RD Sharma Class 12 Exercise 27.2 Straight line in space Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 27.2 Straight line in space Solutions Maths - Download PDF Free Online

Edited By Lovekush kumar saini | Updated on Jan 25, 2022 09:47 AM IST

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• The RD Sharma class 12th exercise 27.2 is used by thousands of students and teachers for the practical knowledge of maths. The RD Sharma class 12 solution of Straight line in space exercise 27.2 consists of 36 questions covering all the concepts which are
• Prove the lines are mutually perpendicular

• lines through the points are perpendicular to the line through the points

• lines through the points are parallel to the line through the points

• The angle between the pair of lines given in vector and cartesian form

• Equation of line passing through the points

• Perpendicularity of two lines

## Straight Line in Space Excercise: 27.2

Straight Line in Space Exercise 27.2 Question 1

Answer: three direction cosines i.e. a, b, c are perpendicular to each other
Given: show that three lines with directions cosines $\frac{12}{13},\frac{-3}{13},\frac{-4}{13},\frac{4}{13},\frac{12}{13},\frac{3}{13},\frac{3}{13},\frac{-4}{13},\frac{12}{13}$; are mutually perpendicular
Hint: for mutally perpendicular
$\vec{a}.\vec{b}=0,\vec{b}.\vec{c}=0,\vec{a}.\vec{c}=0$
solution:
\begin{aligned} &1^{\text {st }} \text { vector } \vec{a}=\frac{12}{13} \hat{i}-\frac{3}{13} \hat{j}-\frac{4}{13} \hat{k} \\ \end{aligned}
\begin{aligned} &2^{\text {nd }} \text { vector } \vec{b}=\frac{4}{13} \hat{i}+\frac{12}{13} \hat{j}+\frac{3}{13} \hat{k} \\ \end{aligned}
\begin{aligned} &3^{\text {rd }} \text { vector } \vec{c}=\frac{3}{13} \hat{i}-\frac{4}{13} \hat{j}+\frac{12}{13} \hat{k} \end{aligned}
\begin{aligned} &\vec{a} \cdot \vec{b}=\frac{12}{13} \times \frac{4}{13}-\frac{3}{13} \times \frac{12}{13}-\frac{-4}{13} \times \frac{3}{13}=0 \\ \end{aligned}
\begin{aligned} &\vec{a} \cdot \vec{c}=\frac{36}{13 \times 13}+\frac{-12}{13 \times 13}-\frac{48}{13 \times 13}=0 \\ \end{aligned}
\begin{aligned} &\vec{b} \cdot \vec{c}=\frac{12}{13 \times 13}-\frac{48}{13 \times 13}+\frac{36}{13 \times 13}=0 \\\\ &=\frac{12-48+36}{13 \times 13}=\frac{0}{13 \times 13}=0 \end{aligned}
so all the three dimensions cosines are mutually perpendicular to each other .

Straight Line in Space Exercise 27.2 Question 2

Answer : the lines are perpendicular to the given points showed
Given: show that the line through the points$( 1 ,-1,2) and (3,4,-2)$ is perpendicular to the points $(0,3,2) and (3,5,6)$
hint: sum will solve by the help of the direction ratio
\begin{aligned} &\mathrm{DR}=\left(x_{2}-x_{1}\right) \hat{i}+\left(y_{2}-y_{1}\right) \hat{j}+\left(z_{2}-z_{1}\right) \hat{k}\\ \end{aligned}
Solution: direction ratio of
\begin{aligned} &Line_{1}=(3-1)_{,}(4+1),(-2-2)\\ \end{aligned}
\begin{aligned} &=2 \hat{i}+5 \hat{j}-4 \hat{k}\\ \end{aligned}

Direction ratio of \begin{aligned} &Line_{2}=(3-0),(5-3),(6-2)\\ \end{aligned}
\begin{aligned} &=3 \hat{i}+2 \hat{j}+4 \hat{k}\\ \end{aligned}
For showing perpendicular, \begin{aligned} &\text { } \mathrm{a}_{1} \mathrm{a}_{2}+\mathrm{b}_{1} \mathrm{~b}_{2}+\mathrm{c}_{1} \mathrm{c}_{2}=0\\ \end{aligned}
Now
\begin{aligned} &3.2+10-16=6+10-16\\ \end{aligned}
$=0$
Hence line1 and line2 are perpendicular to the points (showed)

Straight Line in Space Exercise 27.2 Question 3

Answer: The given two lines will be parallel through the two points equal to ‘0’ showed;

Given: Show that the line through the points$(4,7,8) and (2,3,4)$ is parallel to the line through the points $(-1,-2,1) and (1,2,5)$

Hint: for showing parallel v1 and v2 must be equal to ‘0’

Solution:

\begin{aligned} &\vec{v}_{1}=(4-2) \hat{i}+(7-3) \hat{j}+(8-4) \hat{k} \\ &\overrightarrow{v_{1}}=2 \hat{i}+4 \hat{j}+4 \hat{k} \\ &\overline{v_{2}}=(1+1) \hat{i}+(2+2) \hat{j}+(5-1) \hat{k} \\ &\overline{v_{2}}=2 \hat{i}+4 \hat{j}+4 \hat{k} \end{aligned}

