OP Malhotra Class 12 Maths Solutions Chapter 23 Three Dimensional Geometry Ex 23(d)

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S Chand Class 12 ICSE Maths Solutions Chapter 23 Three Dimensional Geometry Ex 23(d)

Question 1.
Find the angle between the following pairs of lines:
(i) \(\frac{x+4}{3}\) = \(\frac{y-1}{5}\) = \(\frac{y+3}{4}\);
\(\frac{x+1}{1}\) = \(\frac{y-4}{1}\) = \(\frac{z-5}{2}\)
(ii) \(\frac{x-2}{3}\) = \(\frac{y+3}{-2}\), z = 5 ; \(\frac{x+1}{1}\) = \(\frac{2 y-3}{3}\) = \(\frac{z-5}{2}\)
Answer:
(i) Given lines are and
\(\frac{x+4}{3}\) = \(\frac{y-1}{5}\) = \(\frac{z+3}{4}\)
\(\frac{x+1}{1}\) = \(\frac{y-4}{1}\) = \(\frac{z-5}{2}\)
Let \(\vec{b}_1\) and \(\overrightarrow{b_2}\) be the vectors || to line (1) and (2).
Then \(\vec{b}_1\) = 3 \(\hat{i}\) + 5 \(\hat{j}\) + 4 \(\hat{k}\);
\(\overrightarrow{b_2}\) = \(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\)
Let θ be the angle between given line
Then
Then cosθ = \(\frac{\overrightarrow{b_1}\overrightarrow{b_2}}{|\overrightarrow{b_1}||\overrightarrow{b_2}|}\)
= \(\frac{(3 \hat{i}+5 \hat{j}+4 \hat{k}) (\hat{i}+\hat{j}+2 \hat{k})}{\sqrt{9+25+16} \sqrt{1+1+4}}\)
⇒ cosθ = \(\frac{3(1)+5(1)+4(2)}{\sqrt{50} \sqrt{6}}\)
∴ cosθ = \(\frac{16}{10 \sqrt{3}}\) = \(\frac{8}{5 \sqrt{3}}\)
∴ θ = cos^-1 (\(\frac{8}{5 \sqrt{3}}\))

(ii) Given lines are \(\frac{x-2}{3}\)
= \(\frac{y+3}{-2}\)
= \(\frac{z-5}{0}\)
and
\(\frac{x+1}{1}\) = \(\frac{y-\frac{3}{2}}{\frac{3}{2}}\)
= \(\frac{z-5}{2}\) i.e. \(\frac{x+1}{2}\) = \(\frac{y-\frac{3}{2}}{3}\) = \(\frac{z-5}{4}\)
Let \(\vec{b}_1\) and \(\overrightarrow{b_2}\) be the vectors parallel to given lines.
then \(\vec{b}_1\) = 3 \(\hat{i}\) – 2 \(\hat{j}\) + 0 \(\hat{k}\);
\(\overrightarrow{b_2}\) = 2 \(\hat{i}\) + 3 \(\hat{j}\) + 4 \(\hat{k}\)
Let θ be the angle between given lines and hence θ be the angle between \(\overrightarrow{b_1}\) and \(\overrightarrow{b_2}\)
∴ cosθ = \(\frac{\overrightarrow{b_1} \cdot \vec{b}_2}{|\overrightarrow{b_1}||\vec{b}_2|}\)
= \(\frac{(3 \hat{i}-2 \hat{j}+0 \hat{k}) (2 \hat{i}+3 \hat{j}+4 \hat{k})}{\sqrt{9+4+0} \sqrt{4+9+16}}\)
= \(\frac{3(2)-2(3)+0}{\sqrt{13} \sqrt{29}}\) = 0
⇒ θ = \(\frac{\pi}{2}\)

