

Question 9.
Find the Cartesian equations of the line which passes through the point (2, 1, 3) and perpendicular
to lines \(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}\) and \(\frac{x}{-3}=\frac{y}{2}=\frac{z}{5}\).
Solution:
Let the required line have direction ratios p, q, r.
It is perpendicular to the vector \(\bar{b}=\hat{i}+2 \hat{j}+\hat{k}\) and \(\bar{c}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}\).
∴ it is perpendicular to lines whose direction ratios are 1, 2, 1 and 3, 2, -1.
∴ p + 2q + r = 0, 3 + 2q – r = 0

∴ the required line has direction ratios 2, -7, 4.
The cartesian equations of the line passing through (x₁, y₁, z₁) and having direction ratios a, b, c are
\(\frac{x=x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\)
∴ the cartesian equation of the line passing through the point (2, -7, 4) and having directions ratios 2, -7, 4 are
\(\frac{x-2}{2}=\frac{y-1}{-7}=\frac{z-2}{4}\)

Question 10.
Find the vector equation of the line which passes through the origin and intersect the line x – 1 = y – 2 = z – 3 at right angle.
Solution:
The given line is \(\frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{1}\) = λ … (Say)
∴ coordinates of any point on the line are
x = λ + 1, y = λ + 2, z = λ + 3
∴ position vector of any point on the line is
(λ + 1)\(\hat{i}\) + (λ + 2)\(\hat{j}\) + (λ + 3)\(\hat{k}\) … (1)
If \(\bar{b}\) is parallel to the given line whose direction ratios are 1, 1, 1, then \(\bar{b}=\hat{i}+\hat{j}+\hat{k}\).
Let the required line passing through O meet the given line at M.
∴ position vector of M
= \(\bar{m}\) = (λ + 1)\(\hat{i}\) + (λ + 2)\(\hat{j}\) + (λ + 3)\(\hat{k}\) … [By (1)]
The required line is perpendicular to given line

The vector equation of the line passing through 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 o(\(\bar{o}\)) and M(\(\bar{m}\)) is
\(\bar{r}=\overline{0}+\lambda(\bar{m}-\overline{0})=\lambda \bar{m}=\lambda(-\hat{i}+\hat{k})\) where λ is a scalar.
Hence, vector equation of the required line is \(\).

Question 11.
Find the value of λ so that lines \(\frac{1-x}{3}=\frac{7 y-14}{2 \lambda}=\frac{z-3}{2}\) and \(\frac{7-7 x}{3 \lambda}=\frac{y-5}{1}=\frac{6-z}{5}\) are at right angle.
Solution:
The equations of the given lines are

Question 12.
Find the acute angle between lines \(\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-3}{2}\) and \(\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{1}\).
Solution:

Question 13.
Find the acute angle between lines x = y, z = 0 and x = 0, z = 0.
Solution:
The equations x = y, z = 0 can be written as \(\frac{x}{1}=\frac{y}{1}\), z = 0
∴ the direction ratios of the line are 1, 1, 0.
The direction ratios of the line x = 0, z = 0, i.e., Y-axis J are 0, 1, 0.
∴ its directiton ratios are 0, 1, 0.
Let \(\bar{a}\) and \(\bar{b}\) be the vectors in the direction of the lines x = y, z = 0 and x = 0, z = 0.

If θ is the acute angle between the lines, then

Question 14.
Find the acute angle between lines x = -y, z = 0 and x = 0, z = 0.
Solution:
The equations x = -y, z = 0 can be written as \(\frac{x}{1}=\frac{y}{1}\), z = 0.
∴ the direction ratios of the line are 1, 1, 0.
The direction ratios of the line x = 0, z = 0, i.e., Y-axis are 0, 1, 0.
∴ its direction ratios are 0, 1, 0.
Let \(\bar{a}\) and \(\bar{b}\) be the vectors in the direction of the lines x = y, z = 0 and x = 0, z = 0

Question 15.
Find the co-ordinates of the foot of the perpendicular drawn from the point (0, 2, 3) to the line \(\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}\).
Solution:
Let P = (0, 2, 3)
Let M be the foot of the perpendicular drawn from P to the line \(\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}\) = λ ……(Say)
The coordinates of any point on the line are given by
x = 5λ – 3, y = 2λ + 1, z = 3λ – 4
Let M = (5λ – 3, 2λ + 1, 3λ – 4) …(1)
The direction ratios of PM are
5λ – 3 – 0, 2λ + 1 – 2, 3λ – 4 – 3 i.e. 5λ – 3, 2λ – 1, 3λ – 7
Since, PM is perpendicular to the line whose direcction ratios are 5, 2, 3,
5(5λ – 3) + 2(2λ – 1) + 3(3λ – 7) = 0
25λ – 15 + 4λ – 2 + 9λ – 21 =0
38λ – 38 = 0 ∴ λ = 1
Substituting λ = 1 in (1), we get.
M = (5 – 3, 2 + 1, 3 – 4) = (2, 3, -1).
Hence, the coordinates of the foot of perpendicular are (2, 3, – 1).

Question 16.
By computing the shortest distance determine whether following lines intersect each other.
(i) \(\bar{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(2 \hat{i}-\hat{j}+\hat{k})\) and \(\bar{r}=(2 \hat{i}+2 \hat{j}-3 \hat{k})+\mu(\hat{i}+\hat{j}-2 \hat{k})\)
Solution:
The shortest distance between the lines


Shortest distance between the lines is 0.
∴ the lines intersect each other.

(ii) \(\frac{x-5}{4}=\frac{y-7}{5}=\frac{z+3}{5}\) and x – 6 = y – 8 = z + 2.
Solution:
The shortest distance between the lines

∴ x₁ = 5, y₁ = 7, z₁ = 3, x₂ = 6, y₂ = 8, z₂ = 2,
l₁ = 4, m₁ = 5, n₁ = 1, l₂ = 1, m₂ = -2, n₂ = 1

= 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 + (1 – 7)² + (-14 + 6)²
= 16 + 36 + 64
= 116
Hence, the required shortest distance between the given lines

or
Shortest distance between the lines is 0.
∴ the lines intersect each other.

Question 17.
If lines \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\) and \(\frac{x-2}{1}=\frac{y+m}{2}=\frac{z-2}{1}\) intersect each other then find m.
Solution:

Here, (x₁, y₁, z₁) ≡ (1, -1, 1),
(x₂, y₂, z₂) ≡ (2, -m, 2),
a₁ = 2, b₁ = 3, c₁ = 4,
a₂ = 1, b₂ = 2, c₂ = 1
Substituting these values in (1), we get

∴ 1(3 – 8) – (1 – m)(2 – 4) + 1 (4 – 3) = 0
∴ -5 + 2 – 2m + 1 = 0
∴ -2m = 2
∴ m = -1.

Question 18.
Find the vector and Cartesian equations of the line passing through the point (-1, -1, 2) and parallel to the line 2x – 2 = 3y + 1 = 6z – 2.
Solution:
Let \(\bar{a}\) be the position vector of the point A (-1, -1, 2) w.r.t. the origin.
Then \(\bar{a}=-\hat{i}-\hat{j}+2 \hat{k}\)
The equation of given line is
x – 2 = 3y + 1 = 6z – 2.

The direction ratios of this line are

Question 19.
Find the direction cosines of the line \(\bar{r}=\left(-2 \hat{i}+\frac{5}{2} \hat{j}-\hat{k}\right)+\lambda(2 \hat{i}+3 \hat{j})\).
Solution:

Question 20.
Find the Cartesian equation of the line passing through the origin which is perpendicular to x – 1 = y – 2 = z – 1 and intersects the \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\).
Solution:
Let the required line have direction ratios a, b, c
Since the line passes through the origin, its cartesian equations are
\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\) …(1)
This line is perpendicular to the line
x – 1 = y – 2 = z – 1 whose direction ratios are 1, 1, 1.

∴ 1(4b – 3c) + 1(4a – 2c) + 1(3a – 2b) = 0
∴ 4b – 3c + 4a – 2c + 3a – 2b = 0
∴ 7a + 2b – 5c = 0
From (2) and (3), we get

Question 21.
Write the vector equation of the line whose Cartesian equations are y = 2 and 4x – 3z + 5 = 0.
Solution:
4x – 3z + 5 = 0 can be written as

This line passes through the point A(0, 2, \(\frac{5}{3}\)) position vector is \(\bar{a}=2 \hat{j}+\frac{5}{3} \hat{k}\)
Also the line has direction ratio 3, 0, 4.
If \(\bar{b}\) is a vector parallel to the line, then \(\bar{b}=3 \hat{i}+4 \hat{k}\)
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 \(\bar{a}\) scalar,
∴ the vector equation of the required line is
\(\bar{r}=\left(2 \hat{j}+\frac{5}{3} \hat{k}\right)+\lambda(3 \hat{i}+4 \hat{k})\).

Question 22.
Find the co-ordinates of points on the line \(\frac{x-1}{1}=\frac{y-2}{-2}=\frac{z-3}{2}\) which are at the distance 3 unit from the base point A(1, 2, 3).
Solution:
The cartesian equations of the line are \(\frac{x-1}{1}=\frac{y-2}{-2}=\frac{z-3}{2}\) = λ
The coordinates of any point on this line are given by
x = λ + 1, y = -2λ + 2, z = 2λ + 3
Let M(λ + 1, -2λ + 2, 2λ + 3) … (1)
be the point on the line whose distance from A(1, 2, 3) is 3 units.

When λ = 1, M = (1 + 1, -2 + 2, 2 + 3) … [By (1)]
i. e. M = (2, 0, 5)
When λ = -1, M = (1 – 1, 2 + 2, -2 + 3) … [By (1)]
i. e. M = (0, 4, 1)
Hence, the coordinates of the required points are (2, 0, 5) and (0, 4, 1).

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Line and Plane Ex 6.4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Line and Plane Ex 6.4

Question 1.
Find the angle between planes \(\bar{r} \cdot(\hat{i}+\hat{j}+2 \hat{k})\) = 13 and \(\bar{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})\) = 31 .
Solution:
The acute angle θ between the planes \(\bar{r} \cdot \bar{n}_{1}\) = d₁ and \(\bar{r} \cdot \bar{n}_{2}\) = d₂ is given by

Question 2.
Find the acute angle between the line \(\bar{r} \cdot(\hat{i}+2 \hat{j}+2 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}-6 \hat{k})\) and the plane \(\bar{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})\) = 0
Solution:
The acute angle θ between the line \(\bar{r}=\bar{a}+\lambda \bar{b}\) and the plane \(\bar{r} \cdot \bar{n}\) = d is given by

Question 3.
Show that lines \(\bar{r}=(2 \hat{j}-3 \hat{k})+\lambda(\hat{i}+2 \hat{j}+3 \hat{k})\) and \(\bar{r}=(2 \hat{i}+6 \hat{j}+3 \hat{k})+\mu(2 \hat{i}+3 \hat{j}+4 \hat{k})\) are coplanar. Find the equation of the plane determined by them.
Solution:

= 2(-1) + 6(2) + 3(-1)
= -2 + 12 – 3 = 7
∴ \(\bar{a}_{1} \cdot\left(\bar{b}_{1} \times \bar{b}_{2}\right)=\bar{a}_{2} \cdot\left(\bar{b}_{1} \times \bar{b}_{2}\right)\)
Hence, the given lines are coplanar.
The plane determined by these lines is given by
∴ \(\bar{r} \cdot\left(\overline{b_{1}} \times \overline{b_{2}}\right)=\overline{a_{1}} \cdot\left(\overline{b_{1}} \times \overline{b_{2}}\right)\)
i.e. \(\bar{r} \cdot(-\hat{i}+2 \hat{j}-\hat{k})\)
Hence, the given lines are coplanar and the equation of the plane determined by these lines is
\(\bar{r} \cdot(-\hat{i}+2 \hat{j}-\hat{k})\) = 7