Now, \begin{aligned} \cos \theta=\cos \theta &=\frac{\overrightarrow{v_{1}} \overline{v_{2}}}{\left|\overrightarrow{v_{1}}\right|\left|\overrightarrow{v_{2}}\right|} \\ \end{aligned}

\begin{aligned} &=\frac{4+16+16}{\sqrt{36} \sqrt{36}} \\ \end{aligned}

\begin{aligned} =& \frac{36}{36} \\ \end{aligned}

\begin{aligned} &=1 \\ \end{aligned}

\begin{aligned} \cos \theta &=1 \\ \end{aligned}

\begin{aligned} \cos \theta=\cos 0^{\circ} \\ \end{aligned}

\begin{aligned} \theta=0^{\circ} \\ \end{aligned}

\begin{aligned} \mathrm{v}_{1} \| \mathrm{v}_{2} \text { (proved) } \end{aligned}

So the two lines are parallel to each other

Straight Line in Space Exercise 27.2 Question 4

Answer : the Cartesian equation which is parallel to given lines will be $\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}$
Given: find the Cartesian equation of the line which passes through the points $(-2,4,-5)$and parallel to the given line by: $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$
Hint: let the Cartesian equation will be :- $\frac{x+x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z+z_{1}}{c}$
Solution: $\frac{x+x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z+z_{1}}{c}$ $[here a,b,c\; are 3,-5,6 and\; x,y,z \; is -2,4,-5]$
now, $\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}$
so the above line $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$ is parallel to $\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}$ and passes through the given points.

Straight Line in Space Exercise 27.2 Question 5

Answer : the given lines $\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}$ and $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ are perpendicular to each other
Given: show that lines $\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}$ and $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ are perpendicular to each other.
Hint: For showing perpendicular a.b must be equal to ‘ 0’
Solution: $\frac{x-5}{7}=\frac{y+2}{-5}=\frac{Z}{1}$
Direction: ratio $= 7,-5,1$
Vector $\vec{a}=7\hat{i}-5\hat{j}+\hat{k}$
Again,
$\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$
\begin{aligned} &\therefore \text { Direction ratios }=1,2,3 \\ \end{aligned}
\begin{aligned} &\therefore \vec{b}=\hat{i}+2 \hat{j}+3 \hat{k} \\ \end{aligned}
\begin{aligned} &\vec{a} \cdot \vec{b}=7-10+3 \\ &=0 \end{aligned}
So the given two lines are perpendicular to each other (showed)

Straight Line in Space Exercise 27.2 Question 6

Answer: the line joining by the point $(2,1,1)$ is perpendicular to the line determined by the other two points
Given: show that the line joining to the point $(2,1,1)$ is perpendicular to the line determined by the points $(3,5,-1) and (4,3,-1)$
Hint:
Direction ratio$\left (x _{2}-0 \right ),\left ( y_{2}-0 \right ),\left ( z_{2}-0 \right )$because line passes through origin
Solution:
\begin{aligned} &v_{1}=(2-0) \hat{i}+(1-0) \hat{j}+(1-0) \hat{k} \\ &v_{1}=2 \hat{i}+\hat{j}+\hat{k} \\ &v_{2}=(4-3) \hat{i}+(3-5) \hat{j}+(-1+1) \hat{k} \\ &v_{2}=\hat{i}-2 \hat{j} \end{aligned}
Now from \begin{aligned} v_{1} \cdot v_{2} &=0 \\ \end{aligned}
\begin{aligned} v_{1} v_{2} &=2-2 \\ &=0 \end{aligned}

So the line joining to the point $(2,1,1)$ is perpendicular to the line joining by the determined points $(3,5,-1) and (4,3,-1)$

Straight Line in Space Exercise 27.2 Question 7

Answer: the equation parallel to x-axis will be $\frac{x}{1}=\frac{y}{0}=\frac{3}{0}$
Given: find the equation of the line parallel to x-axis and passing through the origin
Hint: the line parallel to x-axis so the direction ratio will be $(1,0,0);(0,1,0);(0,0,1)$
Solution:
Equation:
\begin{aligned} &\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c} \\ &\frac{x-0}{1}=\frac{y-0}{0}=\frac{z-0}{0} \\ &\frac{x}{1}=\frac{y}{0}=\frac{z}{0} \end{aligned}
So the equation will be $=\frac{x}{1}=\frac{y}{0}=\frac{z}{0}$

Straight Line in Space Exercise 27.2 Question 8(i)

Answer: the angles between the lines will be $0^{0}$
Given : find the angle between the points of lines
$\vec{r}=(4 \hat{i} - \hat{j})+8(\hat{i}+2 \hat{j}-2 \hat{k}) and$
$\vec{r}=(\hat{i}-\hat{j}+2 \hat{k})-\mu[2 \hat{i}+4 \hat{j}-4 \hat{k}] \\$
Hint: $\cos \theta= \frac{\overline{b_{1}} \cdot \overline{b_{2}}}{\left|b_{1}\right| \cdot\left|b_{2}\right|}$
Solution:
The 1st line is parallel to $b1= (i+2j-2k)$ and the 2nd line is parallel to $b2 = (2i+4j-4k)$
If $\theta$ be an angle between the lines , so $\theta$ will be the angle between b1&b2
So,\begin{aligned} &\cos \theta=\frac{2+8+8}{\sqrt{2^{2}+2^{2}(-2)} \sqrt{2^{2}+4^{2}+\left(-4^{2}\right)}} \\ \end{aligned}
\begin{aligned} &=\frac{18}{\sqrt{9 \mid \sqrt{36}}} \\ &=\frac{18}{3 \times 6} \\ &\quad=1 \\ &\cos \theta=\cos 0 \\ &\theta=0^{\circ} \end{aligned}

The angle between the lines will be $0^{0}$

Straight Line in Space Exercise 27.2 Question 8(ii)