Question 2.
Find the angle between the following pairs of lines:
(i) \(\vec{r}\) = 4 \(\hat{i}\) – \(\hat{j}\) + λ(\(\hat{i}\) + 2 \(\hat{j}\) – 2 \(\hat{k}\)) ; \(\vec{r}\) = \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\) – µ(2 \(\hat{i}\) + 4 \(\hat{j}\) – 4 \(\hat{k}\))
(ii) \(\vec{r}\) = 3 \(\hat{i}\) + 2 \(\hat{j}\) – 4 \(\hat{k}\) + λ(\(\hat{i}\) + 2 \(\hat{j}\) + 2 \(\hat{k}\)); \(\vec{r}\) = 5 \(\hat{i}\) – 2 \(\hat{j}\) + µ(3 \(\hat{i}\) + 2 \(\hat{j}\) + 6 \(\hat{k}\))
(iii) \(\vec{r}\) = λ(\(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\)) ; \(\vec{r}\) = 2 \(\hat{j}\) + µ[(\(\sqrt{3}\) – 1) \(\hat{i}\) – (\(\sqrt{3}\) + 1) \(\hat{j}\) + 4 \(\hat{k}\)]
Answer:
(i) Given eqns. of lines are
\(\vec{r}\) = (4 \(\hat{i}\) – \(\hat{j}\)) + λ(\(\hat{i}\) + 2 \(\hat{j}\) – 2 \(\hat{k}\))
\(\vec{r}\) = \(\hat{i}\) – \(\hat{j}\) + 2 \(\hat{k}\) – µ(2 \(\hat{i}\) + 4 \(\hat{j}\) – 4 \(\hat{k}\))
Let \(\overrightarrow{m_1}\) and \(\overrightarrow{m_2}\) vectors parallel to line (1) and (2).
∴ \(\vec{m}_1\) = \(\hat{i}\) + 2 \(\hat{j}\) – 2 \(\hat{k}\) ;
\(\overrightarrow{m_2}\) = 2 \(\hat{i}\) + 4 \(\hat{j}\) – 4 \(\hat{k}\)
Let θ be the angle between given line
∴ θ be also the angle between \(\vec{m}_1\) and \(\vec{m}_2\)
Then
Then cosθ = \(\frac{\overrightarrow{m_1} \cdot \overrightarrow{m_2}}{|\overrightarrow{m_1}||\overrightarrow{m_2}|}\)
= \(\frac{(\hat{i}+2 \hat{j}-2 \hat{k})(2 \hat{i}+4 \hat{j}-4 \hat{k})}{\sqrt{1+4+4} \sqrt{4+16+16}}\)
⇒ cosθ = \(\frac{1(2)+2(4)-2(-4)}{3 \times 6}\)
= \(\frac{18}{18}\) = 1
Thus θ = 0° ∴ given pair of lines are parallel.

(ii) The given pair of lines are
and
\(\vec{r}\) = 3 \(\hat{i}\) + 2 \(\hat{j}\) – 4 \(\hat{k}\) + λ(\(\hat{i}\) + 2 \(\hat{j}\) + 2 \(\hat{k}\))
and \(\vec{r}\) = 5 \(\hat{j}\) – 2 \(\hat{k}\) + µ(3 \(\hat{i}\) + 2 \(\hat{j}\) + 6 \(\hat{k}\))
Here, \(\vec{b}\) = \(\hat{i}\) + 2 \(\hat{j}\) + 2 \(\hat{k}\)
∴\(|\vec{b}|\) = \(\sqrt{1^2+2^2+2^2}\)
= 3 and \(\vec{d}\) = 3 \(\hat{i}\) + 2 \(\hat{j}\) + 6 \(\hat{k}\)
∴ \(|\vec{d}|\) = \(\sqrt{3^2+2^2+6^2}\) = 7 and \(\vec{b}\) \(\vec{d}\) = 13 + 2 2 + 2 6
= 3 + 3 + 12 = 19
If θ be the required angle then
cosθ = \(\frac{\vec{b} \cdot \vec{d}}{|\vec{b}||\vec{d}|}\)
= \(\frac{19}{3 \times 7}\) = \(\frac{19}{21}\)
∴ θ = cos^-1 (\(\frac{19}{21}\))

