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

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

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

Lovekush kumar sainiUpdated on 25 Jan 2022, 09:48 AM IST

Planning a schedule that every 12th grader must follow to keep in touch with their ongoing syllabus and preparing for the board exams is essential. This task can be accomplished easily only with the help of the right reference books. Many students struggle with mathematics to arrive at the right solution in various concepts. If any student faces challenges in the Straight Line in Space topic, they can refer to the RD Sharma Class 12th Chapter 27 VSA book.

RD Sharma Class 12 Solutions Chapter27VSA Straight line in space - Other Exercise

Straight Line in Space Excercise: 27 VSA

Straight Line in Space Exercise Very Short Answer Question 1

Answer:
Required answer is $\lambda \hat{i}$
Hint:
Use properties of vector
Given:
Cartesian and vector equation of x-axis
Solution:
Since x-axis passes through the point $\left ( 0,0,0 \right )$ having position vector
$\vec{a}=a \hat{i}+0 \hat{j}+0 \hat{k}$ and parallel to the vector $\vec{b}=1 \hat{i}+0 \hat{j}+0 \hat{k}$ having direction ratios proportional to 1, 0, 0.
The Cartesian equation of x-axis is
$\begin{aligned} &\frac{x-0}{1}=\frac{y-0}{0}=\frac{z-0}{0} \\ & \end{aligned}$
$\frac{x}{1}=\frac{y}{0}=\frac{z}{0}$
Also its vector equation is
$\begin{aligned} &\vec{r}=\vec{a}+\lambda \vec{b} \\ & \end{aligned}$
$=0 \hat{i}+0 \hat{j}+0 \hat{k}+\lambda(\hat{i}+0 \hat{j}+0 \hat{k}) \\$
$=\lambda \hat{i}$

Straight Line in Space Exercise Very Short Answer Question 2

Answer:
Required answer is $\lambda \hat{j}$
Hint:
Use properties of vector
Given:
Cartesian and vector equation of y-axis
Solution:
Since y-axis passes through the point $\left ( 0,0,0 \right )$ having position vector
$\vec{a}=0 \hat{i}+0 \hat{j}+0 \hat{k}$ and is parallel to the vector $\vec{b}=0 \hat{i}+1 \hat{j}+0 \hat{k}$ having direction ratios proportional to 0, 1, 0
The Cartesian equation of y-axis is
$\begin{aligned} &\frac{x-0}{0}=\frac{y-0}{1}=\frac{z-0}{0} \\ & \end{aligned}$
$\frac{x}{0}=\frac{y}{1}=\frac{z}{0}$
Also the vector equation is
$\begin{aligned} &\vec{r}=\vec{a}+\lambda \vec{b} \\ \end{aligned}$
$=0 \hat{i}+0 \hat{j}+0 \hat{k}+\lambda(0 \hat{i}+\hat{j}+0 \hat{k}) \\$
$=\lambda \hat{j}$

Straight Line in Space Exercise Very Short Answer Question 3

Answer:
Required answer is $\lambda \hat{k}$
Hint:
Use properties of vector
Given:
Cartesian and vector equation of z-axis
Solution:
Since z-axis passes through the point $\left ( 0,0,0 \right )$ having position vector
$\vec{a}=0 \hat{i}+0 \hat{j}+0 \hat{k}$ and is parallel to the vector $\vec{b}=0 \hat{i}+0 \hat{j}+1 \hat{k}$ having direction ratios proportional to 0, 0, 1
The Cartesian of z-axis is
$\begin{aligned} &\frac{x-0}{0}=\frac{y-0}{0}=\frac{z-0}{1} \\ & \end{aligned}$
$\frac{x}{0}=\frac{y}{0}=\frac{z}{1}$
Also its vector equation
$\begin{aligned} &\vec{r}=\vec{a}+\lambda \vec{b} \\ & \end{aligned}$
$=0 \hat{i}+0 \hat{j}+0 \hat{k}+\lambda(0 \hat{i}+0 \hat{j}+\hat{k}) \\$
$=\lambda \hat{k}$

Striaght Line in Space Exercise Very Short Answer question 4

Answer:
Required answer is $\vec{r}=\vec{\alpha}+\lambda \vec{\beta}$
Hint:
Use properties of vector
Given:
$\overrightarrow{\alpha }$ Vector is parallel to $\overrightarrow{\beta }$ vector
Solution:
The vector equation of the line passing through the point having position vector $\overrightarrow{\alpha }$ and parallel to the vector $\overrightarrow{\beta }$ is $\vec{r}=\vec{\alpha}+\lambda \vec{\beta}$