Answer: the angle between the points of lines will be $\cos ^{-1}\frac{19}{21}$
Given: find the angle between the points of lines
$\vec{r}=(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-4 \hat{\mathrm{k}})+8(\hat{\mathrm{i}}+2 \hat{j}+2 \hat{k}) and$
$\vec{r}=(5 \hat{j}-2 \hat{k})+\mu(3 \hat{i}+2 \hat{j}+6 \hat{k})$
Hint: $\cos \theta=\frac{\overrightarrow{b_{1}} \cdot \overrightarrow{b_{2}}}{\left|\vec{b}_{1}\right|\left|\overrightarrow{b_{2}}\right|}$
Solution:
Let ‘$\theta$’ be the angle between points of lines
\begin{aligned} &\vec{b}_{1}=\hat{i}+2 \hat{j}+2 \hat{k} \\ &\overrightarrow{b_{2}}=3 \hat{i}+2 \hat{j}+6 \hat{k} \\ \end{aligned}
\begin{aligned} &\cos \theta=\frac{3+4+12}{\left|\sqrt{1^{2}+2^{2}+2^{2}} \| \sqrt{3^{2}+2^{2}+6^{2}}\right|} \\ \end{aligned}
\begin{aligned} &=\frac{19}{|\sqrt{9} \| \sqrt{49}|} \\ \end{aligned}
\begin{aligned} &=\frac{19}{3 \times 7} \\ \end{aligned}\begin{aligned} &=\frac{19}{3 \times 7} \\ \end{aligned}
\begin{aligned} &\cos \theta=\frac{19}{21} \\ \end{aligned}
\begin{aligned} &\theta=\cos ^{-1} \frac{19}{21} \end{aligned}
So the answer will be $\cos ^{-1}\frac{19}{21}$

Straight Line in Space Exercise 27.2 Question 8(iii)

Answer: the angle between the points of lines will be $\frac{\pi }{3}$
Given: find the angle between the points of lines
\begin{aligned} &\vec{r}=\lambda(\hat{i}+\hat{j}+2 \hat{k}) \\ &\vec{r}=2 \hat{j}+\mu(\sqrt{3}-1) \hat{\mathrm{i}}-(\sqrt{3}+1) \hat{j}+4 \hat{k} \end{aligned}
Hint: $\cos \theta=\frac{\overrightarrow{b_{1}} \cdot \overline{b_{2}}}{\left|\vec{b}_{1}\right|\left|\overrightarrow{b_{2}}\right|}$
Solution:
\begin{aligned} &\vec{r}=2 \hat{j}+\mu[(-1) \hat{i}-(1) \hat{j}+4 \hat{k}] \\\\ &\overrightarrow{b_{1}}=\sqrt{1^{2}+1^{2}+2^{2}}=\sqrt{6} \\\\ &\overrightarrow{b_{2}}=\sqrt{(\sqrt{3}-1)^{2}+(\sqrt{3}+1)^{2}+4^{2}}=\sqrt{24}=2 \sqrt{6} \\\\ &\therefore \cos \theta=\frac{(\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}})[(\sqrt{3-1})-(\sqrt{3+1})+4 \hat{k}}{\sqrt{6} \times 2 \sqrt{6}} \\ &\qquad \begin{array}{c} \end{array} \end{aligned}
\begin{aligned} \cos \theta=\frac{1}{2} \therefore \theta=\frac{\pi}{3} \\\\ \Rightarrow \cos \theta=\cos 60^{\circ} \end{aligned}

Straight Line in Space Exercise 27.2 Question 9(i)

Answer: the angle between the given pairs of lines will be $\cos ^{-1}=\frac{8}{5\sqrt{3}}$
Given: find the angle between the following pairs of lines:
$\frac{x+y}{3}=\frac{y-1}{5}=\frac{z+3}{4} and \frac{x+1}{1}=\frac{y-4}{1}=\frac{z-5}{2}$
Hints: $\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$
Solution:
Directon ratio $= 3,5,4$
Direction vector $\vec{a}=3\hat{i}+5\hat{j}+4\hat{k}$
Again $DR=1,1,2$
Direction vector $\vec{b}=\hat{i}+\hat{j}+2\hat{k}$
\begin{aligned} \cos \theta &=\frac{3+5+8}{\left|\sqrt{3^{2}+5^{2}+4^{2}}\right| \mid \sqrt{1^{2}+1^{2}+2^{2}}} \mid \\ &=\frac{16}{|\sqrt{50}| \sqrt{6} \mid} \\ =& \frac{8}{5 \sqrt{3}} \\ & \theta=\cos ^{-1} \frac{8}{5 \sqrt{3}} \end{aligned}
So the answer will be $\theta =\cos ^{-1}\frac{8}{5\sqrt{3}}$

Straight Line in Space Exercise 27.2 Question 9(ii)

Answer: the angle between the given pairs of a line is $\cos ^{-1}=\frac{10}{9\sqrt{22}}$
Given: find the angle between the pairs of lines
$\begin{gathered} \frac{x-1}{2}=\frac{y-1}{3}=\frac{z-3}{-3} \text { and } \frac{x+3}{-1}=\frac{y-5}{8}=\frac{z-1}{4} \\ \end{gathered}$
Hint:$\begin{gathered} \cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \end{gathered}$
Solution: $Direction \: ratio \: of$$line_{1}=2,3,-3$
$Direction \; vector \vec{a}=2 \hat{i}+3 \hat{j}-3 \hat{k}\\\\$
$Direction ratio of line _{2}=-1,8,4\\\\\$
$Direction \: vector \vec{b}=-\hat{i}+8 \hat{j}+4 \hat{k}$
\begin{aligned} \cos \theta &=\frac{-2+24-12}{\left|\sqrt{2^{2}+3^{2}-(3)^{2}} \| \sqrt{(-1)^{2}+8^{2}+4^{2}}\right|} \\ &=\frac{24-14}{|\sqrt{22} \| \sqrt{81}|} \\ &=\frac{10}{9 \sqrt{22}} \\ & \theta=\cos ^{-1}=\frac{10}{9 \sqrt{22}} \end{aligned}
The angle between the points will be $\cos ^{-1}=\frac{10}{9\sqrt{22}}$