(iii) Given lines are
\(\vec{r}\) = λ(\(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\))
\(\vec{r}\) = 2 \(\hat{j}\) + µ{(\(\sqrt{3}\) – 1) \(\hat{i}\) – (\(\sqrt{3}\) + 1) \(\hat{j}\) + 4 \(\hat{k}\)
Let \(\vec{b}_1\) and \(\overrightarrow{b_2}\) be the vectors parallel to line (1) and line (2)
∴ \(\vec{b}_1\) = \(\hat{i}\) + \(\hat{j}\) + 2 \(\hat{k}\);
\(\overrightarrow{b_2}\) = (\(\sqrt{3}\) – 1) \(\hat{i}\) – (\(\sqrt{3}\) + 1) \(\hat{j}\) + 4 \(\hat{k}\)
Let θ be the angle between given lines then θ be the angle between \(\vec{b}_1\) and \(\overrightarrow{b_2}\)
∴ cos θ = \(\frac{\overrightarrow{b_1}\vec{b}_2}{|\overrightarrow{b_1}||\overrightarrow{b_2}|}\)
= \(\frac{(\hat{i}+\hat{j}+2 \hat{k}) \cdot[(\sqrt{3}-1) \hat{i}-(\sqrt{3}+1) \hat{j}+4 \hat{k}]}{\sqrt{1+1+4} \sqrt{(\sqrt{3}-1)^2+(\sqrt{3}+1)^2+16}}\)
= \(\frac{\sqrt{3}-1-\sqrt{3}-1+8}{\sqrt{6} \sqrt{3+1-2 \sqrt{3}+3+1+2 \sqrt{3}+16}}\)
∴ cosθ = \(\frac{6}{\sqrt{6} \sqrt{24}}\) = \(\frac{6}{\sqrt{144}}\) = \(\frac{6}{12}\) = \(\frac{1}{2}\)
Thus
θ = \(\frac{\pi}{3}\)

Question 3.
Find the angle between the pairs of lines with direction ratios:
(i) 2, 2, 1 and 4, 1, 8
(ii) 1, 2, -2 and -2, 2, 1.
Answer:
(i) Here a₁ = 2 ; b₁ = 2 ; c₁ = 1 and a₂ = 4 ; b₂ = 1 ; c₂ = 8
Let θ be the angle between given pair of lines.
Then
cosθ = \(\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\)
= \(\frac{2(4)+2(1)+1(8)}{\sqrt{4+4+1} \sqrt{16+1+64}}\) = \(\frac{18}{3 \times 9}\)
= \(\frac{2}{3}\)
θ = cos^-1 (\(\frac{2}{3}\))

(ii) Here a₁ = 1 ; b₁ = 2 ; c₁ = -2 ; a₂ = -2 ; b₂ = 2 ; c₂ = 1
Let θ be the angle between given pairs of lines.
Then
cosθ = \(\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\)
= \(\frac{1(-2)+2(2)-2(1)}{\sqrt{1+4+4} \sqrt{4+4+1}}\) = 0
Thus, θ = \(\frac{\pi}{2}\)

Question 4.
Prove that the two lines \(\frac{x}{1}\) = \(\frac{y}{2}\) = \(\frac{z}{1}\) and \(\frac{x}{1}\) = \(\frac{y}{-1}\) = \(\frac{z}{1}\) are at right angles.
Answer:
Given lines are
\(\frac{x}{1}\) = \(\frac{y}{2}\) = \(\frac{z}{1}\)
\(\frac{x}{1}\) = \(\frac{y}{-1}\) = \(\frac{z}{1}\)
and
∴ D ratios of line (1) and (2) are < 1, 2, 1> and < 1, -1, 1 >
Here, a₁ = 1 ; b₁ = 2 ; c₁ = 1 and a₂ = 1 ; b₂ = -1 ; c₂ = 1
Now, a₁ a₂ + b₁ b₂ + c₁ c₂ = 1(1) + 2(-1) + 1 1 = 0
Thus both lines (1) and (2) are at right angles.

Question 5.
P, Q, R, S are the points (-2, 3, 4), (-4, 4, 6), (4, 3, 5) and (0, 1, 2). Show that P Q is perpendicular to R S.
Answer:
Here, D ratios of line PQ be < -4 + 2, 4 – 3, 6 – 4 >
i.e., < -2, 1, 2 > and D ratios of line RS be < 0 – 4, 1 – 3, 2 – 5 >
i.e., < -4, -2, -3 >
Here, a₁ a₂ + b₁ b₂ + c₁ c₂ = (-2)(-4) + 1(-2) + 2(-3) = 0
Thus line PQ is ⊥ to line RS.