Question 4.
Find the distance of the point \(4 \hat{i}-3 \hat{j}+\hat{k}\) from the plane \(\bar{r} \cdot(2 \hat{i}+3 \hat{j}-6 \hat{k})\) = 21 .
Solution:
The distance of the point A(\(\bar{a}\)) from the plane \(\bar{r} \cdot \bar{n}=p\) is given by d = \(\frac{|\bar{a} \cdot \bar{n}-p|}{|n|}\) …(1)
Here, \(\bar{a}=4 \hat{i}-3 \hat{j}+\hat{k}\), \(\bar{n}=2 \hat{i}+3 \hat{j}-6 \hat{k}\), p = 21
∴ \(\bar{a} \cdot \bar{n}\) = \((4 \hat{i}-3 \hat{j}+\hat{k}) \cdot(2 \hat{i}+3 \hat{j}-6 \hat{k})\)
= (4)(2) + (-3)(3) + (1)(-6)
= 8 – 9 – 6 = -7
Also, \(\sqrt{2^{2}+3^{2}+(-6)^{2}}=\sqrt{49}\) = 7
∴ from (1), the required distance
= \(\frac{|-7-21|}{7}\) = 4units

Question 5.
Find the distance of the point (1, 1, -1) from the plane 3x + 4y – 12z + 20 = 0.
Solution:
The distance of the point (x₁, y₁, z₁) from the plane ax + by + cz + d = 0 is \(\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|\)
∴ the distance of the point (1, 1, -1) from the plane 3x + 4y – 12z + 20 = 0 is \(\left|\frac{3(1)+4(1)-12(-1)+20}{\sqrt{3^{2}+4^{2}+(-12)^{2}}}\right|\)
= \(\left|\frac{3+4+12+20}{\sqrt{9+16+144}}\right|=\frac{39}{\sqrt{169}}\)
= \(\frac{39}{13}\) = 3units

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Pair of Straight Lines Ex 4.3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Ex 4.3

Question 1.
Find the joint equation of the pair of lines:
(i) Through the point (2, -1) and parallel to lines represented by 2x² + 3xy – 9y² = 0
Solution:
The combined equation of the given lines is
2x² + 3 xy – 9y² = 0
i.e. 2x² + 6xy – 3xy – 9y² = 0
i.e. 2x(x + 3y) – 3y(x + 3y) = 0
i.e. (x + 3y)(2x – 3y) = 0
∴ their separate equations are
x + 3y = 0 and 2x – 3y = 0
∴ their slopes are m₁ = \(\frac{-1}{3}\) and m₂ = \(\frac{-2}{-3}=\frac{2}{3}\).
The slopes of the lines parallel to these lines are m₁ and m₂, i.e. \(-\frac{1}{3}\) and \(\frac{2}{3}\).
∴ the equations of the lines with these slopes and through the point (2, -1) are
y + 1 = \(-\frac{1}{3}\) (x – 2) and y + 1 = \(\frac{2}{3}\)(x – 2)
i.e. 3y + 3= -x + 2 and 3y + 3 = 2x – 4
i.e. x + 3y + 1 = 0 and 2x – 3y – 7 = 0
∴ the joint equation of these lines is
(x + 3y + 1)(2x – 3y – 7) = 0
∴ 2x² – 3xy – 7x + 6xy – 9y² – 21y + 2x – 3y – 7 = 0
∴ 2x² + 3xy – 9y² – 5x – 24y – 7 = 0.

(ii) Through the point (2, -3) and parallel to lines represented by x² + xy – y² = 0
Solution:
Comparing the equation
x² + xy – y² = 0 … (1)
with ax² + 2hxy + by² = 0, we get,
a = 1, 2h = 1, b = -1
Let m₁ and m₂ be the slopes of the lines represented by (1).

The slopes of the lines parallel to these lines are m₁ and m₂.
∴ the equations of the lines with these slopes and through the point (2, -3) are
y + 3 = m₁(x – 2) and y + 3 = m₂(x – 2)
i.e. m₁(x – 2) – (y + 3) = 0 and m₂(x – 2) – (y + 3) = 0
∴ the joint equation of these lines is
[m₁(x – 2) – (y + 3)][m₂(x – 2) – (y + 3)] = 0
∴ m₁m₂(x – 2)² – m₁(x – 2)(y + 3) – m₂(x – 2)(y + 3) + (y + 3)² = o
∴ m₁m₂(x – 2)² – (m₁ + m₂)(x – 2)(y + 3) + (y + 3)³ = 0
∴ -(x – 2)² – (x – 2)(y + 3) + (y + 3)² = 0 …… [By (2)]
∴ (x – 2)² + (x – 2)(y + 3) – (y + 3)² = 0
∴ (x² – 4x + 4) + (xy + 3x – 2y – 6) – (y² + 6y + 9) = 0
∴ x² – 4x + 4 + xy + 3x – 2y – 6 – y² – 6y – 9 = 0
∴ x² + xy – y² – x – 8y – 11 = 0.

Question 2.
Show that equation x² + 2xy+ 2y² + 2x + 2y + 1 = 0 does not represent a pair of lines.
Solution:
Comparing the equation
x² + 2xy + 2y² + 2x + 2y + 1 = 0 with
ax² + 2hxy + by² + 2gx + 2fy + c = 0, we get,
a = 1, h = 1, b = 2, g = 1, f = 1, c = 1.
The given equation represents a pair of lines, if
D = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|\) = 0 and h² – ab ≥ 0
Now, D = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 1
\end{array}\right|\)
= 1 (2 – 1) – 1(1 – 1) + 1 (1 – 2)
= 1 – 0 – 1 = 0
and h² – ab = (1)² – 1(2) = -1 < 0
∴ given equation does not represent a pair of lines.

Question 3.
Show that equation 2x² – xy – 3y² – 6x + 19y – 20 = 0 represents a pair of lines.
Solution:
Comparing the equation
2x² – xy – 3y² – 6x + 19y – 20 = 0
with ax² + 2hxy + by² + 2gx + 2fy + c = 0, we get,
a = 2, h = \(-\frac{1}{2}\), b = -3, g = -3, f = \(\frac{19}{2}\), c = -20.
∴ D = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=\left|\begin{array}{rrr}
2 & -\frac{1}{2} & -3 \\
-\frac{1}{2} & -3 & \frac{19}{2} \\
-3 & \frac{19}{2} & -20
\end{array}\right|\)
Taking \(\frac{1}{2}\) common from each row, we get,
D = \(\frac{1}{8}\left|\begin{array}{rrr}
4 & -1 & -6 \\
-1 & -6 & 19 \\
-6 & 19 & -40
\end{array}\right|\)
= \(\frac{1}{8}\)[4(240 – 361) + 1(40 + 114) – 6(-19 – 36)]
= \(\frac{1}{8}\)[4(-121) + 154 – 6(-55)]
= \(\frac{11}{8}\)[4(-11) + 14 – 6(-5)]
= \(\frac{1}{8}\)(-44 + 14 + 30) = 0
Also h² – ab = \(\left(-\frac{1}{2}\right)^{2}\) – 2(-3) = \(\frac{1}{4}\) + 6 = \(\frac{25}{4}\) > 0
∴ the given equation represents a pair of lines.

Question 4.
Show the equation 2x² + xy – y² + x + 4y – 3 = 0 represents a pair of lines. Also find the acute angle between them.
Solution:
Comparing the equation
2x² + xy — y² + x + 4y — 3 = 0 with
ax² + 2hxy + by² + 2gx + 2fy + c — 0, we get,
a = 2, h = \(\frac{1}{2}\), b = -1, g = \(\frac{1}{2}\), f = 2, c = – 3.
∴ D = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=\left|\begin{array}{lrr}
2 & \frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & -1 & 2 \\
\frac{1}{2} & 2 & -3
\end{array}\right|\)
Taking \(\frac{1}{2}\) common from each row, we get,
D = \(\frac{1}{8}\left|\begin{array}{rrr}
4 & 1 & 1 \\
1 & -2 & 4 \\
1 & 4 & -6
\end{array}\right|\)
= \(\frac{1}{8}\)[4(12 —16) — 1( —6 — 4) + 1(4 + 2)]
= \(\frac{1}{8}\)[4( – 4) – 1(-10) + 1(6)]
= \(\frac{1}{8}\)(—16 + 10 + 6) = 0
Also, h² – ab = \(\left(\frac{1}{2}\right)^{2}\) – 2(-1) = \(\frac{1}{4}\) + 2 = \(\frac{9}{4}\) > 0
∴ the given equation represents a pair of lines. Let θ be the acute angle between the lines
∴ tan θ = \(\left|\frac{2 \sqrt{h^{2}-a b}}{a+b}\right|\)

Question 5.
Find the separate equation of the lines represented by the following equations :
(i) (x – 2)² – 3(x – 2)(y + 1) + 2(y + 1)² = 0
Solution:
(x – 2)² – 3(x – 2)(y + 1) + 2(y + 1)2 = 0
∴ (x – 2)² – 2(x – 2)(y + 1) – (x – 2)(y + 1) + 2(y + 1)² = 0
∴ (x – 2) [(x – 2) – 2(y + 1)] – (y + 1)[(x – 2) – 2(y + 1)] = 0
∴ (x – 2)(x – 2 – 2y – 2) – (y + 1)(x – 2 – 2y – 2) = 0
∴ (x – 2)(x – 2y – 4) – (y + 1)(x – 2y – 4) = 0
∴ (x – 2y – 4)(x – 2 – y – 1) = 0
∴ (x – 2y – 4)(x – y – 3) = 0
∴ the separate equations of the lines are
x – 2y – 4 = 0 and x – y – 3 = 0.
Alternative Method :
(x – 2)² – 3(x – 2)(y + 1) + 2(y + 1)² = 0 … (1)
Put x – 2 = X and y + 1 = Y
∴ (1) becomes,
X² – 3XY + 2Y² = 0
∴ X² – 2XY – XY + 2Y² = 0
∴ X(X – 2Y) – Y(X – 2Y) = 0
∴ (X – 2Y)(X – Y) = 0
∴ the separate equations of the lines are
∴ X – 2Y = 0 and X – Y = 0
∴ (x – 2) – 2(y + 1) = 0 and (x – 2) – (y +1) = 0
∴ x – 2y – 4 = 0 and x – y – 3 = 0.

(ii) 10(x + 1)² + (x + 1)( y – 2) – 3(y – 2)² = 0
Solution:
10(x + 1)² + (x + 1)( y – 2) – 3(y – 2)² = 0 …(1)
Put x + 1 = X and y – 2 = Y
∴ (1) becomes
10x² + xy – 3y² = 0
10x² + 6xy – 5xy – 3y² = 0
2x(5x + 3y) – y(5x + 3y) = 0
(2x – y)(5x + 3y) = 0
5x + 3y = 0 and 2x – y = 0
5x + 3y = 0
5(x + 1) + 3(y – 2) = 0
5x + 5 + 3y – 6 = 0
∴ 5x + 3y – 1 = 0
2x – y = 0
2(x + 1) – (y – 2) = 0
2x + 2 – y + 2 = 0
∴ 2x – y + 4 = 0

Question 6.
Find the value of k if the following equations represent a pair of lines :
(i) 3x² + 10xy + 3y² + 16y + k = 0
Solution:
Comparing the given equation with
ax² + 2hxy + by² + 2gx + 2fy + c = 0,
we get, a = 3, h = 5, b = 3, g = 0, f= 8, c = k.
Now, given equation represents a pair of lines.
∴ abc + 2fgh – af² – bg² – ch² = 0
∴ (3)(3)(k) + 2(8)(0)(5) – 3(8)² – 3(0)² – k(5)² = 0
∴ 9k + 0 – 192 – 0 – 25k = 0
∴ -16k – 192 = 0
∴ – 16k = 192
∴ k= -12.