Straight Line in Space Exercise 27.2 Question 9(iii)

Answer: the angle between the lines given in the question is $\cos ^{-1}\frac{11}{14}$
Given: find the angle between the pairs of lines;
\begin{aligned} &\frac{5-x}{-2}=\frac{y+3}{1}=\frac{1-z}{3} \text { and } \frac{x}{3}=\frac{1-y}{-2}=\frac{z+5}{-1} \\ \end{aligned}
Hint:
\begin{aligned} &\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \end{aligned}

Solution:

\begin{aligned} &\frac{-(x-5)}{-2}=\frac{y-(-3)}{1}=\frac{-(3-1)}{3} \\ &\frac{x-b}{2}=\frac{y+3}{1}=\frac{(3+1)}{3} \end{aligned}

$Direction \: \: vector \vec{a}=2 \hat{i}+\hat{j}-3 \hat{k}$

Again:

$\frac{x-0}{3}=\frac{(y-1)}{ 2}=\frac{(z+5)}{-1}$

\begin{aligned} Direction \: \: vector \vec{b}=3 \hat{i}+2 \hat{j}-\hat{k} \end{aligned}

\begin{aligned} &\cos \theta=\frac{6+2+3}{\left|\sqrt{2^{2}+1^{2}+(-3)^{2}}\right| \sqrt{3^{2}+(-2)^{2}+(-1)^{2}} \mid} \\ \end{aligned}

$\frac{11}{\sqrt{14} \sqrt{14}}=\frac{11}{14} \\$

$\theta=\cos ^{-1} \frac{11}{14}$

So the angle between the lines will be$\cos ^{-1}\frac{11}{14}$

Straight Line in Space Exercise 27.2 Question 9(iv)

Answer: the angle between the pairs of lines will be $\frac{\lambda }{2}$
Given: find the angle between the points of lines
\begin{aligned} &\frac{x-2}{3}=\frac{y+3}{-2}=z=5 \\ &\frac{x+1}{1}=\frac{2 y-3}{3}=\frac{z-5}{2} \end{aligned}
Hint:$\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
Solution:$\frac{x-2}{3}=\frac{y-(-3)}{-2}=\frac{z-5}{0} \\$
$\vec{A}=3 \hat{i}-2 \hat{j}+0 \hat{k}$
$Secondly: \frac{x-(-1)}{1}=\frac{y-3 / 2}{3 / 2}=\frac{z-5}{2}$
$\cos \theta=\frac{(3)-3+0}{\sqrt{3^{2}+(-2)^{2}} \sqrt{1^{2}+(3 / 2)^{2}+(2)^{2}}} \\$
$\cos \theta=0$
$\cos \theta=\cos \frac{\lambda}{2}$
$\theta=\frac{\pi}{2}$
So the angle will be $\theta =\frac{\lambda }{2}$

Straight Line in Space Exercise 27.2 Question 9(v)

Answer: the angle between the pairs of given line is $\cos ^{-1}\frac{4}{5\sqrt{6}}$
Given: find the angle between the pairs of lines
\begin{aligned} &\frac{x-5}{1}=\frac{2 y+6}{-2}=\frac{z-3}{1} \text { and } \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-6}{5} \\ \end{aligned}
Hint:\begin{aligned} &\cos \theta=\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}}} \\ \end{aligned}
Solution:
\begin{aligned} &\frac{x-5}{1}=\frac{y+3}{-1}=\frac{z-3}{1} \\\\ &a_{1}, b_{1}, c_{1}=1,-1,1 \\ \end{aligned}
again,
$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-6}{5} \\\\$
$\cdot a_{2}, b_{2}, c_{2}=3,4,5 \\\\$
$\cos \theta=\frac{3-4+5}{\sqrt{1^{2}+(-1)^{2}+1^{2}} \sqrt{3^{2}+4^{2}+5^{2}}}$
\begin{aligned} &\quad=\frac{8-4}{\sqrt{3} \sqrt{50}}=\frac{4}{5 \sqrt{6}} \\ &\cos \theta=\cos ^{-1} \frac{4}{5 \sqrt{6}} \end{aligned}
So the angle will be$\cos ^{-1}\frac{4}{5\sqrt{6}}$

Straight Line in Space Exercise 27.2 Question 10(i)

Answer: the angle between the pairs of lines will be $\cos ^{-1}\frac{1}{65}$
Given: find the angle between the pairs of lines with direction ratios proportional to
$5,-12,13 and -3,4,5$
Hint: \begin{aligned} &\cos \theta=\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}}}} \\ \end{aligned}

Solution:
\begin{aligned} &\cos \theta=\frac{-15-48+65}{\sqrt{5^{2}+(-12)^{2}}+13^{2} \sqrt{(-3)^{2}+4^{2}}+5^{2}} \\ &=\frac{-63+65}{\sqrt{338} \sqrt{50}} \\ &=\frac{2}{13 \sqrt{2} \mid 5 \sqrt{2}} \\ &\therefore \quad \cos \theta=\frac{2}{65 \times 2}=\frac{1}{65} \\ &\therefore \quad \theta=\cos ^{-1} \frac{1}{65} \end{aligned}
The angle between the pairs of lines will be $\cos ^{-1}\frac{1}{65}$