Question 6.
(i) Determine the value of k so that the line joining the points A(k, -1, -1), B(2, 0, 2 k) is perpendicular to the line joining the points C(4, 2 k, 1) and D(2, 3, 2).
(ii) For what value of p will the line through (4, 1, 2) and (5, p, 0) be perpendicular to the line through (2, 1, 1) and (3, 3, -1) ?
Answer:
(i) D ratios of line joining the points A(k, -1, -1) and B(2, 0, 2 k) be < 2 – k, 0 + 1, 2 k + 1 > and D ratios of line joining the points C(4, 2 k, 1) and D(2, 3, 2) be < 2 – 4, 3 – 2 k, 2 – 1 > i.e., < -2, 3 – 2 k, 1 >
Here, a₁ = 2 – k ; b₁ = 1 ; c₁ = 2 k + 1 and a₂ = -2 ; b₂ = 3 – 2 k ; c₂ = 1 Since, it is given that, line AB is ⊥ to line CD.
∴ a₁ a₂ + b₁ b₂ + c₁ c₂ = 0
⇒ (2 – k)(-2) + 1(3 – 2 k) + (2 k + 1) 1 = 0
⇒ -4 + 2 k + 3 – 2 k + 2 k + 1 = 0
⇒ 2 k = 0
⇒ k = 0

(ii) Direction ratios of the line through A(4, 1, 2) and B(5, p, 0) be given by < 5 – 4, p – 1, 0 – 2 > i.e., < 1, p – 1, -2 >
and direction ratios of the line through the points C(2, 1, 1) and D(3, 3, -1) be < 3 – 2, 3 – 1, -1 – 1 > i.e., < 1, 2, -2 >
Since it is given that, line AB is ⊥ to line CD.
∴ a₁ a₂ + b₁ b₂ + c₁ c₂ = 0 ⇒ 1(1)+(p-1) 2+(-2)(-2) = 0
⇒ 1 + 2 p – 2 + 4 = 0
⇒ 2p = -3
⇒ p =\( -\frac{3}{2}\)

Question 7.
Find the value of A, so that the following lines are perpendicular to each other :
(i) \(\frac{x-5}{5 \lambda+2}\) = \(\frac{2-y}{5}\) = \(\frac{1-z}{-1}\);
\(\frac{x}{1}\) = \(\frac{2 y+1}{4 \lambda}\) = \(\frac{1-z}{-3}\)
(ii) \(\frac{1-x}{3}\) = \(\frac{7 y-14}{2 \lambda}\) = \(\frac{5 z-10}{11}\);
\(\frac{7-7 x}{3 \lambda}\) = \(\frac{y-5}{1}\) = \(\frac{6-z}{5}\)
Answer:
(i) Given lines can be written as
\(\frac{x-5}{5 \lambda+2}\) = \(\frac{+y-2}{-5}\) = \(\frac{z-1}{1}\)
\(\frac{x}{1}\) = \(\frac{y+1 / 2}{2 \lambda}\) = \(\frac{z-1}{3}\)
and
\(\frac{x}{1}\) = \(\frac{y+1 / 2}{2 \lambda}\) = \(\frac{z-1}{3}\)
Direction ratios of line (1) are proportional to, < 5 λ + 2, -5, 1 >
i.e. a₁ = 5λ + 2 ; b₁ = -5 ; c₁ = 1
and D’ ratios of line (2) be proportional to < 1, 2λ, 3 >
Here a₂ = 1 ; b₂ = 2λ ; c₂ = 3 lines (1) and (2) are perpendicular to eachother
∴ a₁ a₂ + b₁ b₂+ c₁c₂=0
⇒ (5λ + 2) 1 – 5(2λ) + 1(3) = 0
⇒5λ + 2 – 10λ + 3 = 0
⇒ -5λ + 5 = 0 ⇒ λ = 1

(ii) eqn. of given lines are; \(\frac{1-x}{3}\) = \(\frac{7 y-14}{2 \lambda}\)
= \(\frac{5 z-10}{11}\)
i.e., \(\frac{x-1}{-3}\) = \(\frac{7(y-2)}{2 \lambda}\) = \(\frac{5(z-2)}{11}\)
i.e., \(\frac{x-1}{-3}\) = \(\frac{y-2}{\frac{2 \lambda}{7}}\) = \(\frac{z-2}{\frac{11}{5}}\)
∴ Direction ratios of line (1) are < -3, \(\frac{2 \lambda}{5}\), \(\frac{11}{5}\) >
and eqn. of given line be \(\frac{7-7 x}{3 \lambda}\) = \(\frac{y-5}{1}\)
= \(\frac{6-z}{5}\)
⇒ \(\frac{-7(x-1)}{3 \lambda}\) = \(\frac{y-5}{1}\) = \(\frac{z-6}{-5}\)
i.e., \(\frac{x-1}{\frac{3 \lambda}{-7}}\) = \(\frac{y-5}{1}\) = \(\frac{z-6}{-5}\)
∴ D ratios of line (2) are < \(\frac{-3 \lambda}{7}\), 1, -5 >
Now lines (1) and (2) are ⊥ to each other
∴ (-3)(\(\frac{3 \lambda}{7}\)) + (\(\frac{2 \lambda}{7}\)) 1 + \(\frac{11}{5}\)(-5) = 0
⇒ \(\frac{9 \lambda}{7}\) + \(\frac{2 \lambda}{7}\) – 11 = 0
⇒ 11λ – 77 = 0
⇒ λ = 7