(ii) kxy + 10x + 6y + 4 = 0
Solution:
Comparing the given equation with
ax² + 2 hxy + by² + 2gx + 2fy + c = 0,
we get, a = 0, h = \(\frac{k}{2}\), b = 0, g = 5, f = 3, c = 4
Now, given equation represents a pair of lines.
∴ abc + 2fgh – af² – bg² – ch² = 0
∴ (0)(0)(4) + 2(3)(5)\(\left(\frac{k}{2}\right)\) – 0(3)² – 0(5)² – 4\(\left(\frac{k}{2}\right)^{2}\) = 0
∴ 0 + 15k – 0 – 0 – k² = 0
∴ 15k – k² = 0
∴ -k(k – 15) = 0
∴ k = 0 or k = 15.
If k = 0, then the given equation becomes
10x + 6y + 4 = 0 which does not represent a pair of lines.
∴ k ≠ o
Hence, k = 15.

(iii) x² + 3xy + 2y² + x – y + k = 0
Solution:
Comparing the given equation with
ax² + 2hxy + by² + 2gx + 2fy + c = 0,
we get, a = 1, h = \(\frac{3}{2}\), b = 2, g = \(\frac{1}{2}\), f= \(-\frac{1}{2}\), c = k.
Now, given equation represents a pair of lines.


∴ 2(8k – 1) – 3(6k + 1) + 1(-3 – 4) = 0
∴ 16k – 2 – 18k – 3 – 7 = 0
∴ -2k – 12 = 0
∴ -2k = 12 ∴ k = -6.

Question 7.
Find p and q if the equation px² – 8xy + 3y² + 14x + 2y + q = 0 represents a pair of perpendicular lines.
Solution:
The given equation represents a pair of lines perpendicular to each other
∴ (coefficient of x²) + (coefficient of y²) = 0
∴ p + 3 = 0 p = -3
With this value of p, the given equation is
– 3x² – 8xy + 3y² + 14x + 2y + q = 0.
Comparing this equation with
ax² + 2hxy + by² + 2gx + 2fy + c = 0, we have,
a = -3, h = -4, b = 3, g = 7, f = 1 and c = q.
D = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=\left|\begin{array}{rrr}
-3 & -4 & 7 \\
-4 & 3 & 1 \\
7 & 1 & q
\end{array}\right|\)
= -3(3q – 1) + 4(-4q – 7) + 7(-4 – 21)
= -9q + 3 – 16q – 28 – 175
= -25q – 200= -25(q + 8)
Since the given equation represents a pair of lines, D = 0
∴ -25(q + 8) = 0 ∴ q= -8.
Hence, p = -3 and q = -8.

Question 8.
Find p and q if the equation 2x² + 8xy + py² + qx + 2y – 15 = 0 represents a pair of parallel lines.
Solution:
The given equation is
2x² + 8xy + py² + qx + 2y – 15 = 0
Comparing it with ax² + 2hxy + by² + 2gx + 2fy + c = 0, we get,
a = 2, h = 4, b = p, g = \(\frac{q}{2}\), f = 1, c = – 15
Since the lines are parallel, h² = ab
∴ (4)² = 2p ∴ P = 8
Since the given equation represents a pair of lines

i.e. – 242 + 240 + 2q + 2q – 2q² = 0
i.e. -2q² + 4q – 2 = 0
i.e. q² – 2q + 1 = 0
i.e. (q – 1)² = 0 ∴ q – 1 = 0 ∴ q = 1.
Hence, p = 8 and q = 1.

Question 9.
Equations of pairs of opposite sides of a parallelogram are x² – 7x+ 6 = 0 and y² – 14y + 40 = 0. Find the joint equation of its diagonals.
Solution:
Let ABCD be the parallelogram such that the combined equation of sides AB and CD is x² – 7x + 6 = 0 and the combined equation of sides BC and AD is y² – 14y + 40 = 0.
The separate equations of the lines represented by x² – 7x + 6 = 0, i.e. (x – 1)(x – 6) = 0 are x – 1 = 0 and x – 6 = 0.
Let equation of the side AB be x – 1 = 0 and equation of side CD be x – 6 = 0.
The separate equations of the lines represented by y² – 14y + 40 = 0, i.e. (y – 4)(y – 10) = 0 are y – 4 = 0 and y – 10 = 0.
Let equation of the side BC be y – 4 = 0 and equation of side AD be y – 10 = 0.

Coordinates of the vertices of the parallelogram are A(1, 10), B(1, 4), C(6, 4) and D(6, 10).
∴ equation of the diagonal AC is
\(\frac{y-10}{x-1}\) = \(\frac{10-4}{1-6}\) = \(\frac{6}{-5}\)
∴ -5y + 50 = 6x – 6
∴ 6x + 5y – 56 = 0
and equation of the diagonal BD is
\(\frac{y-4}{x-1}\) = \(\frac{4-10}{1-6}\) = \(\frac{-6}{-5}\) = \(\frac{6}{5}\)
∴ 5y – 20 = 6x – 6
∴ 6x – 5y + 14 = 0
Hence, the equations of the diagonals are 6x + 5y – 56 = 0 and 6x – 5y + 14 = 0.
∴ the joint equation of the diagonals is (6x + 5y – 56)(6x – 5y + 14) = 0
∴ 36x² – 30xy + 84x + 30xy – 25y² + 70y – 336x + 280y – 784 = 0
∴ 36x² – 25y² – 252x + 350y – 784 = 0.

Question 10.
∆OAB is formed by lines x² – 4xy + y² = 0 and the line 2x + 3y – 1 = 0. Find the equation of the median of the triangle drawn from O.
Solution:

Let D be the midpoint of seg AB where A is (x₁, y₁) and B is (x₂, y₂).
Then D has coordinates \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\).
The joint (combined) equation of the lines OA and OB is x² – 4xy + y² = 0 and the equation of the line AB is 2x + 3y – 1 = 0.
∴ points A and B satisfy the equations 2x + 3y – 1 = 0
and x² – 4xy + y² = 0 simultaneously.
We eliminate x from the above equations, i.e.,
put x = \(\frac{1-3 y}{2}\) in the equation x² – 4xy + y² = 0, we get,
∴ \(\left(\frac{1-3 y}{2}\right)^{2}\) – 4\(\left(\frac{1-3 y}{2}\right)\)y + y² = 0
∴ (1 – 3y)² – 8(1 – 3y)y + 4y² = 0
∴1 – 6y + 9y² – 8y + 24y² + 4y² = 0
∴ 37y² – 14y + 1 = 0
The roots y₁ and y₂ of the above quadratic equation are the y-coordinates of the points A and B.
∴ y₁ + y₂ = \(\frac{-b}{a}=\frac{14}{37}\)
∴ y-coordinate of D = \(\frac{y_{1}+y_{2}}{2}=\frac{7}{37}\).
Since D lies on the line AB, we can find the x-coordinate of D as
2x + 3\(\left(\frac{7}{37}\right)\) – 1 = 0
∴ 2x = 1 – \(\frac{21}{37}=\frac{16}{37}\)
∴ x = \(\frac{8}{37}\)
∴ D is (8/37, 7/37)
∴ equation of the median OD is \(\frac{x}{8 / 37}=\frac{y}{7 / 37}\),
i.e., 7x – 8y = 0.

Question 11.
Find the co-ordinates of the points of intersection of the lines represented by x² – y² – 2x + 1 = 0.
Solution:
Consider, x² – y² – 2x + 1 = 0
∴ (x² – 2x + 1) – y² = 0
∴ (x – 1)² – y² = 0
∴ (x – 1 + y)(x – 1 – y) = 0
∴ (x + y – 1)(x – y – 1) = 0
∴ separate equations of the lines are
x + y – 1 = 0 and x – y +1 = 0.
To find the point of intersection of the lines, we have to solve
x + y – 1 = 0 … (1)
and x – y + 1 = 0 … (2)
Adding (1) and (2), we get,
2x = 0 ∴ x = 0
Substituting x = 0 in (1), we get,
0 + y – 1 = 0 ∴ y = 1
∴ coordinates of the point of intersection of the lines are (0, 1).

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Pair of Straight Lines Ex 4.2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Ex 4.2

Question 1.
Show that lines represented by 3x² – 4xy – 3y² = 0 are perpendicular to each other.
Solution:
Comparing the equation 3x² – 4 xy – 3y² = 0 with ax² + 2hxy + by² = 0, we get, a = 3, 2h = -4, b = -3 Since a + b = 3 + (-3) = 0, the lines represented by 3x² – 4xy – 3y² = 0 are perpendicular to each other.

Question 2.
Show that lines represented by x² + 6xy + gy²= 0 are coincident.
The question is modified.
Show that lines represented by x² + 6xy + 9y²= 0 are coincident.
Solution:
Comparing the equation x² + 6xy + 9y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 1, 2h = 6, i.e. h = 3 and b = 9
Since h² – ab = (3)² – 1(9)
= 9 – 9 = 0, .
the lines represented by x² + 6xy + 9y² = 0 are coincident.

Question 3.
Find the value of k if lines represented by kx² + 4xy – 4y² = 0 are perpendicular to each other.
Solution:
Comparing the equation kx² + 4xy – 4y² = 0 with ax² + 2hxy + by² = 0, we get,
a = k, 2h = 4, b = -4
Since lines represented by kx² + 4xy – 4y² = 0 are perpendicular to each other,
a + b = 0
∴ k – 4 = 0 ∴ k = 4.

Question 4.
Find the measure of the acute angle between the lines represented by:
(i) 3x² – 4\(\sqrt {3}\)xy + 3y² = 0
Solution:
Comparing the equation 3x² – 4\(\sqrt {3}\)xy + 3y² = 0 with
ax² + 2hxy + by² = 0, we get,
a = 3, 2h = -4\(\sqrt {3}\), i.e. h = -24\(\sqrt {3}\) and b = 3
Let θ be the acute angle between the lines.

∴ θ = 30°.

(ii) 4x² + 5xy + y² = 0
Solution:
Comparing the equation 4x² + 5xy + y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 4, 2h = 5, i.e. h = \(\frac{5}{2}\) and b = 1.
Let θ be the acute angle between the lines.

(iii) 2x² + 7xy + 3y² = 0
Solution:
Comparing the equation
2x² + 7xy + 3y² = 0 with
ax² + 2hxy + by² = 0, we get,
a = 2, 2h = 7 i.e. h = \(\frac{7}{2}\) and b = 3
Let θ be the acute angle between the lines.


tanθ = 1
∴ θ = tan 1 = 45°
∴ θ = 45°

(iv) (a² – 3b²)x² + 8abxy + (b² – 3a²)y² = 0
Solution:
Comparing the equation
(a² – 3b²)x² + 8abxy + (b² – 3a²)y² = 0, with
Ax² + 2Hxy + By² = 0, we have,
A = a² – 3b², H = 4ab, B = b² – 3a².
∴ H² – AB = 16a²b² – (a² – 3b²)(b² – 3a²)
= 16a²b² + (a² – 3b²)(3a² – b²)
= 16a²b² + 3a⁴ – 10a²b² + 3b⁴
= 3a⁴ + 6a²b² + 3b⁴
= 3(a⁴ + 2a²b² + b⁴)
= 3 (a² + b²)²
∴ \(\sqrt{H^{2}-A B}\) = \(\sqrt {3}\) (a² + b²)
Also, A + B = (a² – 3b²) + (b² – 3a²)
= -2 (a² + b²)
If θ is the acute angle between the lines, then
tan θ = \(\left|\frac{2 \sqrt{H^{2}-A B}}{A+B}\right|=\left|\frac{2 \sqrt{3}\left(a^{2}+b^{2}\right)}{-2\left(a^{2}+b^{2}\right)}\right|\)
= \(\sqrt {3}\) = tan 60°
∴ θ = 60°

Question 5.
Find the combined equation of lines passing through the origin each of which making an angle of 30° with the line 3x + 2y – 11 = 0
Solution:
The slope of the line 3x + 2y – 11 = 0 is m₁ = \(-\frac{3}{2}\) .
Let m be the slope of one of the lines making an angle of 30° with the line 3x + 2y – 11 = 0.
The angle between the lines having slopes m and m1 is 30°.