Straight Line in Space Exercise 27.2 Question 10(ii)

Answer: the angle between the given pairs of lines will be $\cos ^{-1}\frac{2}{3}$
Given: find the angle between the pairs of lines with direction ratios proportion to$2,2,1 and 4,1,8$
Hint:$\begin{gathered} \cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}} \\ \end{gathered}$
Solution:$\begin{gathered} \cos \theta=\frac{8+2+8}{\sqrt{2^{2}+2^{2}+1^{2}} \sqrt{4^{2}+1^{2}+8^{2}}} \\ \end{gathered}$
$\begin{gathered} =\frac{18}{\sqrt{9} \sqrt{81}} \\ =\cos \theta=\frac{2}{3} \\ \therefore \theta=\cos ^{-1} \frac{2}{3} \end{gathered}$
So the angle between the pairs of lines will be $\cos ^{-1}\frac{2}{3}$

Straight Line in Space Exercise 27.2 Question 10(iii)

Answer: the angle between the given points of lines will be $90^{0}$
Given: find the angle between the pairs of lines with direction ratio proportional to $(1,2,-2)$ and
$(-2,2,1)$
Hint: \begin{aligned} &\cos \theta=\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}}}} \\ \end{aligned}
Solution:\begin{aligned} &\cos \theta=\frac{-2+4-2}{\sqrt{1^{2}+2^{2}+(-2)^{2}} \sqrt{(-2)^{2}+2^{2}+1^{2}}} \\ \end{aligned}
\begin{aligned} &=\frac{0}{\sqrt{1+4+4} \sqrt{4+4+1}} \\ \end{aligned}
\begin{aligned} &\cos \theta=0 \\ \end{aligned}
\begin{aligned} \cos \theta=90^{\circ} \\ \therefore \theta=90^{\circ} \end{aligned}
The angle between the pairs of lines will be $\theta =90^{0}$

Straight Line in Space Exercise 27.2 Question 10(iv)

Answer: the angle between the given pairs of lines will be $90^{0}$
Given: find the angle between the pairs of lines with direction ratio proportional to$(a,b,c)$and
$(b-c), (c-a) (a-b)$
Hint: \begin{aligned} \cos \theta=& \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}}}} \\ \end{aligned}
Solution:\begin{aligned} & a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=a b-a c+b c-b a+a c-b c=0 \\ \end{aligned}
\begin{aligned} & \therefore \cos \theta=\frac{0}{\sqrt{a^{2}+b^{2}+c^{2}} \sqrt{(b-c)^{2}+(c-a)^{2}+(a-b)^{2}}} \\ \end{aligned}
\begin{aligned} \cos \theta=0 \\ \end{aligned}
\begin{aligned} & \cos \theta=90^{\circ} \\ \end{aligned}
\begin{aligned} \therefore \theta=90^{\circ} \end{aligned}
The angle between the pairs of lines will be $\theta=90^{0}$

Straight Line in Space Exercise 27.2 Question 11

Answer: the angle between the pairs given in the question $\cos ^{-1}\frac{2}{3}$
Given: find the angle between two lines on of which has direction ratio $(2,2,1)$ while other one is obtained by joining the points $(3,1,4) and (7,2,12)$
Hint: \begin{aligned} &\cos \theta=\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}}}} \\ \end{aligned}
Solution: now $\left ( a_{1},b_{1},c_{1}\right )=2,2,1$
While \begin{aligned} &a_{2}=\left(x_{2}-x_{1}\right)=4 \\ \end{aligned}
\begin{aligned} &b_{2}=\left(y_{2}-y_{1}\right)=1 \\ &c_{2}=\left(z_{2}-z_{1}\right)=8 \\ \end{aligned}
\begin{aligned} &\cos \theta=\frac{8+2+8}{\sqrt{2^{2}+2^{2}+1^{2} \sqrt{1^{2}+1^{2}+8^{2}}}} \\ \end{aligned}\begin{aligned} &=\frac{18}{\sqrt{9 \sqrt{81}}} \\ &=\frac{2}{3} \\ &\theta=\cos ^{-1} \frac{2}{3} \end{aligned}

So the angle between the pairs will be $\cos ^{-1}\frac{2}{3}$

Straight Line in Space Exercise 27.2 Question 12

Answer: the equation will be:$\frac{x-1}{4}=\frac{y-2}{2}=\frac{z+4}{3}$
Given: find the equation of line passing through the point $(1,2,-4)$ and parallel to line $\frac{x-3}{4}=\frac{y-5}{2}=\frac{z+1}{3}$
Hint: equation will be $\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}$
Solution: Direction ratio of lines $(4,2,3)$ points $(1,2,-4)$
$\frac{x-1}{4}=\frac{y-2}{2}=\frac{z+4}{3}$
So the equation of the line passing through the point is $\frac{x-1}{4}=\frac{y-2}{2}=\frac{z+4}{3}$