Question 8.
Find the values ofp and q so that the line joining the points (7, p, 2) and (q, -2, 5) may be parallel to the line joining the points (2, -3, 5) and (-6, -15, 11).
Answer:
Given direction ratios of the line joining the point A(7, p, 2) and B(q, -2, 5) be
< q – 7, -2 – p, 5 – 2 > i.e., < q – 7, -2 – p, 3 >
and direction ratios of the line joining the points C(2, -3, 5) and D(-6, -15, 11) are
< -6 – 2, -15 + 3, 11 – 5 >
i.e., < -8, -12, 6 > i.e., < 4, 6, -3 >
Since line AB is || to line CD.
∴ direction ratios of both lines are proportional
∴ \(\frac{q-7}{4}\) = \(\frac{-2-p}{6}\)
= \(\frac{3}{-3}\)
⇒ \(\frac{q-7}{4}\) = \(\frac{p+2}{-6}\) = -1
⇒ \(\frac{q-7}{4}\) = -1
⇒ q – 7 = -4 ⇒ q = 3
and \(\frac{p+2}{-6}\) = -1
⇒ p + 2 = 6
⇒ p = 4

Question 9.
Find the equations of the line passing through the point (2, 1, 3) and perpendicular to the lines \(\frac{x-1}{1}\) = \(\frac{y-2}{2}\) = \(\frac{z-3}{3}\) and \(\frac{x}{-3}\) = \(\frac{y}{2}\) = \(\frac{z}{5}\)
Answer:
Let the direction ratios of the required line be proportional to a, b, c.
Since the required line is ⊥ to two given lines
a + 2 b + 3 c = 0
-3 a + 2 b + 5 c = 0
On solving eqn. (1) and (2) by cross multiplication method, we have
\(\frac{a}{10-6}\) = \(\frac{b}{-9-5}\) = \(\frac{c}{2+6}\)
⇒ \(\frac{a}{4}\)
= \(\frac{b}{-14}\) = \(\frac{c}{8}\)
i.e. \(\frac{a}{2}\) = \(\frac{b}{-7}\) = \(\frac{c}{4}\)
∴ direction ratio’s of required line are proportional to 2, -7, 4.
Hence the eqn. of required line which passes through (2, 1, 3) and having direction ratios proportional to < 2, -7, 4 > is given by
\(\frac{x-2}{2}\)
= \(\frac{y-1}{-7}\)
= \(\frac{z-3}{4}\)

Question 10.
Find the equations of the line passing through the point A(-1, 3, 7) and perpendicular to the lines \(\vec{r}\) = 2 \(\hat{i}\) – 3 \(\hat{j}\) + λ(2\(\hat{i}\) + 3 \(\hat{j}\) + \(\hat{k}\)) and \(\vec{r}\) = \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\) + µ(7 \(\hat{j}\) – 5 \(\hat{k}\)).
Answer:
The required eqn. of line through the point A(-1, 3, 7) and having direction ratios < a, b, c > be given by
\(\frac{x+1}{a}\) = \(\frac{y-3}{b}\) = \(\frac{z-7}{c}\)
and equs of given lines are (in cartesian form)
\(\frac{x-2}{2}\) = \(\frac{y+3}{3}\) = \(\frac{z-0}{1}\)
and
\(\frac{x-1}{0}\) = \(\frac{y+1}{7}\) = \(\frac{z-1}{-5}\)
Now, line (1) is ⊥ to lines (2) and (3).
Then
2 a + 3 b + c = 0
0 a + 7 b – 5 c = 0
and
0 a + 7 b – 5 c = 0
on solving eqn. (4) and eqn. (5) ; by using cross-multiplication method, we have
\(\frac{a}{-15-7}\) = \(\frac{b}{0+10}\) = \(\frac{c}{14-0}\)
i.e., \(\frac{a}{-22}\) = \(\frac{b}{10}\) = \(\frac{c}{14}\)
i.e., \(\frac{a}{+11}\) = \(\frac{b}{-5}\) = \(\frac{c}{-7}\)
∴ from (1); required eqn. of line be ; \(\frac{x+1}{11}\) = \(\frac{y-3}{-5}\) = \(\frac{z-7}{-7}\)