Straight Line in Space Exercise Very Short Answer question 5

Answer:
Required answer is $1,-7,2$
Hint:
Use properties of vector
Given:
$\frac{2 x-1}{2}=\frac{4-y}{7}=\frac{z+1}{2}$
Solution:
We have
$\frac{2 x-1}{2}=\frac{4-y}{7}=\frac{z+1}{2}$
Now the equation of the line AB can be rewritten as,
$\frac{x-\frac{1}{2}}{1}=\frac{y-4}{-7}=\frac{z+1}{2}$
The direction ratios of the line parallel to AB proportional to $1,-7,2$

Straight Line in Space Exercise Very Short Answer question 6

Answer:
Required answer is $\left(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\right)$
Hint:
Use properties of vector
Given:
$6 x-2=3 y+1=2 z-4$
Solution:
We have
$6 x-2=3 y+1=2 z-4$
The equation of the given line can be rewritten as,
$\begin{aligned} &\frac{x-\frac{1}{3}}{\frac{1}{6}}=\frac{y+\frac{1}{3}}{\frac{1}{3}}=\frac{z-2}{\frac{1}{2}} \\ & \end{aligned}$
$=\frac{x-\frac{1}{3}}{1}=\frac{y+\frac{1}{3}}{2}=\frac{z-2}{3}$
The direction ratios of the line parallel to AB are proportional to $1,2,3$
The direction cosines of the line parallel to AB are proportional to
$\begin{aligned} &\frac{1}{\sqrt{1^{2}+2^{2}+3^{2}}}, \frac{2}{\sqrt{1^{2}+2^{2}+3^{2}}}, \frac{3}{\sqrt{1^{2}+2^{2}+3^{2}}} \\ & \end{aligned}$
$=\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}$

Straight Line in Space Exercise Very Short Answer Question 7

Answer:
Required answer is $\frac{4}{\sqrt{29}}, \frac{-3}{\sqrt{29}}, \frac{2}{\sqrt{29}}$
Hint:
Use properties of vector
Given:
$\frac{x-2}{2}=\frac{2 y-5}{-3}, z=2$
Solution:
We have,
$\frac{x-2}{2}=\frac{2 y-5}{-3}, z=2$
The equation of the given line can be rewritten as,
$\begin{aligned} &\frac{x-2}{2}=\frac{y-\frac{5}{2}}{-\frac{3}{2}}=\frac{z-2}{1} \\ & \end{aligned}$
$\frac{x-2}{4}=\frac{y-\frac{5}{2}}{-3}=\frac{z-2}{2}$
The direction ratios of the line parallel to AB are proportional to 4, -3, 1
The direction cosines of the line parallel to AB proportional to
$\begin{aligned} &\frac{4}{\sqrt{4^{2}+(-3)^{2}+2^{2}}}, \frac{-3}{\sqrt{4^{2}+(-3)^{2}+2^{2}}}, \frac{2}{\sqrt{4^{2}+(-3)^{2}+2^{2}}} \\ & \end{aligned}$
$=\frac{4}{\sqrt{29}}, \frac{-3}{\sqrt{29}}, \frac{2}{\sqrt{29}}$

Straight Line in Space Exercise Very Short Answer Question 8

Answer:
The given line is perpendicular to z-axis
Hint:
Use properties of vector
Given:
$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-1}{0}$
Solution:
We have
$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-1}{0}$
The given line is parallel to the vector $\vec{b}=3 \hat{i}+4 \hat{j}+0 \hat{k}$
Let, $x \hat{i}+y \hat{j}+z \hat{k}$ be perpendicular to the given line
Now, $3 x+4 y+0 z=0$
Direction ratios are $\left(a_{1}, b_{1}, c_{1}\right)=(3,4,0)$
Check perpendicular condition with x axis
Direction ratio of x axis is $\left(a_{2}, b_{2}, c_{2}\right)=(1,0,0)$
$\begin{aligned} &a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2} \\ & \end{aligned}$
$=3(1)+4(0)+0(0) \\$
$=3$
$\neq 0$ Hence x axis is not perpendicular
Check perpendicular condition with y axis
Direction ratio of y axis is $\left(a_{2}, b_{2}, c_{2}\right)=(0,1,0)$
$\begin{aligned} &a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2} \\ & \end{aligned}$
$=3(0)+4(1)+0(0) \\$
$=4 \neq 0$
Hence y axis is not perpendicular
Check perpendicular condition with z axis
Direction ratio of z axis is $\left(a_{2}, b_{2}, c_{2}\right)=(0,0,1)$
$\begin{aligned} &a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2} \\ & \end{aligned}$
$=3(0)+4(0)+0(1) \\$
$=0$
Hence, the given line is perpendicular to the z-axis