Straight Line in Space Exercise 27.2 Question 13

Answer: the equation of the lines will be $\vec{r}=(-\hat{i}+2 \hat{j}+\hat{k})+\lambda(2 \hat{i}+2 / 3 \hat{j}-3 \hat{k})$
Given : Find the equation of the line passing through the point $(-1,2,1)$ and parallel to the line
$\frac{2x-1}{2}=\frac{3y+5}{2}=\frac{2-z}{3}$
Hint:$\vec{r}=\vec{a}+\lambda \vec{b}$
Solution:$\vec{a}=\vec{-i}+2\hat{j}+\hat{k}$
Now: $\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c} \\$
$=\frac{x-1 / 2}{2}=\frac{y-(-5 / 3)}{2 / 3}=\frac{z-2}{-3} \\$
$\vec{R}=(-\hat{i}+2 \hat{j}+\hat{k})+\lambda(2 \hat{i}+2 / 3 \hat{j}-3 \hat{k})$
So the equation will be $\vec{R}=(-\hat{i}+2 \hat{j}+\hat{k})+\lambda\left(2 \hat{i}+\frac{2}{3} \hat{j}-3 \hat{k}\right)$

Straight Line in Space Exercise 27.2 Question 14

Answer: the equation of the line passing through the point will be $\therefore \vec{r}=(2 \hat{i}-\hat{j}+3 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}-5 \hat{k})$
Given: find the equation of the line passing through the point$(2,-1,-3)$and parallel to the line $\therefore \vec{r}=(2 \hat{i}-\hat{j}+3 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}-5 \hat{k})$
Hint: for parallel equation will be: $\vec{r}=\vec{a}+\lambda \vec{b}$
Solution: direction ratio $= 2,3,-5$
$\therefore \vec{r}=(2 \hat{i}-\hat{j}+3 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}-5 \hat{k})$
∴the equation will be;
$\therefore \vec{r}=(2 \hat{i}-\hat{j}+3 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}-5 \hat{k})$

Straight Line in Space Exercise 27.2 Question 15

Answer: the equation of the line will be $\vec{r}=(2 \hat{i}+\hat{j}+3 \hat{k})+\lambda(4 \hat{i}-14 \hat{j}+8 \hat{k})$
Given: find the equation of the line passing through the point $(2,1,3)$ and perpendicular to the
\begin{aligned} &\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{-3} \text { and } \frac{x}{-3}=\frac{y}{2}=\frac{z}{5} \\ \end{aligned}
Hint:\begin{aligned} &\vec{r}=\vec{a}+\lambda \vec{b} \\ \end{aligned}
Solution:
\begin{aligned} &\vec{b}=\vec{b}_{1} \times \overline{b_{2}} \\ &\vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -3 & 2 & 5 \end{array}\right| \\ &=\hat{i}(10-6)-\hat{j}(5+9)+\hat{k}(2+6) \\ &=4 \hat{i}-14 \hat{j}+8 \hat{k} \\ &\vec{r}=(2 \hat{i}+\hat{j}+3 \hat{k})+\lambda(4 \hat{i}-14 \hat{j}+8 \hat{k}) \end{aligned}
So, the equation of the line will be $\vec{r}=(2 \hat{i}+\hat{j}+3 \hat{k})+\lambda(4 \hat{i}-14 \hat{j}+8 \hat{k})$

Straight Line in Space Exercise 27.2 Question 16

Answer: The equation will be \begin{aligned} &\vec{r}=(\hat{i}+\hat{j}-3 \hat{k})+\lambda(4 \hat{i}-5 \hat{j}+\hat{k}) \\ \end{aligned}
Given: Find the equation of the line passing through the point$i+j-3k$ and perpendicular to the lines \begin{aligned} &\vec{r}=\hat{i}+\lambda(2 \hat{i}+\hat{j}-3 \hat{k}) \& \vec{r}=(2 \hat{i}+\hat{j}-\hat{k})+\mu(\hat{i}+\hat{j}+\hat{k}) \\ \end{aligned}
Hint: \begin{aligned} &\vec{r}=\vec{a}+\lambda \vec{b} \end{aligned}
Solution:
$\vec{b}=\vec{b}_{1} \times \overrightarrow{b_{2}} \rightarrow b_{1}=2 \hat{i}+\hat{j}-3 \hat{k} \& b_{2}=\hat{i}+\hat{j}+\hat{k}$
$\vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{array}\right| \\$
$=\hat{i}(1+3)+\hat{j}(-3-2)+\hat{k}(2-1)$
$= 4 \hat{i}-5 \hat{j}+\hat{k}$
$\\ \therefore \vec{r}=(\hat{i}+\hat{j}-3 \hat{k})+\lambda(4 \hat{i}-5 \hat{j}+\hat{k)}$

Straight Line in Space Exercise 27.2 Question 17

Answer : The equation will be $\frac{x-1}{10}=\frac{y+1}{y}=\frac{3}{7}$
Given: Find the equation of the line passing through the point$(1, -1, 1)$ and perpendicular to the lines joining the points $(4,3,2)(1,-1,0) and (1,2,-1), (2,1,1$
Hint: equation will be in the form of
$\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}$
Solution:
Direction ratio of $a_{1}b_{1}c_{1}=3,4,2$
Direction ratio of $a_{2}b_{2}c_{2}=\left ( -1,1,-2 \right )$
Now
\begin{aligned} &3 a+4 b+2 c=0 \\ &-a+b-2 c=0 \\ &\frac{a}{10}=\frac{b}{y}=\frac{c}{7} \\ &\frac{x-1}{10}=\frac{y+1}{4}=\frac{3-0}{7} \end{aligned}
The equation will be$\frac{x-1}{10}=\frac{y+1}{y}=\frac{3}{7}$