Question 11.
Find the equations of the line passing through the point (1, -1, 1) and perpendicular to the lines joining the points (4, 3, 2),(1, -1, 0) and (1, 2, -1),(2, 1, 1).
Answer:
Let the direction ratios of the required line be proportional to < a, b, c >.
Since the required line is ⊥ to given two line whose direction ratios are proportional to
< 1 – 4, -1 – 3, 0 – 2 > and < 2 – 1, 1 – 2, 1 + 1 >
i.e. < -3, -4, -2 > and < 1, -1, 2 >
Thus -3 a – 4 b – 2 c = 0
⇒ 3 a + 4 b + 2 c = 0
and
a – b + 2 c = b
By cross multiplication method, we have
\(\frac{a}{10}\) = \(\frac{b}{-4}\) = \(\frac{c}{-7}\)
Hence the required line passing through the point (1, -1, 1) and having direction ratio proportional to < 10, -4, -7 > is given by
\(\frac{x-1}{10}\) = \(\frac{y+1}{-4}\) = \(\frac{z-1}{-7}\)

Question 12.
Find the coordinates of the foot of the perpendicular from (1, 1, 1) on the line joining (5, 4, 4) and (1, 4, 6).
Answer:

Here eqn. of line AB} be
\(\frac{x-5}{1-5}\) = \(\frac{y-4}{4-4}\) = \(\frac{z-4}{6-4}\)
i.e., \(\frac{x-5}{-4}\)
= \(\frac{y-4}{0}\) = \(\frac{z-4}{2}\) = t (say)
any point on line (1) be M(-4 t + 5, 4, 2 t + 4)
Let this point M be the foot of ⊥ drawn from P on A B. Then P M ⊥ A B
Now direction ratios of line PM are
< -4 t + 5 – 1, 4 – 1, 2 t + 4 – 1 >
i.e., < -4 t + 4, 3, 2 t + 3 >
and direction ratios of line AB are < 1 – 5, 4 – 4, 6 – 4 >
i.e., < -4, 0, 2 >.
∴ from (2); we have
(-4 t + 4)(-4) + 3 × 0 + 2(2 t + 3) = 0
⇒ 16 t – 16 + 4 t + 6 = 0
⇒ 20 t – 10 = 0
⇒ t = \(\frac{1}{2}\)
Thus the coordinates of point are (-2 + 5, 4, 1 + 4) i.e., M(3, 4, 5)
Hence the required coordinates of foot of ⊥ are (3, 4, 5).

Question 13.
Find the foot of the perpendicular from (0, 2, 7) on the line \(\frac{x+2}{-1}\) = \(\frac{y-1}{3}\) = \(\frac{z-3}{-2}\)
Answer:
Let L be the foot of perpendicular drawn from P(0, 2, 7) on given line.

So any point on given line be given by
\(\frac{x+2}{-1}\) = \(\frac{y-1}{3}\)
= \(\frac{z-2}{-2}\) = t (say)
i.e. x = -t – 2 ; y = 3 t + 1 ; z = -2 t + 3
Thus the coordinates of point L are (-t – 2, 3 t + 1, -2 t + 3)
∴ The direction ratios of line PL are proportional to
-t – 2 – 0, 3 t + 1 – 2, -2 t + 3 – 7
i.e. -t – 2, 3 t – 1,-2 t – 4
Also the D’ ratios of given line are proportional to < -1, 3, -2 >
Since line PL is ⊥ to given line.
∴(-t – 2)(-1) + (3 t – 1) 3 + (-2 t – 4)(-2) = 0
⇒ t + 2 + 9 t – 3 + 4 t + 8 = 0
⇒ 14 t = -7
⇒ t = \(-\frac{1}{2}\)
Thus, the coordinates of L i.e. foot of ⊥ are
(\(-\frac{1}{2}\) – 2, \(-\frac{3}{2}\) + 1, (-2) (\(-\frac{1}{2}\)) + 3)
i.e. (\(-\frac{5}{2}\), \(-\frac{1}{2}\), 4)