Straight Line in Space Exercise Very Short Answer Question 9

Answer:
Required answer is
Hint:
Use properties of vector
Given:
$\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z-2}{1} \text { and } \frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{3}$
Solution:
We have,
$\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z-2}{1} \text { and } \frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{3}$
The given lines are parallel to the vectors
$\begin{aligned} &\vec{b}_{1}=7 \hat{i}-5 \hat{j}+\hat{k} \\ \end{aligned}$
$\overline{b_{2}}=\hat{i}+2 \hat{j}+3 \hat{k}$
Let $\theta$ be the angle between the given lines
Now,
$\begin{aligned} &\cos \theta=\frac{\vec{b}_{1}\cdot \overline{b_{2}}}{\left|\vec{b}_{1}\right|\left|\overrightarrow{b_{2}}\right|} \\ & \end{aligned}$$\begin{aligned} &\cos \theta=\frac{\vec{b}_{1}\cdot \overline{b_{2}}}{\left|\vec{b}_{1}\right|\left|\overrightarrow{b_{2}}\right|} \\ & \end{aligned}$
$=\frac{(7 \hat{i}-5 \hat{j}+\hat{k}) \cdot(\hat{i}+2 \hat{j}+3 \hat{k})}{\sqrt{7^{2}+(-5)^{2}+1^{2}} \cdot \sqrt{1^{2}+2^{2}+3^{2}}}$
$\begin{aligned} &=\frac{7-10+3}{\sqrt{49+25+1} \cdot \sqrt{1+4+9}} \\ & \end{aligned}$
$=0 \\$
$\therefore \theta=\frac{\pi}{2}$

Straight Line of Space Excercise Very Short Anwer Question 10

Answer:
Required answer is $\frac{3}{7}, \frac{2}{7}, \frac{-6}{7}$
Hint:
Use equation of a line in space.
Given:
Cartesian equations are $2 x=3 y=-z$
Solution:
We have,
$2 x=3 y=-z$
The equation of the given line can be rewritten as,
$\begin{aligned} &\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{-1} \\ & \end{aligned}$
$\frac{x}{3}=\frac{y}{2}=\frac{z}{-6}$
The direction ratios of the line parallel to AB are proportional to $3,2,-6$
The direction cosines of the line parallel to AB are proportional to
$\begin{aligned} &\frac{3}{\sqrt{3^{2}+2^{2}+(-6)^{2}}}, \frac{2}{\sqrt{3^{2}+2^{2}+(-6)^{2}}}, \frac{-6}{\sqrt{3^{2}+2^{2}+(-6)^{2}}} \\ & \end{aligned}$
$=\frac{3}{7}, \frac{2}{7}, \frac{-6}{7}$

Straight Line of Space Excercise Very Short Anwer Question 11

Answer:
Required answer is $\theta = \frac{\pi}{2}$
Hint:
Use formula $\cos \theta=\frac{\vec{b}_{1}\cdot \vec{b}_{2}}{\left|\vec{b}_{1}\right|\left|\vec{b}_{2}\right|}$
Given:
$2 x=3 y=-z \text { and } 6 x=-y=-4 z$
Solution:
We have,
$2 x=3 y=-z \text { and } 6 x=-y=-4 z$
The given lines can be rewritten as,
$\frac{x}{3}=\frac{y}{2}=\frac{z}{-6} \text { and } \frac{x}{2}=\frac{y}{-12}=\frac{z}{-3}$
These lines are parallel to vectors $\vec{b}_{1}=3 \hat{i}+2 \hat{j}-6 \hat{k} \text { and } \overline{b_{2}}=2 \hat{i}-12 \hat{j}-3 \hat{k}$
Let $\theta$ be the angle between these lines.
Now,
$\begin{aligned} &\cos \theta=\frac{\vec{b}_{1} \cdot\overline{b_{2}}}{\left|\vec{b}_{1}\right|\left|\overline{b_{2}}\right|} \\ & \end{aligned}$
$\begin{aligned} &=\frac{6-24+18}{\sqrt{9+4+36} \cdot \sqrt{4+144+9}}=0 \\ & \end{aligned}$
$\theta=\frac{\pi}{2}$