Straight Line in Space Exercise 27.2 Question 18

Answer: The equation of the given question will be $\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})$
Given: Determine the equation of the line passing through the point (1,2,-4) and perpendicular to the two lines;
$\frac{x-8}{3}=\frac{y+9}{-16}=\frac{3-0}{7}$ &$\frac{x-15}{3}=\frac{y-29}{8}=\frac{3-5}{5}$
Hint:$\vec{r}=\vec{a}+\lambda \vec{b}$(for showing parallel lines)
Solution:
$Direction\: ratio _{1}=3 i-16 \hat{j}+7 \hat{k}$
$Direction \: ratio _{2}=3 i-8 \hat{j}-5 \hat{k}$
\begin{aligned} &\therefore D R_{1} \times D R_{2}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & -8 & -5 \end{array}\right| \\ &=24 \hat{i}+36 \hat{j}+72 \hat{k} \\ &=12(2 \hat{i}+3 \hat{j}+6 \hat{k}) \\ &\therefore D R=2,3,6 \\ &\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k}) \end{aligned}
The equation will be $\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})$

Straight Line in Space Exercise 27.2 Question 19

Answer: The two lines are perpendicular to each other(shown)
Given: show that the lines $\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}$ and $\frac{x}{1}=\frac{y}{2}=\frac{3}{2}$are perpendicular to each other
Hint: for showing perpendicular $\vec{b}.\vec{c}=0$
Solution:
Direction $ratio_{1}=7\hat{i}-5\hat{j}+\hat{k}$
Direction $ratio_{2}=\hat{i}-2\hat{j}+3\hat{k}$
$\vec{b}.\vec{c}=0$
\begin{aligned} &\Rightarrow(7 \hat{i}-5 \hat{j}+\hat{k}) \cdot(\hat{i}+2 \hat{j}+3 \hat{k}) \\ &\Rightarrow 7-10+3=0 \end{aligned}
Therefore the two lines are perpendicular

Straight Line in Space Exercise 27.4 Question 12

Answer: the equation will be $\frac{x-2}{1}=\frac{y+1}{2}=\frac{z+1}{3}$ which is parallel to $\vec{r}=(2 \hat{i}-\hat{j}-\hat{k})+\lambda(\hat{i}+2 \hat{j}+3 \hat{k})$
Given: find the vector equation of the line passing through the point $(2,-1,-1)$ parallel to line $6x-2=3y+1=2z-2$
Hint: for vector equation, $\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}$
Solution: $6x-2=3y+1=2z-2$
$6(x-1 / 3)=3(y+1 / 3)=2(z-1) \\$
$=\frac{x-1 / 3}{1 / 6}=\frac{y+1 / 3}{1 / 3}=\frac{z-1}{1 / 2} \\$
$D R=1 / 6,1 / 3,1 / 2=1,2,3$
Now $\frac{x-2}{1} =\frac{y+1}{2}=\frac{z+1}{3} \\$
$\therefore \vec{b}=\hat{i}+2 \hat{j}+3 \hat{k} \\$
$\vec{r}=(2 \hat{i}-\hat{j}-\hat{k})+\lambda(\hat{i}+2 \hat{j}+3 \hat{k}) \\$

So, the vector equation will be
$\frac{x-2}{1}= \frac{y+1}{2}=\frac{z+1}{3}$

Straight Line in Space Exercise 27.2 Question 21

Answer : the value of the $\lambda$ will be $\frac{10}{7}$
Given: if the lines $\frac{x-1}{-3}=\frac{y-2}{2 \lambda}=\frac{z-3}{2} \text { and } \frac{x-1}{3 \lambda}=\frac{y-1}{1}=\frac{z-6}{5}$ are perpendicular find the value of $\lambda$
Hint: Because lines are perpendicular
So $\vec{a}.\vec{b}=0$
Solution: $\frac{x-1}{-3}=\frac{y-2}{2 \lambda}=\frac{z-3}{2} \text { and } \frac{x-1}{3 \lambda}=\frac{y-1}{1}=\frac{z-6}{5}$ are perpendicular that means they =’$0$
So, $\cos \theta =0$
\begin{aligned} &=(3 \lambda)(-3)+(2 \lambda)(1)+(2 \times 5)=0 \\\\ &=-9 \lambda+2 \lambda+10=0 \\\\ &=-7 \lambda=-10 \\\\ &\lambda=\frac{10}{7} \end{aligned}
so the value will be $\frac{10}{7}$

Straight Line in Space Exercise 27.2 Question 22

Answer : the angle cannot be made between AB and CD because both are parallel to each other
Given: the coordinates of the points A,B,C,D be $(1,2,3);(4,5,7);(-4,3,-3);(2,9,2)$ respectively. Find angles between the lines AB and CD
Hint: $\vec{CD}=2\vec{AB}$
Solution:
\begin{aligned} &\vec{A}=(\hat{i}+2 \hat{j}+3 \hat{k}) \\ &\vec{B}=(4 \hat{i}+5 \hat{j}+7 \hat{k}) \\ &\vec{C}=(-4 \hat{i}+3 \hat{j}-6 \hat{k}) \\ \end{aligned}\begin{aligned} &\vec{D}=(2 \hat{i}+9 \hat{j}+2 \hat{k}) \\ &\overrightarrow{A B}=3 \hat{i}+3 \hat{j}+4 \hat{k} \\ &\overrightarrow{C D}=6 \hat{i}+6 \hat{j}+8 \hat{k} \\ &\therefore C D=2 \overrightarrow{A B} \\ &=2(3 \hat{i}+3 \hat{j}+4 \hat{k}) \\ &=6 \hat{i}+6 \hat{j}+8 \hat{k} \end{aligned}