Question 14.
A(0, 6, -9), B(-3, -6, 3) and C(7, 4, -1) are three points. Find the equa ion of the line A B. If D is the foot of perpendicular drawn from C to the line A B, find coordinate of the point D.
Answer:
Now D’ratios of given line A B are proportional to < -3 – 0, -6 – 6, 3 + 9 >
i.e. < -3, -12, 12 >
i.e. < 1, -4, -4 >
Thus eqn. of line AB which passes through A(0, 6, -9) and B(-3, -6, 3) is given by

\(\frac{x-0}{1}\) = \(\frac{y-6}{4}\) = \(\frac{z+9}{-4}\)
Any point on given line be given by
\(\frac{x-0}{1}\) = \(\frac{y-6}{4}\) = \(\frac{z+9}{-4}\) = t
i.e. x = t, y = 4 t + 6 ; z = -4 t – 9
∴ The coordinates of point D are (t, 4 t + 6, -4 t – 9)
Thus the direction ratios of line CD are proportional to < t – 7, 4 t + 2, -4 t – 8 >
also the direction ratios of given line are proportional to, < 1, 4, -4 >
Since C D ⊥ A B.
∴(t – 7) + (4 t + 2) 4 + (-4 t – 8)(-4) = 0
⇒ t – 7 + 16 t + 8 + 16 t + 32 = 0
⇒ 33 t = -33
⇒ t = -1
Thus the required coordinates of D are (-1, 2, -5) and the required eqn. of C D be given by
\(\frac{x-7}{-1-7}\) = \(\frac{y-4}{2-4}\)
= \(\frac{z+1}{-5+1}\)
i.e. \(\frac{x-7}{-8}\) = \(\frac{y-4}{-2}\) = \(\frac{z+1}{-4}\)
i.e. \(\frac{x-7}{4}\) = \(\frac{y-4}{1}\)
= \(\frac{z+1}{2}\)

Question 15.
Find the distance of (1, 0, 0) from the line through (1, -1, -10) whose d.c’s are proportional to 2, -3, 8.
Answer:
Let L be the foot of ⊥ drawn from P(1, 0, 0) on given line. any point on given line is given by
\(\frac{x-1}{2}\) = \(\frac{y+1}{-3}\) = \(\frac{z+10}{8}\) = t

i.e. x = 2 t + 1 ; y = -3 t – 1 ; z = 8 t – 10
∴ The coordinates of point L be (2 t + 1, -3 t + 1, 8 t – 10)
Now the direction ratios of line PL are proportional to
< 2 t + 1 – 1, -3 t – 1, 8 t – 10 > i.e. < 2 t, -3 t – 1, 8 t – 10 >
Now direction ratio of given line be proportional to < 2, -3, 8 >
Since P L ⊥ to given line.
∴ 2(2 t) + (-3 t – 1)(-3) + (8 t – 10) 8 = 0
⇒ 4 t + 9 t + 3 + 64 t – 80 = 0
⇒ 77 t – 77 = 0
⇒ t = 1
∴ The coordinates of point L i.e. foot of ⊥ are (3, -4, -2) and required ⊥ distance = |PL|
= \(\sqrt{(3-1)^2+(-4-0)^2+(-2-0)^2}\)
= \(\sqrt{4+16+4}\) = 2 \(\sqrt{6}\) units

Question 16.
Find the perpendicular distance of the point (2, 3, 4) from the line \(\frac{4-x}{2}\) = \(\frac{y}{6}\) = \(\frac{1-z}{3}\). Also find the coordinates of the foot of the perpendicular.
Answer:
Let P(2, 3, 4) be the given point and M be the foot of ⊥ from P on given line AB.
\(\frac{x-4}{-2}\) = \(\frac{y}{6}\) = \(\frac{z-1}{-3}\) = t (say)