On squaring both sides, we get,
\(\frac{1}{3}=\frac{(2 m+3)^{2}}{(2-3 m)^{2}}\)
∴ (2 – 3m)² = 3 (2m + 3)²
∴ 4 – 12m + 9m² = 3(4m² + 12m + 9)
∴ 4 – 12m + 9m² = 12m² + 36m + 27
3m² + 48m + 23 = 0
This is the auxiliary equation of the two lines and their joint equation is obtained by putting m = \(\frac{y}{x}\).
∴ the combined equation of the two lines is
3\(\left(\frac{y}{x}\right)^{2}\) + 48\(\left(\frac{y}{x}\right)\) + 23 = 0
∴ \(\frac{3 y^{2}}{x^{2}}+\frac{48 y}{x}\) + 23 = 0
∴ 3y² + 48xy + 23x² = 0
∴ 23x² + 48xy + 3y² = 0.

Question 6.
If the angle between lines represented by ax² + 2hxy + by² = 0 is equal to the angle between lines represented by 2x² – 5xy + 3y² = 0 then show that 100(h² – ab) = (a + b)².
Solution:
The acute angle θ between the lines ax² + 2hxy + by² = 0 is given by
tan θ = \(\left|\frac{2 \sqrt{h^{2}-a b}}{a+b}\right|\) ..(1)
Comparing the equation 2x² – 5xy + 3y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 2, 2h= -5, i.e. h = \(-\frac{5}{2}\) and b = 3
Let ∝ be the acute angle between the lines 2x² – 5xy + 3y² = 0.

This is the required condition.

Question 7.
Find the combined equation of lines passing through the origin and each of which making angle 60° with the Y- axis.
Solution:

Let OA and OB be the lines through the origin making an angle of 60° with the Y-axis.
Then OA and OB make an angle of 30° and 150° with the positive direction of X-axis.
∴ slope of OA = tan 30° = \(\frac{1}{\sqrt{3}}\)
∴ equation of the line OA is
y = \(\frac{1}{\sqrt{3}}\) = x, i.e. x – \(\sqrt {3}\)y = 0
Slope of OB = tan 150° = tan (180° – 30°)
= tan 30° = \(-\frac{1}{\sqrt{3}}\)
∴ equation of the line OB is
y = \(-\frac{1}{\sqrt{3}}\)x, i.e. x + \(\sqrt {3}\) y = 0
∴ required combined equation is
(x – \(\sqrt {3}\)y)(x + \(\sqrt {3}\)y) = 0
i.e. x² – 3y² = 0.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Pair of Straight Lines Ex 4.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Ex 4.1

Question 1.
Find the combined equation of the following pairs of lines:
(i) 2x + y = 0 and 3x – y = 0
Solution:
The combined equation of the lines 2x + y = 0 and 3x – y = 0 is
(2x + y)( 3x – y) = 0
∴ 6x² – 2xy + 3xy – y² = 0
∴ 6x² – xy – y² = 0.

(ii) x + 2y – 1 = 0 and x – 3y + 2 = 0
Solution:
The combined equation of the lines x + 2y – 1 = 0 and x – 3y + 2 = 0 is
(x + 2y – 1)(x – 3y + 2) = 0
∴ x² – 3xy + 2x + 2xy – 6y² + 4y – x + 3y – 2 = 0
∴ x² – xy – 6y² + x + 7y – 2 = 0.

(iii) Passing through (2, 3) and parallel to the co-ordinate axes.
Solution:
Equations of the coordinate axes are x = 0 and y = 0.
∴ the equations of the lines passing through (2, 3) and parallel to the coordinate axes are x = 2 and
i.e. x – 2 = 0 and y – 3 = 0.
∴ their combined equation is
(x – 2)(y – 3) = 0.
∴ xy – 3x – 2y + 6 = 0.

(iv) Passing through (2, 3) and perpendicular to lines 3x + 2y – 1 = 0 and x – 3y + 2 = 0
Solution:
Let L₁ and L₂ be the lines passing through the point (2, 3) and perpendicular to the lines 3x + 2y – 1 = 0 and x – 3y + 2 = 0 respectively.
Slopes of the lines 3x + 2y – 1 = 0 and x – 3y + 2 = 0 are \(\frac{-3}{2}\) and \(\frac{-1}{-3}=\frac{1}{3}\) respectively.
∴ slopes of the lines L₁ and L₂ are \(\frac{2}{3}\) and -3 respectively.
Since the lines L₁ and L₂ pass through the point (2, 3), their equations are
y – 3 = \(\frac{2}{3}\)(x – 2) and y – 3 = -3 (x – 2)
∴ 3y – 9 = 2x – 4 and y – 3= -3x + 6
∴ 2x – 3y + 5 = 0 and 3x – y – 9 = 0
∴ their combined equation is
(2x – 3y + 5)(3x + y – 9) = 0
∴ 6x² + 2xy – 18x – 9xy – 3y² + 27y + 15x + 5y – 45 = 0
∴ 6x² – 7xy – 3y² – 3x + 32y – 45 = 0.

(v) Passsing through (-1, 2),one is parallel to x + 3y – 1 = 0 and the other is perpendicular to 2x – 3y – 1 = 0.
Solution:
Let L₁ be the line passing through (-1, 2) and parallel to the line x + 3y – 1 = 0 whose slope is –\(\frac{1}{3}\).
∴ slope of the line L₁ is –\(\frac{1}{3}\)
∴ equation of the line L₁ is
y – 2 = –\(\frac{1}{3}\)(x + 1)
∴ 3y – 6 = -x – 1
∴ x + 3y – 5 = 0
Let L₂ be the line passing through (-1, 2) and perpendicular to the line 2x – 3y – 1 = 0
whose slope is \(\frac{-2}{-3}=\frac{2}{3}\).
∴ slope of the line L₂ is –\(\frac{3}{2}\)
∴ equation of the line L₂ is
y – 2= –\(\frac{3}{2}\)(x + 1)
∴ 2y – 4 = -3x – 3
∴ 3x + 2y – 1 = 0
Hence, the equations of the required lines are
x + 3y – 5 = 0 and 3x + 2y – 1 = 0
∴ their combined equation is
(x + 3y – 5)(3x + 2y – 1) = 0
∴ 3x² + 2xy – x + 9xy + 6y² – 3y – 15x – 10y + 5 = 0
∴ 3x² + 11xy + 6y² – 16x – 13y + 5 = 0

Question 2.
Find the separate equations of the lines represented by following equations:
(i) 3y² + 7xy = 0
Solution:
3y² + 7xy = 0
∴ y(3y + 7x) = 0
∴ the separate equations of the lines are y = 0 and 7x + 3y = 0.

(ii) 5x² – 9y² = 0
Solution:
5x² – 9y² = 0
∴ (\(\sqrt {5}\) x)² – (3y)² = 0
∴ (\(\sqrt {5}\)x + 3y)(\(\sqrt {5}\)x – 3y) = 0
∴ the separate equations of the lines are
\(\sqrt {5}\)x + 3y = 0 and \(\sqrt {5}\)x – 3y = 0.

(iii) x² – 4xy = 0
Solution:
x² – 4xy = 0
∴ x(x – 4y) = 0
∴ the separate equations of the lines are x = 0 and x – 4y = 0

(iv) 3x² – 10xy – 8y² = 0
Solution:
3x² – 10xy – 8y² = 0
∴ 3x² – 12xy + 2xy – 8y² = 0
∴ 3x(x – 4y) + 2y(x – 4y) = 0
∴ (x – 4y)(3x +2y) = 0
∴ the separate equations of the lines are x – 4y = 0 and 3x + 2y = 0.

(v) 3x² – \(2 \sqrt{3}\) xy – 3y² = 0
Solution:
3x² – 2\(\sqrt {3}\)xy – 3y2 = 0
∴ 3x² – 3\(\sqrt {3}\)xy + \(\sqrt {3}\)xy – 3y² = 0
∴ 3x(x – \(\sqrt {3}\)y) + \(\sqrt {3}\)y(x – \(\sqrt {3}\)y) = 0
∴ (x – \(\sqrt {3}\)y)(3x + \(\sqrt {3}\)y) = 0
∴ the separate equations of the lines are
∴ x – \(\sqrt {3}\)y = 0 and 3x + \(\sqrt {3}\)y = 0.

(vi) x² + 2(cosec ∝)xy + y² = 0
Solution:
x² + 2 (cosec ∝)xy – y² = 0
i.e. y² + 2(cosec∝)xy + x² = 0
Dividing by x², we get,

∴ the separate equations of the lines are
(cosec ∝ – cot ∝)x + y = 0 and (cosec ∝ + cot ∝)x + y = 0.

(vii) x² + 2xy tan ∝ – y² = 0
Solution:
x² + 2xy tan ∝ – y² = 0
Dividind by y²

The separate equations of the lines are
(sec∝ – tan ∝)x + y = 0 and (sec ∝ + tan ∝)x – y = 0

Question 3.
Find the combined equation of a pair of lines passing through the origin and perpendicular
to the lines represented by following equations :
(i) 5x² – 8xy + 3y² = 0
Solution:
Comparing the equation 5x² – 8xy + 3y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 5, 2h = -8, b = 3
Let m₁ and m₂ be the slopes of the lines represented by 5x² – 8xy + 3y² = 0.
∴ m₁ + m₂ = \(\frac{-2 h}{b}=\frac{8}{3}\)
amd m₁m₂ = \(\frac{a}{b}=\frac{5}{3}\) …(1)
Now required lines are perpendicular to these lines
∴ their slopes are -1 /m₁ and -1/m₂ Since these lines are passing through the origin, their separate equations are
y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x
i.e. m₁y = -x and m₂y = -x
i.e. x + m₁y = 0 and x + m₂y = 0
∴ their combined equation is
(x + m₁y) (x + m₂y) = 0
∴ x² + (m₁ + m₂)xy + m₁m₂y² = 0
∴ x² + \(\frac{8}{3}\)xy + \(\frac{5}{3}\)y² = 0 … [By (1)]
∴ x² + 8xy + 5y\(\frac{8}{3}\) = 0

(ii) 5x² + 2xy – 3y² = 0
Solution:
Comparing the equation 5x² + 2xy – 3y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 5, 2h = 2, b = -3
Let m₁ and m₂ be the slopes of the lines represented by 5x² + 2xy – 3y² = 0
∴ m₁ + m₂ = \(\frac{-2 h}{b}=\frac{-2}{-3}=\frac{2}{3}\) and m₁m₂ = \(\frac{a}{b}=\frac{5}{-3}\) ..(1)
Now required lines are perpendicular to these lines
∴ their slopes are \(\frac{-1}{m_{1}}\) and \(\frac{-1}{m_{2}}\)
Since these lines are passing through the origin, their separate equations are
y = \(\frac{-1}{\mathrm{~m}_{1}}\)x and y = \(\frac{-1}{\mathrm{~m}_{2}}\)x
i.e. m₁y = -x amd m₂y = -x
i.e. x + m₁y = 0 and x + m₂y = 0
∴ their combined equation is
∴ (x + m₁y)(x + m₂y) = 0
x² + (m₁ + m₂)xy + m₁m₂y² = 0
∴ x² + \(\frac{2}{3}\)xy – \(\frac{5}{3}\)y = 0 …[By (1)]
∴ 3x² + 2xy – 5y² = 0

(iii) xy + y² = 0
Solution:
Comparing the equation xy + y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 0, 2h = 1, b = 1
Let m₁ and m₂ be the slopes of the lines represented by xy + y² = 0

Now required lines are perpendicular to these lines
∴ their slopes are \(\frac{-1}{m_{1}}\) and \(\frac{-1}{m_{2}}\).
Since these lines are passing through the origin, their separate equations are
y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x
i.e. m₁y = -x and m₂y = -x
i.e. x + m₁y = 0 and x + m₂y = 0
∴ their combined equation is
(x + m₁y) (x + m₂y) = 0
∴ x² + (m₁ + m₂)xy + m₁m₂y² = 0
∴ x² – xy = 0.y² = 0 … [By (1)]
∴ x² – xy = 0.
Alternative Method :
Consider xy + y² = 0
∴ y(x + y) = 0
∴ separate equations of the lines are y = 0 and
3x² + 8xy + 5y² = 0.
x + y = 0.
Let m₁ and m₂ be the slopes of these lines.
Then m₁ = 0 and m₂ = -1
Now, required lines are perpendicular to these lines.
∴ their slopes are \(-\frac{1}{m_{1}}\) and \(-\frac{1}{m_{2}}\)
Since, m₁ = 0, \(-\frac{1}{m_{1}}\) does not exist.
Also, m₂ = -1, \(-\frac{1}{m_{2}}\) = 1
Since these lines are passing through the origin, their separate equations are x = 0 and y = x,
i.e. x – y = 0
∴ their combined equation is
x(x – y) = 0
x² – xy = 0.