Straight Line of Space Excercise Very Short Anwer Question 12

Answer:
Required answer is $\lambda = \frac{-10}{7}$
Hint:
Use equation of line in space
Given:
$\frac{x-3}{-3}=\frac{y+2}{2 \lambda}=\frac{z+4}{2} \text { And } \frac{x+1}{3 \lambda}=\frac{y-2}{1}=\frac{z+6}{-5}$
Solution:
We have,
$\frac{x-3}{-3}=\frac{y+2}{2 \lambda}=\frac{z+4}{2} \text { And } \frac{x+1}{3 \lambda}=\frac{y-2}{1}=\frac{z+6}{-5}$
The given lines are parallel to vectors
$\vec{b}_{1}=-3 \hat{i}+2 \lambda \hat{j}+2 \hat{k} \text { and } \vec{b_{2}}=3 \lambda \hat{i}+\hat{j}-5 \hat{k}$
Now, for $\vec{b}_{1} \perp \vec{b_{2}}$ we must have,
$\begin{aligned} &\vec{b}_{1} \cdot \vec{b_{2}}=0 \\ & \end{aligned}$
$(-3 \hat{i}+2 \lambda \hat{j}+2 \hat{k}) \cdot(3 \lambda \hat{i}+\hat{j}-5 \hat{k})=0 \\$
$-7 \lambda-10=0 \\$
$\lambda=-\frac{10}{7}$

Straight Line in Space Exercise Very Short Answer Question 13

Answer:
Required answer is $d=\left|\frac{\left(\vec{a}_{2}-\vec{a}_{1}\right) \cdot \vec{b}}{|\vec{b}|}\right|$
Hint:
Use equation of line in space
Given:
$\vec{r}=\vec{a}_{1}+\lambda \vec{b} \text { and } \vec{r}=\vec{a}_{2}+\mu \vec{b}$
Solution:
The shortest distance between the parallel lines
$\begin{aligned} &\vec{r}=\vec{a}_{1}+\lambda \vec{b} \\ & \end{aligned}$
$\vec{r}=\vec{a}_{2}+\mu \vec{b}$ is given by,
$d=\left|\frac{\left(\vec{a}_{2}-\vec{a}_{1}\right) \cdot \vec{b}}{|\vec{b}|}\right|$

Straight Line in Space Exercise Very Short Answer Question 14

Answer:
Required answer is $\left(\vec{a}_{2}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)=0$
Hint:
Use formula $d=\left|\frac{\left(\vec{a}_{2}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)}{\left|\vec{b}_{1} \times \vec{b}_{2}\right|}\right|$
Given:
$\vec{r}=\vec{a}_{1}+\lambda \vec{b}_{1} \text { And } \vec{r}=\vec{a}_{2}+\mu \vec{b}_{2}$
Solution:
The shortest distance between the lines $\vec{r}=\vec{a}_{1}+\lambda \vec{b}_{1} \text { And } \vec{r}=\vec{a}_{2}+\mu \vec{b}_{2}$ is given by
$d=\left|\frac{\left(\vec{a}_{2}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)}{\left|\vec{b}_{1} \times \vec{b}_{2}\right|}\right|$
For the lines to be intersecting, $d=0$
$\left|\frac{\left(\vec{a}_{2}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)}{\left|\vec{b}_{1} \times \vec{b}_{2}\right|}\right|= 0$
$\left(\vec{a}_{2}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)=0$

Straight Line in Space Exercise Very Short Answer Question 15

Answer:
Required answer is $\frac{\sqrt{3}}{\sqrt{55}}, \frac{4}{\sqrt{55}}, \frac{6}{\sqrt{55}}$
Hint:
Use equation of line in space
Given:
$\frac{2 x-1}{\sqrt{3}}=\frac{y+2}{2}=\frac{z-3}{3}$
Solution:
We have,
$\frac{2 x-1}{\sqrt{3}}=\frac{y+2}{2}=\frac{z-3}{3}$
The equation of the given line can be rewritten as,
$\begin{aligned} &\frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{y+2}{2}=\frac{z-3}{3} \\ & \end{aligned}$
$\frac{x-\frac{1}{2}}{\sqrt{3}}=\frac{y+2}{4}=\frac{z-3}{6}$
The direction ratios of the line parallel to AB are proportional to $\sqrt{3},4,6$
The direction cosines of the line parallel to AB are proportional to,
$\begin{aligned} &\frac{\sqrt{3}}{\sqrt{(\sqrt{3})^{2}+4^{2}+6^{2}}}, \frac{4}{\sqrt{(\sqrt{3})^{2}+4^{2}+6^{2}}}, \frac{6}{\sqrt{(\sqrt{3})^{2}+4^{2}+6^{2}}} \\ & \end{aligned}$
$=\frac{\sqrt{3}}{\sqrt{55}}, \frac{4}{\sqrt{55}}, \frac{6}{\sqrt{55}}$