Any point on given line be M(-2 t + 4, 6 t, -3 t + 1)
∴ D’ratios of line PM are < -2 t + 2, 6 t – 3, -3 t – 3 > Now line AB is ⊥ to line PM.
∴ (-2 t + 2)(-2) + (6 t – 3) 6 + (-3 t – 3)(-3) = 0
⇒ 4 t – 4 + 36 t – 18 + 9 t + 9 = 0
⇒ 49 t = 13 ⇒ t = \(\frac{13}{49}\)
∴ foot of ⊥ is given by M ( \(\frac{-26}{49}\) + 4, \(\frac{78}{49}\), \(\frac{-39}{49}\) + 1)
i.e. M (\(\frac{170}{49}\), \(\frac{78}{49}\), \(\frac{10}{49}\))
∴ length of ⊥=|PM| = \(\sqrt{(\frac{170}{49}-2)^2+(\frac{78}{49}-3)^2+(\frac{10}{49}-4)^2}\)
= \(\frac{\sqrt{9(576+529+3844)}}{49}\)
= \(\frac{3}{49} \sqrt{49 \times 101}\)
= \(\frac{3}{7}\) \(\sqrt{101}\) units

Question 17.
Find the image of the point (3, 5, 3) in the line \(\frac{x}{1}\) = \(\frac{y-1}{2}\) = \(\frac{z-2}{3}\)
Answer:
Let P be the given point, i.e., P}(3,5,3) and AB} be the given line,
i.e., \(\frac{x}{1}\) = \(\frac{y-1}{2}\) = \(\frac{z-2}{3}\) = t (say)

So, any point on line (1) be M(t, 2 t + 1, 3 t + 2)
Now, let M be the foot of ⊥ drawn from P on AB ∴ PM ⊥ AB
Further produe PM to P
S.t PM = MP then P be the image of P in AB.
D ratios of line PM are
< t – 3, 2 t + 1 – 5, 3 t + 2 – 3 >
i.e., < t – 3, 2 t – 4, 3 t – 1 >
D ratios of line AB are < 1, 2, 3 >
Since, PM ⊥ AB
∴(t – 3) 1 + (2 t – 4) 2 + (3 t – 1) 3 = 0
⇒ t – 3 + 4 t – 8 + 9 t – 3 = 0
⇒ 14 t – 14 = 0 ⇒ t = 1
∴ Coordinates of point M be (1, 3, 5)
Also, M be the mid point of PP and let coordinates of P be (α, β, γ).
∴ \(\frac{\alpha+3}{2}\) = 1;
\(\frac{\beta+5}{2}\) = 3 and \(\frac{\gamma+3}{2}\) = 5
i.e., α = -1 ; β = 1 ; γ = 7
Hence required coordinates of image of P(3, 5, 3) in given line are (-1, 1, 7)

Question 18.
Find the image of the point (2, -1, 5) in the line \(\vec{r}\) = (11 \(\hat{i}\) – 2 \(\hat{j}\) – 8 \(\hat{k}\)) + λ(10 \(\hat{i}\) – 4 \(\hat{j}\) – 11 \(\hat{k}\)).
Answer:
Let P(2, -1, 5) be the given point and eqn. of line in cartesian form be given by

\(\frac{x-11}{10}\) – \(\frac{y+2}{-4}\) = \(\frac{z+8}{-11}\) = t (say)
From P, drawn PM ⊥ AB and produce PM to P then P be the image of P in AB when PM = MP
any point on line (1) be M(10 t + 11, -4 t – 2, -11 t – 8)
∴ direction ratios of line PM are
< 10 t + 11 – 2, -4 t – 2 + 1, -11 t – 8 – 5 >
i.e., < 10 t + 9, -4 t – 1, -11 t – 13 >
Also direction ratios of line AB are < 10, -4, -11 >
Since PM is ⊥ to AB.
∴(10 t + 9) 10 + (-4 t – 1)(-4) + (-11 t – 13)(-11) = 0
⇒ 100 t + 90 + 16 t + 4+ 121 t + 143 = 0
⇒ 237 t + 237 = 0 ⇒ t = -1
Thus the coordinates of M are (-10 + 11, 4 – 2, 11 – 8)
i.e., M(1, 2, 3)
Let P(α, β, γ) be the image P then M be mid point of PP.
Thus, \(\frac{\alpha+2}{2}\) = 1 ;
\(\frac{\beta-1}{2}\) = 2 ;
\(\frac{\gamma+5}{2}\) = 3
i.e, α = 0 ; β = 5 and γ = 1
Hence, coordinates of image of point (2, -1, 5) in given line be (0, 5, 1).