(iv) 3x² – 4xy = 0
Solution:
Consider 3x² – 4xy = 0
∴ x(3x – 4y) = 0
∴ separate equations of the lines are x = 0 and 3x – 4y = 0.
Let m₁ and m₂ be the slopes of these lines.
Then m₁ does not exist and and m₁ = \(\frac{3}{4}\).
Now, required lines are perpendicular to these lines.
∴ their slopes are \(-\frac{1}{m_{1}}\) and \(-\frac{1}{m_{2}}\).
Since m₁ does not exist, \(-\frac{1}{m_{1}}\) = 0
Also m₂ = \(\frac{3}{4^{\prime}}-\frac{1}{m_{2}}=-\frac{4}{3}\)
Since these lines are passing through the origin, their separate equations are y = 0 and y = \(-\frac{4}{3}\)x,
i.e.   4x + 3y = 0
∴ their combined equation is
y(4x + 3y) = 0
∴ 4xy + 3y² = 0.

Question 4.
Find k if,
(i) the sum of the slopes of the lines represented by x² + kxy – 3y² = 0 is twice their product.
Solution:
Comparing the equation x² + kxy – 3y² = 0 with ax² + 2hxy + by² = 0, we get, a = 1, 2h = k, b = -3.
Let m₁ and m₂ be the slopes of the lines represented by x² + kxy – 3y² = 0.
∴ m₁ + m₂ = \(\frac{-2 h}{b}=-\frac{k}{(-3)}=\frac{k}{3}\)
and m₁m₂ = \(\frac{a}{b}=\frac{1}{(-3)}=-\frac{1}{3}\)
Now, m₁ + m₂ = 2(m₁m₂) ..(Given)
∴ \(\frac{k}{3}=2\left(-\frac{1}{3}\right)\) ∴ k = -2

(ii) slopes of lines represent by 3x² + kxy – y² = 0 differ by 4.
Solution:
(ii) Comparing the equation 3x² + kxy – y² = 0 with ax² + 2hxy + by² = 0, we get, a = 3, 2h = k, b = -1.
Let m₁ and m₂ be the slopes of the lines represented by 3x² + kxy – y² = 0.
∴ m₁ + m₂ = \(\frac{-2 h}{b}=-\frac{k}{-1}\) = k
and m₁₂ = \(\frac{a}{b}=\frac{3}{-1}\) = -3
∴ (m₁ – m₂)² = (m₁ + m₂)² – 4m₁m₂
= k² – 4 (-3)
= k² + 12 … (1)
But |m₁ – m₂| =4
∴ (m₁ – m₂)² = 16 … (2)
∴ from (1) and (2), k² + 12 = 16
∴ k² = 4 ∴ k= ±2.

(iii) slope of one of the lines given by kx² + 4xy – y² = 0 exceeds the slope of the other by 8.
Solution:
Comparing the equation kx² + 4xy – y² = 0 with ² + 2hxy + by² = 0, we get, a = k, 2h = 4, b = -1. Let m₁ and m₂ be the slopes of the lines represented by kx² + 4xy – y² = 0.
∴ m₁ + m₂ = \(\frac{-2 h}{b}=\frac{-4}{-1}\) = 4
and m₁m₂ = \(\frac{a}{b}=\frac{k}{-1}\) = -k
We are given that m₂ = m₁ + 8
m₁ + m₁ + 8 = 4
∴ 2m₁ = -4 ∴ m₁ = -2 … (1)
Also, m₁(m₁ + 8) = -k
(-2)(-2 + 8) = -k … [By(1)]
∴ (-2)(6) = -k
∴ -12= -k ∴ k = 12.

Question 5.
Find the condition that :
(i) the line 4x + 5y = 0 coincides with one of the lines given by ax² + 2hxy + by² = 0.
Solution:
The auxiliary equation of the lines represented by ax² + 2hxy + by² = 0 is bm² + 2hm + a = 0.
Given that 4x + 5y = 0 is one of the lines represented by ax² + 2hxy + by² = 0.
The slope of the line 4x + 5y = 0 is \(-\frac{4}{5}\).
∴ m = \(-\frac{4}{5}\) is a root of the auxiliary equation bm² + 2hm + a = 0.
∴ b\(\left(-\frac{4}{5}\right)^{2}\) + 2h\(\left(-\frac{4}{5}\right)\) + a = 0
∴ \(\frac{16 b}{25}-\frac{8 h}{5}\) + a = 0
∴ 16b – 40h + 25a = 0
∴ 25a + 16b = 40k.
This is the required condition.

(ii) the line 3x + y = 0 may be perpendicular to one of the lines given by ax² + 2hxy + by² = 0.
Solution:
The auxiliary equation of the lines represented by ax² + 2hxy + by² = 0 is bm² + 2hm + a = 0.
Since one line is perpendicular to the line 3x + y = 0
whose slope is \(-\frac{3}{1}\) = -3
∴ slope of that line = m = \(\frac{1}{3}\)
∴ m = \(\frac{1}{3}\)is the root of the auxiliary equation bm² + 2hm + a = 0.
∴ b\(\left(\frac{1}{3}\right)^{2}\) + 2h\(\left(\frac{1}{3}\right)\) + a = 0
∴ \(\frac{b}{9}+\frac{2 h}{3}\) + a = 0
∴ b + 6h + 9a = 0
∴ 9a + b + 6h = 0
This is the required condition.

Question 6.
If one of the lines given by ax² + 2hxy + by² = 0 is perpendicular to px + qy = 0 then show that ap² + 2hpq + bq² = 0.
Solution:
To prove ap² + 2hpq + bq² = 0.
Let the slope of the pair of straight lines ax² + 2hxy + by² = 0 be m₁ and m₂
Then, m₁ + m₂ = \(\frac{-2 h}{b}\) and m₁m₂ = \(\frac{a}{b}\)
Slope of the line px + qy = 0 is \(\frac{-p}{q}\)
But one of the lines of ax² + 2hxy + by² = 0 is perpendicular to px + qy = 0

⇒ bq² + ap² = -2hpq
⇒ ap² + 2hpq + bq² = 0

Question 7.
Find the combined equation of the pair of lines passing through the origin and making an equilateral triangle with the line y = 3.
Solution:
Let OA and OB be the lines through the origin making.an angle of 60° with the line y = 3.
∴ OA and OB make an angle of 60° and 120° with the positive direction of X-axis.
∴ slope of OA = tan60° = \(\sqrt {3}\)
∴ equation of the line OA is
y = \(\sqrt {3}\) x, i.e. \(\sqrt {3}\) x – y = 0

Slope of OB = tan 120° = tan (180° – 60°)
= -tan 60°= –\(\sqrt {3}\)
∴ equation of the line OB is
y = –\(\sqrt {3}\) x, i.e. \(\sqrt {3}\) x + y = 0
∴ required joint equation of the lines is
(\(\sqrt {3}\) x – y)(\(\sqrt {3}\) x + y) = 0
i.e. 3x² – y² = 0.

Question 8.
If slope of one of the lines given by ax² + 2hxy + by² = 0 is four times the other then show that 16h² = 25ab.
Solution:
Let m₁ and m₂ be the slopes of the lines given by ax² + 2hxy + by² = 0.
∴ m₁ + m₂ = \(-\frac{2 h}{b}\)
and m₁m₂ = \(\frac{a}{b}\)
We are given that m₂ = 4m₁

∴ 16h² = 25ab
This is the required condition.

Question 9.
If one of the lines given by ax² + 2hxy + by² = 0 bisects an angle between co-ordinate axes then show that (a + b) ² = 4h².
Solution:
The auxiliary equation of the lines given by ax² + 2hxy + by² = 0 is bm² + 2hm + a = 0.
Since one of the line bisects an angle between the coordinate axes, that line makes an angle of 45° or 135° with the positive direction of X-axis.
∴ slope of that line = tan45° or tan 135°
∴ m = tan45° = 1
or m = tan 135° = tan (180° – 45°)
= -tan 45°= -1
∴ m = ±1 are the roots of the auxiliary equation bm² + 2hm + a = 0.
∴ b(±1)² + 2h(±1) + a = 0
∴ b ± 2h + a = 0
∴ a + b = ±2h
∴ (a + b)² = 4h²
This is the required condition.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Trigonometric Functions Miscellaneous Exercise 3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 3 Trigonometric Functions Miscellaneous Exercise 3

I) Select the correct option from the given alternatives.
Question 1.
The principal of solutions equation sinθ = \(\frac{-1}{2}\) are ________.

Solution:
(b) \(\frac{7 \pi}{6}, \frac{11 \pi}{6}\)

Question 2.
The principal solution of equation cot θ = \(\sqrt {3}\) ___________.

Solution:
(a) \(\frac{\pi}{6}, \frac{7 \pi}{6}\)

Question 3.
The general solution of sec x = \(\sqrt {2}\) is __________.
(a) 2nπ ± \(\frac{\pi}{4}\), n ∈ Z
(b) 2nπ ± \(\frac{\pi}{2}\), n ∈ Z
(c) nπ ± \(\frac{\pi}{2}\), n ∈ Z
(d) 2nπ ± \(\frac{\pi}{3}\), n ∈ Z
Solution:
(a) 2nπ ± \(\frac{\pi}{4}\), n ∈ Z

Question 4.
If cos pθ = cosqθ, p ≠ q rhen ________.
(a) θ = \(\frac{2 n \pi}{p \pm q}\)
(b) θ = 2nπ
(c) θ = 2nπ ± p
(d) nπ ± q
Solution:
(a) θ = \(\frac{2 n \pi}{p \pm q}\)

Question 5.
If polar co-ordinates of a point are \(\left(2, \frac{\pi}{4}\right)\) then its cartesian co-ordinates are ______.
(a) (2, \(\sqrt {2}\) )
(b) (\(\sqrt {2}\), 2)
(c) (2, 2)
(d) (\(\sqrt {2}\) , \(\sqrt {2}\))
Solution:
(d) (\(\sqrt {2}\) , \(\sqrt {2}\))

Question 6.
If \(\sqrt {3}\) cosx – sin x = 1, then general value of x is _________.
(a) 2nπ ± \(\frac{\pi}{3}\)
(b) 2nπ ± \(\frac{\pi}{6}\)
(c) 2nπ ± \(\frac{\pi}{3}-\frac{\pi}{6}\)
(d) nπ + (-1)ⁿ\(\frac{\pi}{3}\)
Solution:
(c) 2nπ ± \(\frac{\pi}{3}-\frac{\pi}{6}\)

Question 7.
In ∆ABC if ∠A = 45°, ∠B = 60° then the ratio of its sides are _________.
(a) 2 : \(\frac{\pi}{2}\) : \(\frac{\pi}{3}\) + 1
(b) \(\frac{\pi}{2}\) : 2 : \(\frac{\pi}{3}\) + 1
(c) 2 \(\frac{\pi}{2}\) : \(\frac{\pi}{2}\) : \(\frac{\pi}{3}\)
(d) 2 : 2 \(\frac{\pi}{2}\) : \(\frac{\pi}{3}\) + 1
Solution:
(a) 2 : \(\frac{\pi}{2}\) : \(\frac{\pi}{3}\) + 1

Question 8.
In ∆ABC, if c² + a² – b² = ac, then ∠B = __________.
(a) \(\frac{\pi}{4}\)
(b) \(\frac{\pi}{3}\)
(c) \(\frac{\pi}{2}\)
(d) \(\frac{\pi}{6}\)
Solution:
(b) \(\frac{\pi}{3}\)

Question 9.
In ABC, ac cos B – bc cos A = ____________.
(a) a² – b²
(b) b² – c²
(c) c² – a²
(d) a² – b² – c²
Solution:
(a) a² – b²

Question 10.
If in a triangle, the are in A.P. and b : c = \(\sqrt {3}\) : \(\sqrt {2}\) then A is equal to __________.
(a) 30°
(b) 60°
(c) 75°
(d) 45°
Solution:
(c) 75°

Question 11.
cos^-1\(\left(\cos \frac{7 \pi}{6}\right)\) = ________.

