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

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

Question 1.

Find the length of the perpendicular from (2, -3, 1) to the line \(\frac{x+1}{2}=\frac{y-3}{3}=\frac{z+1}{-1}\)

Solution:

Let PM be the perpendicular drawn from the point P (2, -3, 1) to the line \(\frac{x+1}{2}=\frac{y-3}{3}=\frac{z+1}{-1}\) = λ …(Say)

The coordinates of any point on the line are given by x = -1 + 2λ, y = 3 + 3λ, z = -1 – λ

Let the coordinates of M be

(-1 + 2λ, 3 + 3λ, -1 – λ) … (1)

The direction ratios of PM are

-1 + 2λ – 2, 3 + 3λ + 3, -1 – λ – 1

i.e. 2λ – 3, 3λ + 6, -λ – 2

The direction ratios of the given line are 2, 3, -1.

Since PM is perpendicular to the given line, we get

2(2λ – 3) + 3(3λ + 6) – 1(-λ – 2) = 0

∴ 4λ – 6 + 9λ + 18 + λ + 2 = 0

∴ 14λ + 14 = 0 ∴ λ = -1.

Put λ = -1 in (1), the coordinats of M are

(-1 – 2, 3 – 3, -1 + 1) i.e. (-3, 0,0).

∴ length of perpendicular from P to the given line

Alternative Method:

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

Substituting these values in (1), we get

length of perpendicular from P to given line

Question 2.

Find the co-ordinates of the foot of the perpendicular drawn from the point \(2 \hat{i}-\hat{j}+5 \hat{k}\) to the line \(\bar{r}=(11 \hat{i}-2 \hat{j}-8 \hat{k})+\lambda(10 \hat{i}-4 \hat{j}-11 \hat{k})\). Also find the length of the perpendicular.

Solution:

Let M be the foot of perpendicular drawn from the point P (\(2 \hat{i}-\hat{j}+5 \hat{k}\)) on the line

\(\bar{r}=(11 \hat{i}-2 \hat{j}-8 \hat{k})+\lambda(10 \hat{i}-4 \hat{j}-11 \hat{k})\).

Let the position vector of the point M be

Then \(\overline{\mathrm{PM}}\) = Position vector of M – Position vector of P

= [(11 + 10λ)\(\hat{i}\) + (-2 – 4λ)\(\hat{j}\) + -8 – 11λ) \(\hat{k}\)] – (2\(\hat{i}\) – \(\hat{j}\) + 5\(\hat{k}\))

= (9 + 10λ)\(\hat{i}\) + (-1 – 4λ)\(\hat{j}\) + (-13 – 11λ)\(\hat{k}\)

Since PM is perpendicular to the given line which is parallel to \(\bar{b}=10 \hat{i}-4 \hat{j}-11 \hat{k}\),

\(\overline{\mathrm{PM}}\) ⊥^{r}\(\bar{b}\) ∴ \(\overline{\mathrm{PM}} \cdot \bar{b}\) = 0

∴ [(9 + 10λ)\(\hat{i}\) + ( – 1 – 4λ)\(\hat{j}\) + (-13 – 11λ)\(\hat{k}\)]-(10\(\hat{i}\) – 4\(\hat{j}\) – 11\(\hat{k}\)) = 0

∴ 10(9 +10λ) – 4( -1 – 4λ) – 11( -13 – 11λ) = 0

∴ 90 + 100λ + 4 + 16λ + 143 +121λ = 0

∴ 237λ + 237 = 0

∴ λ = -1

Putting this value of λ, we get the position vector of M as \(\hat{i}+2 \hat{j}+3 \hat{k}\).

∴ coordinates of the foot of perpendicular M are (1, 2, 3).

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

Question 3.

Find the shortest distance between the lines \(\bar{r}=(4 \hat{i}-\hat{j})+\lambda(\hat{i}+2 \hat{j}-3 \hat{k})\) and \(\bar{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(\hat{i}+4 \hat{j}-5 \hat{k})\)

Solution:

We know that the shortest distance between the skew lines \(\bar{r}=\overline{a_{1}}+\lambda \overline{b_{1}}\) and \(\bar{r}=\overline{a_{2}}+\mu \overline{b_{2}}\) is given by

Question 4.

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

Solution:

The shortest distance between the lines

= 4(-6 + 2) – 6(7 – 1) + 8(-14 + 6)

= -16 – 36 – 64 = -116

and (m_{1}n_{2} – m_{2}n_{1})^{2} + (l_{2}n_{1} – l_{1}n_{2})^{2} + (l_{1}m_{2} – l_{2}m_{1})^{2}

= (-6 + 2)^{2} + (1 – 7)^{2} + (-14 + 6)^{2}

= 16 + 36 + 64 = 116

Hence, the required shortest distance between the given lines = \(\left|\frac{-116}{\sqrt{116}}\right|\) = \(\sqrt{116}\) = \(2 \sqrt{29}\) units

Question 5.

Find the perpendicular distance of the point (1, 0, 0) from the line \(\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}\) Also find the co-ordinates of the foot of the perpendicular.