CD and AB are parallel. So it cannot make any angle

Straight Line in Space Exercise 27.2 Question 23

Answer: the value $\lambda$ will be 1
Given: find the value of $\lambda$ so that the following lines are perpendicular to each other;
\begin{aligned} &\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} \\ \end{aligned}
Hint: \begin{aligned} &\cos \theta=0 \\ \end{aligned} because lines are perpendicular.
Solution: \begin{aligned} &\frac{x-5}{5 \lambda+2}=\frac{y-2}{-5}=\frac{z-1}{1} \\ \end{aligned}
Therefore \begin{aligned} &\cos \theta=0 \\ \end{aligned}
\begin{aligned} &\quad=(5 \lambda+2)(1)+(-5)(2 \lambda)+(3)(1)=0 \\ \end{aligned}
\begin{aligned} &=5 \lambda+2-10 \lambda+3=0 \\ &=5 \lambda=5 \\ &=\lambda=1 \end{aligned}
so the answer will be ‘$1$

Straight Line in Space Exercise 27.2 Question 24

Answer : the direction cosines will be $\frac{2}{7},\frac{3}{7},\frac{-6}{7}$ and the equation will be $\vec{r}=\left ( \hat{i}-2\hat{j}+3\hat{k} \right )+\lambda \left ( 2\hat{i}+3\hat{j}-6\hat{k} \right )$
Given: find the direction cosines of the lines $\frac{x+2}{2}=\frac{2y-7}{6}=\frac{5-z}{6}$ and also find the vector equation of the line through the point $(-1,2,3)$ and parallel to the given line
Hint:
Solution: $\frac{x+2}{2}=\frac{y-7/2}{3}=\frac{z-5}{-6}$
Direction ratio $=2,3,-6$
Direction cosines = $\frac{2}{\sqrt{49}}=\frac{3}{\sqrt{49}}=\frac{6}{\sqrt{49}}$
$\mathrm{DC}=\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, \frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}},=\frac{2}{7}, \frac{3}{7}, \frac{-6}{7}, \frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}$
Now, vector along the line; $2\hat{i}+3\hat{j}-6\hat{k}$$\vec{r}=\hat{i}-2\hat{j}+3\hat{k} +\lambda \left ( 2\hat{i}+3\hat{j}-6\hat{k} \right )$

$\vec{r}=\hat{i}-2\hat{j}+3\hat{k} +\lambda \left ( 2\hat{i}+3\hat{j}-6\hat{k} \right )$ and cosines will be $\frac{2}{7},\frac{3}{7},\frac{-6}{7}$

Straight Line in Space Exercise 27.2 Question 25

Answer: the value of k will be ‘$2$
Given: find the value of k so that the lines $x-y=kz \: and \: \: x-2y=2y+1 = -z+1$ are perpendicular to each other.
Hint: when two lines are perpendicular then dot product will be equal to ‘$0$
Solution: from equ (i)
$x=y=kz=\frac{x}{1}=\frac{y}{-1}=\frac{3}{1/k}$
From equation (ii)
$\frac{x-2}{1}=\frac{y+1/2}{1/2}=\frac{z-1}{-1}$
Since two straight lines are perpendicular to each other then
\begin{aligned} &a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0 \\\\ &(1)+(-1 / 2)+(-1 / k)=0 \\\\ &\frac{2-1}{2}=\frac{1}{k} \\\\ &1 / 2=1 / \mathrm{k} \\\\ &\mathrm{K} \quad=2 \end{aligned}

So the answer will be ‘$2$

Straight Line in Space Exercise 27.2 Question 26

Answer: the required vector equation is $\vec{r}=(\hat{i}+\hat{j}+\hat{k})+\lambda(-4 \hat{i}+4 \hat{j}-\hat{k})$

Given: find the vector and Cartesian equation of the line which is perpendicular to the lines with the equation
$\frac{x+2}{1}=\frac{y-3}{2}=\frac{z+1}{2}and \: \frac{x-1}{2}=\frac{y-2}{3}$
= $\frac{z-3}{4}$ and passes through the points $(1,1,1)$. Also find the angle between the given lines
Hint: for showing perpendicular dot product must be equal to ‘$0$
Solution: let the cartesian equation of the line passing through the point $(1,1,1)$ will be
$\frac{x-1}{a}=\frac{y-1}{b}=\frac{z-1}{c}$
Now parallel vectors are $\bar{b_{1}},\bar{b_{2}},\bar{b_{3}}$
From the question,
\begin{aligned} &\vec{b}_{1}=\hat{a} \hat{i}+b \hat{j}+c \hat{k} \\ &\overrightarrow{b_{2}}=\hat{i}+2 \hat{j}+4 \hat{k} \text { Here } \\ &\overrightarrow{b_{3}}=2 \hat{i}+3 \hat{j}+4 \hat{k} \end{aligned}
\begin{aligned} &a+2 b+4 c=0 \ldots .(\mathrm{lV}) \\ &2 a+3 b+4 c=0 \ldots \ldots(\mathrm{V}) \end{aligned}
Solving (iv) and (v) we get
\begin{aligned} &\frac{a}{8-12}=\frac{b}{8-4}=\frac{c}{3-4} \\ &\therefore \frac{a}{-4}=\frac{b}{4}=\frac{c}{-1}=\lambda \end{aligned}
Therefore $a=4\lambda ,b=4\lambda ,c=-\lambda$
Now putting the above makes in equation;
\begin{aligned} &\frac{x-1}{-4 \lambda}=\frac{y-1}{4 \lambda}=\frac{z-1}{-\lambda} \\ &\frac{x-1}{-4}=\frac{y-1}{4}=\frac{z-1}{-1} \end{aligned} hence the vector equation is $\vec{r}=(\hat{i}+\hat{j}+\hat{k})+\lambda(-4 \hat{i}+4 \hat{j}-\hat{k})$.

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RD Sharma Chapter-wise Solutions

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