Question 12.
The value of cot (tan^-1 2x + cot^-1 2x) is __________.
(a) 0
(b) 2x
(c) π + 2x
(d) π – 2x
Solution:
(a) 0

Question 13.
The principal value of sin^-1\(\left(-\frac{\sqrt{3}}{2}\right)\) is ____________.

Solution:
(d) \(-\frac{\pi}{3}\)

Question 14.
If sin^-1\(\frac{4}{5}\) + cos^-1\(\frac{,12}{13}\) = sin^-1 ∝, then ∝ = _____________.
(a) \(\frac{63}{65}\)
(b) \(\frac{62}{65}\)
(c) \(\frac{61}{65}\)
(d) \(\frac{60}{65}\)
Solution:
(a) \(\frac{63}{65}\)

Question 15.
If tan^-1(2x) + tan^-1(3x) = \(\frac{\pi}{4}\), then x = ________.
(a) -1
(b) \(\frac{1}{6}\)
(c) \(\frac{2}{6}\)
(d) \(\frac{3}{2}\)
Solution:
(b) \(\frac{1}{6}\)

Question 16.
2 tan^-1\(\frac{1}{3}\) + tan^-1\(\frac{1}{7}\) = ______.
(a) tan^-1\(\frac{4}{5}\)
(b) \(\frac{\pi}{2}\)
(c) 1
(d) \(\frac{\pi}{4}\)
Solution:
(d) \(\frac{\pi}{4}\)

Question 17.
tan (2 tan^-1\(\left(\frac{1}{5}\right)-\frac{\pi}{4}\)) = ______.
(a) \(\frac{17}{7}\)
(b) \(-\frac{17}{7}\)
(c) \(\frac{7}{17}\)
(d) \(-\frac{7}{17}\)
Solution:
(d) \(-\frac{7}{17}\)

Question 18.
The principal value branch of sec^-1 x is __________.

Solution:
(b) [0, π] – {\(\frac{\pi}{2}\)}

Question 19.
cos[tan^-1\(\frac{1}{3}\) + tan^-1\(\frac{1}{2}\)] = ________.
(a) \(\frac{1}{\sqrt{2}}\)
(b) \(\frac{\sqrt{3}}{2}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{\pi}{4}\)
Solution:
(a) \(\frac{1}{\sqrt{2}}\)

Question 20.
If tan θ + tan 2θ + tan 3θ = tan θ∙tan 2θ∙tan 3θ, then the general value of the θ is _______.
(a) nπ
(b) \(\frac{n \pi}{6}\)
(c) nπ ± \(\frac{n \pi}{4}\)
(d) \(\frac{n \pi}{2}\)
Solution:
(b) \(\frac{n \pi}{6}\)
[Hint: tan(A + B + C) = \(\frac{\tan A+\tan B+\tan C-\tan A \cdot \tan B \cdot \tan C}{1-\tan A \cdot \tan B-\tan B \cdot \tan C-\tan C \cdot \tan A}\)
Since , tan θ + tan 2θ + tan 3θ = tan θ ∙ tan 2θ ∙ tan 3θ,
we get, tan (θ + 2θ + 3θ) = θ
∴ tan6θ = 0
∴ 6θ = nπ, θ = \(\frac{n \pi}{6}\).]

Question 21.
If any ∆ABC, if a cos B = b cos A, then the triangle is ________.
(a) Equilateral triangle
(b) Isosceles triangle
(c) Scalene
(d) Right angled
Solution:
(b) Isosceles triangle

II: Solve the following
Question 1.
Find the principal solutions of the following equations :
(i) sin2θ = \(-\frac{1}{2}\)
Solution:
sin2θ = \(-\frac{1}{2}\)
Since, θ ∈ (0, 2π), 2∈ ∈ (0, 4π)

(ii) tan3θ = -1
Solution:
Since, θ ∈ (0, 2π), 3∈ ∈ (0, 6π)

… [∵ tan(π – θ) = tan(2π – θ) = tan(3π – θ)
= tan (4π – θ) = tan (5π – θ) = tan (6π – θ) = -tan θ]
∴ tan3θ = tan\(\frac{3 \pi}{4}\) = tan\(\frac{7 \pi}{4}\) = tan\(\frac{11 \pi}{4}\) = tan\(\frac{15 \pi}{4}\)

(iii) cotθ = 0
Solution:
cotθ = 0
Since θ ∈ (0, 2π),
cotθ = 0 = cot \(\frac{\pi}{2}\) = cot (π + \(\frac{\pi}{2}\) …[∵ cos(π + θ) = cotθ]
∴ cotθ = cot\(\frac{\pi}{2}\) = cot\(\frac{3 \pi}{2}\)
∴ θ = \(\frac{\pi}{2}\) or θ = \(\frac{3 \pi}{2}\)
Hence, the required principal solutions are \(\left\{\frac{\pi}{2}, \frac{3 \pi}{2}\right\}\)

Question 2.
Find the principal solutions of the following equations :
(i) sin2θ = \(-\frac{1}{\sqrt{2}}\)
Solution:

(ii) tan5θ = -1
Solution:

(iii) cot2θ = 0
Solution:

Question 3.
Which of the following equations have no solutions ?
(i) cos 2θ = \(\frac{1}{3}\)
Solution:
cos 2θ = \(\frac{1}{3}\)
Since \(\frac{1}{3}\) ≤ cosθ ≤ 1 for any θ
cos2θ = \(\frac{1}{3}\) has solution

(ii) cos² θ = -1
Solution:
cos2θ = -1
This is not possible because cos2θ ≥ 0 for any θ.
∴ cos2θ = -1 does not have any solution.

(iii) 2 sinθ = 3
Solution:
2 sin θ = 3 ∴ sin θ = \(\frac{3}{2}\)
This is not possible because -1 ≤ sin θ ≤ 1 for any θ.
∴ 2 sin θ = 3 does not have any solution.

(iv) 3 sin θ = 5
Solution:
3 sin θ = 5
∴ sin θ = \(\frac{5}{3}\)
This is not possible because -1 ≤ sin θ ≤ 1 for any θ.
∴ 3 sin θ = 5 does not have any solution.

Question 4.
Find the general solutions of the following equations :
(i) tanθ = \(-\sqrt {x}\)
Solution:
The general solution of tan θ = tan ∝ is
θ = nπ + ∝, n ∈ Z.
Now, tanθ = \(-\sqrt {x}\)
∴ tanθ = tan\(\frac{\pi}{3}\) …[∵ tan\(\frac{\pi}{3}\) = \(\sqrt {3}\)]
∴ tanθ = tan\(\left(\pi-\frac{\pi}{3}\right)\) …[∵ tan(π – θ) = -tanθ]
∴ tanθ = tan\(\frac{2 \pi}{3}\)
∴ the required general solution is
θ = nπ + \(\frac{2 \pi}{3}\), n ∈ Z.

(ii) tan²θ = 3
Solution:
The general solution of tan²θ = tan²∝ is
θ = nπ ± ∝, n ∈ Z.
Now, tan²θ = 3 = (\(\sqrt {x}\))²
∴ tan²θ = (tan\(\frac{\pi}{3}\))² …[∵ tan\(\frac{\pi}{3}\) = \(\sqrt {3}\)]
∴ tan²θ = tan²\(\frac{\pi}{3}\)
∴ the required general solution is
θ = nπ ± \(\frac{\pi}{3}\), n ∈ Z.

(iii) sin θ – cosθ = 1
Solution:
∴ cosθ – sin θ = -1

(iv) sin²θ – cos²θ = 1
Solution:
sin²θ – cos²θ = 1
∴ cos²θ – sin²θ = -1
∴ cos2θ = cosπ …(1)
The general solution of cos θ = cos ∝ is
θ = 2nπ ± ∝, n ∈ Z
∴ the general solution of (1) is given by
2θ = 2nπ ± π, n ∈ Z
∴ θ = nπ ± \(\frac{\pi}{2}\), n ∈ Z

Question 5.
In ∆ABC prove that cos \(\left(\frac{A-B}{2}\right)=\left(\frac{a+b}{c}\right)\) sin \(\frac{C}{2}\)
Solution:
By the sine rule,

Question 6.
With usual notations prove that \(\frac{\sin (A-B)}{\sin (A+B)}=\frac{a^{2}-b^{2}}{c^{2}}\).
Solution:
By the sine rule,
\(\frac{a}{\sin \mathrm{A}}\) = \(\frac{b}{\sin \mathrm{B}}\) = \(\frac{c}{\sin \mathrm{C}}\) = k
∴ a = ksinA, b = ksinB, c = ksinC

Question 7.
In ∆ABC prove that (a – b)² 2cos²\(\frac{\mathrm{C}}{2}\) + (a + b)² sin²\(\frac{\mathrm{C}}{2}\) = c².
Solution:
LHS (a – b)² 2cos²\(\frac{\mathrm{C}}{2}\) + (a + b)² sin²\(\frac{\mathrm{C}}{2}\)
= (a² + b² – 2ab) cos²\(\frac{\mathrm{C}}{2}\) + (a² + b² + 2ab) sin\(\frac{\mathrm{C}}{2}\)²
= (a² + b²) cos²\(\frac{\mathrm{C}}{2}\) – 2ab cos²\(\frac{\mathrm{C}}{2}\) + (a² + b²) sin²\(\frac{\mathrm{C}}{2}\) + 2ab sin²\(\frac{\mathrm{C}}{2}\)
= (a² + b²) (cos²\(\frac{\mathrm{C}}{2}\) + sin²\(\frac{\mathrm{C}}{2}\)) – 2ab(cos²\(\frac{\mathrm{C}}{2}\) – sin²\(\frac{\mathrm{C}}{2}\))
= a² + b² – 2ab cos C
= c² = RHS.

Question 8.
In ∆ABC if cosA = sin B – cos C then show that it is a right angled triangle.
Solution:
cos A= sin B – cos C
∴ cos A + cos C = sin B

∴ A – C = B
∴ A = B + C
∴ A + B + C = 180° gives
A + A = 180°
∴ 2A = 180 ∴ A = 90°
∴ ∆ ABC is a rightangled triangle.