Solution:

Let PM be the perpendicular drawn from the point (1, 0, 0) to the line \(\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}\) = λ …(Say)

The coordinates of any point on the line are given by x = -1 + 2λ, y = 3 + 2λ, z = 8 – λ

Let the coordinates of M be

(-1 + 2λ, 3 + 3λ, -1 – λ) …..(1)

The direction ratios of PM are

-1 + 2λ – 2, 3 + 3λ + 3, -1 – λ – 1

i.e. 2λ – 3, 3λ = 6, -λ – 2

The direction ratios of the given line are 2, 3, 8.

Since PM is perpendicular to the given line, we get

2(2λ – 3) + 3(3λ + 6) – 1(-λ – 2) = O

∴ 4λ – 6 + 9λ + 18 + λ + 2 = 0

∴ 14λ + 14 = 0

∴ λ = -1

Put λ in (1), the coordinates of M are

(-1 – 2, 3 – 3, -1 + 1) i.e. (-3, 0, 0).

∴ length of perpendicular from P to the given line

= PM

= \(\sqrt{(-3-2)^{2}+(0+3)^{2}+(0-1)^{2}}\)

= \(\sqrt{(25 + 9 + 1)}\)

= \(\sqrt{35}\)units.

Alternative Method :

We know that the perpendicular distance from the point

Substitutng tese values in (1), w get

length of perpendicular from P to given line

= PM

Question 6.

A(1, 0, 4), B(0, -11, 13), C(2, -3, 1) are three points and D is the foot of the perpendicular from A to BC. Find the co-ordinates of D.

Solution:

Equation of the line passing through the points (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) is

AD is the perpendicular from the point A (1, 0, 4) to the line BC.

The coordinates of any point on the line BC are given by x = 2λ, y = -11 + 8λ, z = 13 – 12λ

Let the coordinates of D be (2λ, -11 + 8λ, 13 – 12λ) … (1)

∴ the direction ratios of AD are

2λ – 1, -1λ + 8λ – 0, 13 – 12λ – 4 i.e.

2λ – 1, -11 + 8λ, 9 – 12λ

The direction ratios of the line BC are 2, 8, -12.

Since AD is perpendicular to BC, we get

2(2λ – 1) + 8(-11 + 8λ) – 12(9 – 12λ) = 0

∴ 42λ – 2 – 88 + 64λ – 108 + 144λ = 0

∴ 212λ – 198 = 0

Question 7.

By computing the shortest distance, determine whether following lines intersect each other.

(i) \(\bar{r}=(\hat{i}-\hat{j})+\lambda(2 \hat{i}+\hat{k})\) and \(\bar{r}=(2 \hat{i}-\hat{j})+\mu(\hat{i}+\hat{j}-\hat{k})\)

Solution:

The shortest distance between the lines

Hence, the given lines do not intersect.

(ii) \(\frac{x-5}{4}=\frac{y-7}{-5}=\frac{z+3}{-5}\) and \(\frac{x-8}{7}=\frac{y-7}{1}=\frac{z-5}{3}\)

Solution:

The shortest distance between the lines

∴ x_{1} = -1, y_{1} = -1, z_{1} = -1, x_{2} = 3, y_{2} = 5, z_{2} = 7,

l_{1} = 7, m_{1} = -6, n_{1} = 1, l_{2} = 1, m_{2} = -2, n_{2} = 1

\(\left|\begin{array}{ccc}

x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\

l_{1} & m_{1} & n_{1} \\

l_{2} & m_{2} & n_{2}

\end{array}\right|\) = \(\left|\begin{array}{ccc}

4 & 6 & 8 \\

4 & -5 & -5 \\

7 & 1 & 3

\end{array}\right|\)

= 4(- 6 + 2) – 6(7 – 1) + 8(-14 + 6)

= -16 – 36 – 64

= -116

and

(m_{1}n_{2} – m_{2}n_{1})^{2} + (l_{2}n_{1} – l_{1}n_{2})^{2} + (l_{1}m_{2} – l_{2}m_{1})^{2}

= (-6 + 2)^{2} + (1 – 7)^{2} + (-14 + 6)^{2}

= 16 + 36 + 64

= 116

Hence, the required shortest distance between the given lines

= \(\left|\frac{-116}{\sqrt{116}}\right|\)

= \(\sqrt{116}\)

=\(2 \sqrt{29}\) units

or

The shortest distance between the lines

= \(\frac{282}{\sqrt{3830}}\)units

Hence, the gives lines do not intersect.

Question 8.

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

Solution:

The lines

∴ x_{1} = 1, y_{1} = -1, z_{1} = 1, x_{2} = 3, y_{2} = k, z_{2} = 0,

l_{1} = 2, m_{1} = 3, n_{1} = 4, l_{2} = 1, m_{2} = 2, n_{2} = 1.

Since these lines intersect, we get

\(\left|\begin{array}{ccc}

2 & k+1 & -1 \\

2 & 3 & 4 \\

1 & 2 & 1

\end{array}\right|\) = 0

∴ 2 (3 – 8) – (k + 1)(2 – 4) – 1 (4 – 3) = 0

∴ -10 + 2(k + 1) – 1 = 0

∴ 2(k + 1) = 11

∴ k + 1 = \(\frac{11}{2}\)

∴ k = \(\frac{9}{2}\)