Question 9.
If \(\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}\) then show that a², b², c², are in A.P.
Solution:
By sine rule,
\(\frac{\sin \mathrm{A}}{a}\) = \(\frac{\sin \mathrm{B}}{b}\) = \(\frac{\sin \mathrm{C}}{c}\) = k
∴ sin A = ka, sin B = kb,sin C = kc
Now, \(\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}\)
∴ sinA∙sin(B – C) = sinC∙sin(A -B)
∴ sin [π – (B + C)] ∙ sin (B – C)
= sin [π – (A + B)]∙sin (A – B) … [∵ A + B + C = π]
∴ sin(B + C) ∙ sin(B – C) = sin (A + B) ∙ sin (A – B)
∴ sin²B – sin²C = sin²A – sin²B
∴ 2 sin²B = sin²A + sin²C
∴ 2k²b² = k²a² + k²c²
∴ 2b² = a² + c²
Hence, a², b², c² are in A.P.

Question 10.
Solve the triangle in which a = (\(\sqrt {3}\) + 1), b = (\(\sqrt {3}\) – 1) and ∠C = 60°.
Solution:
Given : a = \(\sqrt {3}\) + 1, b = \(\sqrt {3}\) – 1 and ∠C = 60°.
By cosine rule,
c² = a² + b² – 2ab cos C
= (\(\sqrt {3}\) + 1)² + (\(\sqrt {3}\) – 1)² – 2(\(\sqrt {3}\) + 1)(\(\sqrt {3}\) – 1)cos60°
= 3 + 1 + 2\(\sqrt {3}\) + 3+ 1 – 2\(\sqrt {3}\) – 2(3 – 1)\(\left(\frac{1}{2}\right)\)
= 8 – 2 = 6
∴ c = \(\sqrt {6}\) …[∵ c > 0)
By sine rule,

∴ sin A = sin 60° cos 45° + cos 60° sin 45°
and sin B = sin 60° cos 45° – cos 60° sin 45°
∴ sin A = sin (60° + 45°) – sin 105°
and sin B = sin (60° – 45°) = sin 15°
∴ A = 105° and B = 15°
Hence, A = 105°, B 15° and C = \(\sqrt {6}\) units.

Question 11.
In ∆ABC prove the following :
(i) a sin A – b sin B = c sin (A – B)
Solution:
By sine rule,
\(\frac{a}{\sin \mathrm{A}}\) = \(\frac{b}{\sin \mathrm{B}}\) = \(\frac{c}{\sin \mathrm{C}}\) = k
∴ a = ksinA, b = ksinB, c = ksinC,
LHS = a sin A – b sinB
= ksinA∙sinA – ksinB∙sinB
= k (sin²A – sin²B)
= k (sin A + sin B)(sin A – sin B)

= k × sin (A + B) × sin (A – B)
= ksin(π – C)∙sin(A – B) … [∵ A + B + C = π]
= k sinC∙sin (A – B)
= c sin (A – B) = RHS.

(ii) \(\frac{c-b \cos A}{b-c \cos A}=\frac{\cos B}{\cos C}\).
Solution:

(iii) a² sin (B – C) = (b² – c²) sinA
Solution:
By sine rule,
\(\frac{a}{\sin \mathrm{A}}\) = \(\frac{b}{\sin \mathrm{B}}\) = \(\frac{c}{\sin \mathrm{C}}\) = k
∴ a = ksinA, b = ksinB, c = ksinC
RHS = (b² – c²) sin A
= (k²sin²B – k²sin²C)sin A
= k²(sin²B – sin²C) sin A
= k²(sin B + sin C)(sin B – sin C) sin A

= k² × sin (B + C) × sin (B – C) × sin A
= k²∙sin(π – A)∙sin(B – C)∙sinA … [∵ A + B + C = π]
= k²sin A∙sin (B – C)∙sin A
= (k sin A)²∙sin (B – C)
= a²sin (B – C) = LHS.

(iv) ac cos B – bc cos A = (a² – b²).
Solution:
LHS = ac cos B – bc cos A
= ac\(\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)\) – bc\(\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)\)
=\(\frac{1}{2}\)(c² + a² – b²) – \(\frac{1}{2}\)(b² + c² – a²)
= \(\frac{1}{2}\)(c² + a² – b² – b² – c² + a²)
= \(\frac{1}{2}\)(2a² – 2b²) = a² – b2 = RHS.

(v) \(\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{a^{2}+b^{2}+c^{2}}{2 a b c}\) .
Solution:

(vi) \(\frac{\cos 2 \mathrm{~A}}{a^{2}}-\frac{\cos 2 \mathrm{~B}}{b^{2}}=\frac{1}{a^{2}}-\frac{1}{b^{2}}\).
Solution:
By sine rule,
\(\frac{\sin \mathrm{A}}{a}=\frac{\sin \mathrm{B}}{b}\)

(vii) \(\frac{b-c}{a}=\frac{\tan \frac{B}{2}-\tan \frac{C}{2}}{\tan \frac{B}{2}+\tan \frac{C}{2}}\)
Solution:
By sine rule,

Question 12.
In ∆ABC if a², b², c², are in A.P. then cot\(\frac{A}{2}\), cot\(\frac{B}{2}\), cot\(\frac{C}{2}\) are also in A.P.
Question is modified
In ∆ABC if a, b, c, are in A.P. then cot\(\frac{A}{2}\), cot\(\frac{B}{2}\), cot\(\frac{C}{2}\) are also in A.P.
Solution:
a, b, c, are in A.P.
∴ 2b = a + c …(1)

Question 13.
In ∆ABC if ∠C = 90º then prove that sin(A – B) = \(\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\)
Solution:
In ∆ABC, if ∠C = 90º
∴ c² = a² + b² …(1)
By sine rule,

Question 14.
In ∆ABC if \(\frac{\cos A}{a}=\frac{\cos B}{b}\), then show that it is an isosceles triangle.
Solution:
Given : \(\frac{\cos A}{a}=\frac{\cos B}{b}\) ….(1)
By sine rule,

∴ sin A cos B = cos A sinB
∴ sinA cosB – cosA sinB = 0
∴ sin (A – B) = 0 = sin0
∴ A – B = 0 ∴ A = B
∴ the triangle is an isosceles triangle.

Question 15.
In ∆ABC if sin²A + sin²B = sin²C then prove that the triangle is a right angled triangle.
Question is modified
In ∆ABC if sin²A + sin²B = sin²C then show that the triangle is a right angled triangle.
Solution:
By sine rule,
\(\frac{\sin \mathrm{A}}{a}\) = \(\frac{\sin \mathrm{B}}{b}\) = \(\frac{\sin \mathrm{C}}{c}\) = k
∴ sin A = ka, sinB = kb, sin C = kc
∴ sin²A + sin²B = sin²C
∴ k²a² + k²b² = k²c²
∴ a² + b² = c²
∴ ∆ABC is a rightangled triangle, rightangled at C.

Question 16.
In ∆ABC prove that a²(cos²B – cos²C) + b²(cos²C – cos²A) + c²(cos²A – cos²B) = 0.
Solution:
By sine rule,
\(\frac{a}{\sin \mathrm{A}}\) = \(\frac{b}{\sin \mathrm{B}}\) = \(\frac{c}{\sin \mathrm{C}}\) = k
LHS = a²(cos²B – cos²C) + b²( cos²C – cos²A) + c²(cos²A – cos²B)
= k²sin²A [(1 – sin²B) – (1 – sin²C)] + k²sin²B [(1 – sin²C) – (1 – sin²A)] + k²sin²C[(1 – sin²A) – (1 – sin²B)]
= k²sin²A (sin²C – sin²B) + k²sin²B(sin²A – sin²C) + k²sin²C (sin²B – sin²A)
= k²(sin²A sin²C – sin²Asin²B + sin²A sin²B – sin²B sin²C + sin²B sin²C – sin²A sin²C)
= k²(0) = 0 = RHS.

Question 17.
With usual notations show that (c² – a² + b²) tan A = (a² – b² + c²) tan B = (b² – c² + a²) tan C.
Solution:
By sine rule,
\(\frac{a}{\sin A}\) = \(\frac{b}{\sin B}\) = \(\frac{c}{\sin C}\) = k
∴ a = fksinA, b = ksinB, c = ksinC


From (1), (2) and (3), we get
(c² – a² + b²) tan A = (a² – b² + c²) tan B
= (b² – c² + a²) tan C.

Question 18.
In ∆ABC, if a cos²\(\frac{C}{2}\) + c cos²\(\frac{A}{2}\) = \(\frac{3 b}{2}\), then prove that a , b ,c are in A.P.
Solution:
a cos²\(\frac{C}{2}\) + c cos²\(\frac{A}{2}\) = \(\frac{3 b}{2}\)

∴ a + c + b = 3b …[∵ a cos C + c cos A = b]
∴ a + c = 2b
Hence, a, b, c are in A.P.

Question 19.
Show that 2 sin^-1\(\left(\frac{3}{5}\right)\) = tan^-1\(\left(\frac{24}{7}\right)\).
Solution:
Let sin²\(\left(\frac{3}{5}\right)\) = x.



∴ tan^-1\(\left(\frac{24}{7}\right)\) = RHS

Question 20.
Show that tan^-1\(\left(\frac{1}{5}\right)\) + tan^-1\(\left(\frac{1}{7}\right)\) + tan^-1\(\left(\frac{1}{3}\right)\) + tan^-1\(\left(\frac{1}{8}\right)\) = \(\frac{\pi}{4}\).
Solution:

Question 21.
Prove that tan^-1\(\sqrt {x}\) = \(\frac{1}{2}\) cos^-1\(\left(\frac{1-x}{1+x}\right)\), if x ∈ [0, 1].
Solution:
Let tan^-1\(\sqrt {x}\) = y
∴ tan y = \(\sqrt {x}\) ∴ x = tan²y

Question 22.
Show that \(\frac{9 \pi}{8}-\frac{9}{4}\) sin^-1\(\frac{1}{3}\) = \(\frac{9}{4}\) sin^-1\(\frac{2 \sqrt{2}}{3}\).
Question is modified
Show that \(\frac{9 \pi}{8}-\frac{9}{4}\) sin^-1\(\left(\frac{1}{3}\right)\) = \(\frac{9}{4}\) sin^-1\(\left(\frac{2 \sqrt{2}}{3}\right)\).
Solution:
We have to show that

Question 23.
Show that

Solution:

Question 24.
If sin(sin^-1\(\frac{1}{5}\) + cos^-1x) = 1, then find the value of x.
Solution:
sin(sin^-1\(\frac{1}{5}\) = 1

Question 25.
If tan^-1\(\left(\frac{x-1}{x-2}\right)\) + tan^-1\(\left(\frac{x+1}{x+2}\right)\) = \(\frac{\pi}{4}\) then find the value of x.
Solution:
tan^-1\(\left(\frac{x-1}{x-2}\right)\) + tan^-1\(\left(\frac{x+1}{x+2}\right)\) = \(\frac{\pi}{4}\)

∴ x = ±\(\frac{1}{\sqrt{2}}\).

Question 26.
If 2 tan^-1(cos x ) = tan^-1(cosec x) then find the value of x.
Solution:
2 tan^-1(cos x ) = tan^-1(cosec x)

Question 27.
Solve: tan^-1\(\left(\frac{1-x}{1+x}\right)\) = \(\frac{1}{2}\)(tan^-1x), for x > 0.
Solution:
tan^-1\(\left(\frac{1-x}{1+x}\right)\) = \(\frac{1}{2}\)(tan^-1x)

Question 28.
If sin^-1(1 – x) – 2sin^-1x = \(\frac{\pi}{2}\), then find the value of x.
Solution:
sin^-1(1 – x) – 2sin^-1x = \(\frac{\pi}{2}\)

Question 29.
If tan^-12x + tan^-13x = \(\frac{\pi}{4}\), then find the value of x.
Question is modified
If tan^-12x + tan^-13x = \(\frac{\pi}{2}\), then find the value of x.
Solution:
tan^-12x + tan^-13x = \(\frac{\pi}{4}\)
∴ tan^-1\(\left(\frac{2 x+3 x}{1-2 x \times 3 x}\right)\) = tan\(\frac{\pi}{4}\), where 2x > 0, 3x > 0
∴ \(\frac{5 x}{1-6 x^{2}}\) = tan\(\frac{\pi}{4}\) = 1
∴ 5x = 1 – 6x²
∴ 6x² + 5x – 1 = 0
∴ 6x² + 6x – x – 1 = 0
∴ 6x(x +1) – 1(x + 1) = 0
∴ (x + 1)(6x – 1) = 0
∴ x = -1 or x = \(\frac{1}{6}\)
But x > 0 ∴ x ≠ -1
Hence, x = \(\frac{1}{6}\)

Question 30.
Show that tan^-1\(\frac{1}{2}\) – tan^-1\(\frac{1}{4}\) = tan^-1\(\frac{2}{9}\).
Solution:
LHS = tan^-1\(\frac{1}{2}\) – tan^-1\(\frac{1}{4}\)

Question 31.
Show that cot^-1\(\frac{1}{3}\) – tan^-1\(\frac{1}{3}\) = cot^-1\(\frac{3}{4}\).
Solution:
LHS = cot^-1\(\frac{1}{3}\) – tan^-1\(\frac{1}{3}\)

Question 32.
Show that tan^-1\(\frac{1}{2}\) = \(\frac{1}{3}\) tan^-1\(\frac{11}{2}\).
Solution:
We have to show that

Question 33.
Show that cos^-1\(\frac{\sqrt{3}}{2}\) + 2sin^-1\(\frac{\sqrt{3}}{2}\) = \(\frac{5 \pi}{6}\)
Solution:

Question 34.
Show that 2cot^-1\(\frac{3}{2}\) + sec^-1\(\frac{13}{12}\) = \(\frac{\pi}{2}\)
Solution:

Question 35.
Prove the following :
(i) cos^-1 x = tan^-1\(\frac{\sqrt{1-x^{2}}}{x}\), if x < 0.
Question is modified
cos^-1 x = tan^-1\(\left(\frac{\sqrt{1-x^{2}}}{x}\right)\), if x > 0.
Solution:

(ii) cos^-1 x = π + tan^-1\(\frac{\sqrt{1-x^{2}}}{x}\), if x < 0.
Solution:

Question 36.
If |x| < 1 , then prove that 2tan^-1 x = tan^-1\(\frac{2 x}{1-x^{2}}\) = sin^-1\(\frac{2 x}{1+x^{2}}\) = cos^-1\(\frac{1-x^{2}}{1+x^{2}}\)
Question is modified
If |x| < 1 , then prove that 2tan^-1 x = tan^-1\(\left(\frac{2 x}{1-x^{2}}\right)\) = sin^-1\(\left(\frac{2 x}{1+x^{2}}\right)\) = cos^-1\(\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Solution:
Let tan^-1x = y
Then, x = tany

Question 37.
If x, y, z, are positive then prove that tan^-1\(\frac{x-y}{1+x y}\) + tan^-1\(\frac{y-z}{1+y z}\) + tan^-1\(\frac{z-x}{1+z x}\) = 0
Solution:

Question 38.
If tan^-1 x + tan^-1 y + tan^-1 z = \(\frac{\pi}{2}\) then, show that xy + yz + zx = 1
Solution:
tan^-1 x + tan^-1 y + tan^-1 z = \(\frac{\pi}{2}\)

∴ 1 – xy – yz – zx = 0
∴ xy + yz + zx = 1.

Question 39.
If cos^-1 x + cos^-1 y + cos^-1 z = π then show that x² + y² + z² + 2xyz = 1.
Solution:
0 ≤ cos^-1x ≤ π and
cos^-1x + cos^-1y+ cos^-1z = 3π
∴ cos^-1x = π, cos^-1y = π and cos^-1z = π
∴ x = y = z = cosπ = -1
∴ x² + y² + z² + 2xyz
= (-1)² + (-1)² + (-1)² + 2(-1)(-1)(-1)
= 1 + 1 + 1 – 2
= 3 – 2 = 1.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Trigonometric Functions Ex 3.3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 3 Trigonometric Functions Ex 3.3

Question 1.
Find the principal values of the following :
(i) sin^-1\(\left(\frac{1}{2}\right)\)
Solution:
The principal value branch of sin^-1x is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
Let sin^-1\(\left(\frac{1}{2}\right)\) = ∝, where \(\frac{-\pi}{2}\) ≤ ∝ ≤ \(\frac{\pi}{2}\)
∴ sin∝ = \(\frac{1}{2}\) = sin\(\frac{\pi}{6}\)
∴ ∝ = \(\frac{\pi}{6}\) …[∵ – \(\frac{\pi}{2}\) ≤ \(\frac{\pi}{6}\) ≤ \(\frac{\pi}{2}\)]
∴ the principal value of sin^-1\(\left(\frac{1}{2}\right)\) is \(\frac{\pi}{6}\).

(ii) cosec^-1(2)
Solution:
The principal value branch of cosec^-1x is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) – {0}.
Let cosec^-1(2) = ∝, where \(\frac{-\pi}{2}\) ≤ ∝ ≤ \(\frac{\pi}{2}\), ∝ ≠ 0
∴ cosec^-1 ∝ = 2 = cosec\(\frac{\pi}{6}\)
∴ ∝ = \(\frac{\pi}{6}\) …[∵ –\(\frac{\pi}{2}\) ≤ \(\frac{\pi}{6}\) ≤ \(\frac{\pi}{2}\)]
∴ the principal value of cosec^-1(2) is \(\frac{\pi}{6}\).

(iii) tan^-1(-1)
Solution:
The principal value branch of tan^-1x is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)
Let tan^-1(-1) = ∝, where \(\frac{-\pi}{2}\) < ∝ < \(\frac{\pi}{2}\)
∴ tan∝ = -1 = -tan\(\frac{\pi}{4}\)
∴ tan∝ = tan\(\left(-\frac{\pi}{4}\right)\) …[∵ tan(-θ) = -tanθ]
∴ ∝ = –\(\frac{\pi}{4}\) …[∵ –\(\frac{\pi}{2}\) < \(\frac{-\pi}{4}\) < \(\frac{\pi}{2}\)]
∴ the principal value of tan^-1(-1) is –\(\frac{\pi}{4}\).

(iv) tan^-1(-\(\sqrt {3}\))
Solution:
The principal value branch of tan^-1x is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).
Let tan^-1(-\(\sqrt {3}\)) = ∝, where \(\frac{-\pi}{2}\) < ∝ < \(\frac{\pi}{2}\)
∴ tan∝ = –\(\sqrt {3}\) = -tan\(\frac{\pi}{3}\)
∴ tan∝ = tan\(\left(-\frac{\pi}{3}\right)\) …[∵ tan(-θ) = -tanθ]
∴ ∝ = –\(\frac{\pi}{3}\) …[∵ –\(\frac{\pi}{2}\) < \(\frac{-\pi}{3}\) < \(\frac{\pi}{2}\)]
∴ the principal value of tan^-1(-\(\sqrt {3}\)) is –\(\frac{\pi}{3}\).

(v) sin^-1 \(\left(\frac{1}{\sqrt{2}}\right)\)
Solution:
The principal value branch of sin^-1x is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
Let sin^-1 \(\left(\frac{1}{\sqrt{2}}\right)\) = ∝, where \(\frac{-\pi}{2}\) < ∝ < \(\frac{\pi}{2}\)
∴ sin∝ = \(\left(\frac{1}{\sqrt{2}}\right)\) = sin\(\frac{\pi}{4}\)
∴ ∝ = \(\frac{\pi}{4}\) …[∵ –\(\frac{\pi}{2}\) ≤ \(\frac{\pi}{4}\) ≤ \(\frac{\pi}{2}\)]
∴ the principal value of sin^-1 \(\left(\frac{1}{\sqrt{2}}\right)\) is \(\frac{\pi}{4}\).

(vi) cos^-1\(\left(-\frac{1}{2}\right)\)
Solution:
The principal value branch of cos^-1x is (0, π).
Let cos^-1\(\left(-\frac{1}{2}\right)\) = ∝, where 0 ≤ ∝ ≤ π
∴ cos∝ = \(-\frac{1}{2}\) = -cos\(\frac{\pi}{3}\)
∴ cos∝ = cos\(\left(\pi-\frac{\pi}{3}\right)\) …[∵ cos(π – θ) = -cosθ)
∴ cos∝ = cos\(\frac{2 \pi}{3}\)
∴ ∝ = \(\frac{2 \pi}{3}\) …[∵ 0 ≤ \(\frac{2 \pi}{3}\) ≤ π]
∴ the principal value of cos^-1\(\left(-\frac{1}{2}\right)\) is \(\frac{2 \pi}{3}\).

Question 2.
Evaluate the following :
(i) tan^-1(1) + cos^-1\(\left(\frac{1}{2}\right)\) + sin^-1\(\left(\frac{1}{2}\right)\)
Solution:

(ii) cos^-1\(\left(\frac{1}{2}\right)\) + 2 sin^-1\(\left(\frac{1}{2}\right)\)
Solution:

(iii) tan^-1\(\sqrt {3}\) – sec^-1(-2)
Solution:

∴ tan^-1\(\sqrt {3}\) – sec^-1(-2)
= \(\frac{\pi}{3}-\frac{2 \pi}{3}\) …[By (1) and (2)]
= –\(\frac{\pi}{3}\).

(iv) cosec^-1( \(-\sqrt{2}\)) + cot^-1(\(\sqrt{3}\))
Solution:

Question 3.
Prove the following :
(i) sin^-1\(\left(\frac{1}{\sqrt{2}}\right)\) – 3sin^-1\(\left(\frac{\sqrt{3}}{2}\right)\) = –\(-\frac{3 \pi}{4}\)
Question is modified.
sin^-1\(\left(\frac{1}{\sqrt{2}}\right)\) – 3sin^-1\(\left(\frac{\sqrt{3}}{2}\right)\) = –\(\frac{3 \pi}{4}\)
Solution:

(ii) sin^-1\(\left(-\frac{1}{2}\right)\) + cos^-1\(\left(-\frac{\sqrt{3}}{2}\right)\) = cos^-1\(\left(-\frac{1}{2}\right)\)
Solution:

(iii) sin^-1\(\left(\frac{3}{5}\right)\) + cos^-1\(\left(\frac{12}{13}\right)\) = sin^-1\(\left(\frac{56}{65}\right)\)
Solution:

(iv) cos^-1\(\left(\frac{3}{5}\right)\) + cos^-1\(\left(\frac{4}{5}\right)\) = \(\frac{\pi}{2}\)
Solution:
Let cos^-1\(\left(\frac{3}{5}\right)\) = x
∴ cosx = \(\left(\frac{3}{5}\right)\), where 0 < x < \(\frac{\pi}{2}\) ∴ sinx > 0

(v) tan^-1\(\left(\frac{1}{2}\right)\) + tan^-1\(\left(\frac{1}{3}\right)\) = \(\frac{\pi}{4}\)
Solution:
LHS = tan^-1\(\left(\frac{1}{2}\right)\) + tan^-1\(\left(\frac{1}{3}\right)\)
= tan^-1\(\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}}\right)\)
= tan^-1\(\left(\frac{3+2}{6-1}\right)\) = tan^-1(1)
= tan^-1\(\left(\tan \frac{\pi}{4}\right)\) = \(\frac{\pi}{4}\)
= RHS.

(vi) 2 tan^-1\(\left(\frac{1}{3}\right)\) = tan^-1\(\left(\frac{3}{4}\right)\)
Solution:

(vii) tan^-1\(\left[\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\right]\) = \(\frac{\pi}{4}\) + θ if θ ∈ \(\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)\)
Solution:

(viii) tan^-1\(\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=\frac{\theta}{2}\), if θ ∈ (0, π)
Solution:

= \(\frac{\theta}{2}\) …[∵ tan^-1(tanθ) = θ]
= RHS.