Straight Line in Space Exercise Very Short Answer question 16

Answer:
Required answer is $-1,-2,4$
Hint:
Use equation of line in space
Given:
$\frac{3-x}{1}=\frac{y+2}{-2}=\frac{z-5}{4}$
Solution:
We have,
$\frac{3-x}{1}=\frac{y+2}{-2}=\frac{z-5}{4}$
The equation of the line AB can be rewritten as,
$\frac{x-3}{-1}=\frac{y+2}{-2}=\frac{z-5}{4}$
Thus, the direction ratios of the line parallel to AB are proportional to $-1,-2,4$

Straight Line in Space Exercise Very Short Answer question 17

Answer:
Required answer is $\vec{r}=5 \hat{i}-4 \hat{j}+6 \hat{k}+\lambda(3 \hat{i}+7 \hat{j}+2 \hat{k})$
Hint:
Use properties of vector
Given:
$\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$
Solution:
We have,
$\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$
The given line passes through the point $\left ( 5,-4,6 \right )$ and has direction ratios proportional to $3,7,2$
Vector equation of the given line passing through the point having position vector
$\vec{a}=5 \hat{i}-4 \hat{j}+6 \hat{k}$ and parallel to a vector $\vec{b}=3 \hat{i}+7 \hat{j}+2 \hat{k}$ is
$\begin{aligned} &\vec{r}=\vec{a}+\lambda \vec{b} \\ & \end{aligned}$
$\vec{r}=5 \hat{i}-4 \hat{j}+6 \hat{k}+\lambda(3 \hat{i}+7 \hat{j}+2 \hat{k})$

Straight Line in Space Exercise Very Short Answer question 18

Answer:
Required answer is $\frac{-1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}$
Hint:
Use the equation of a line in space.
Given:
$\frac{4-x}{3}=\frac{y+3}{3}=\frac{z+2}{6}$
Solution:
We have,
$\frac{4-x}{3}=\frac{y+3}{3}=\frac{z+2}{6}$
The equation of the given line can be rewritten as,
$\frac{x-4}{-3}=\frac{y+3}{3}=\frac{z+2}{6}$
The direction ratios of the line parallel to the given line are proportional to $-3,3,6$
The direction cosines of the line parallel to the given line are proportional to
$\begin{aligned} &\frac{-3}{\sqrt{(-3)^{2}+3^{2}+6^{2}}}, \frac{3}{\sqrt{(-3)^{2}+3^{2}+6^{2}}}, \frac{6}{\sqrt{(-3)^{2}+3^{2}+6^{2}}} \\ & \end{aligned}$
$=\frac{-1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}$

Straight Line in Space Exercise Very Short Answer question 19

Answer: $\left ( -2,4,-5 \right )$
Required answer is
Hint:
Use the equation of a line in space.
Given:
Point $\frac{x+2}{3}=\frac{y-4}{-5}=\frac{z+5}{6}$
Line $\frac{x+3}{3}=\frac{y-4}{-5}=\frac{z+8}{6}$
Solution:
The equation of the given line is
$\frac{x+3}{3}=\frac{y-4}{-5}=\frac{z+8}{6}$
Since the required line is parallel to the given line, the direction ratios of the required line are proportional to $3,-5,6$
Hence the Cartesian equation of the line passing through the point $\left ( -2,4,-5 \right )$ and parallel to a vector having direction ratios proportional to $3,-5,6$ is
$\frac{x+2}{3}=\frac{y-4}{-5}=\frac{z+5}{6}$

Straight Line in Space Exercise Very Short Answer Question 20

Answer:
Required answer is $\cos ^{-1} \frac{19}{21}$
Hint:
Use the equation of a line in space.
Given:
$\begin{aligned} &\vec{r}=(2 \hat{i}-5 \hat{j}+\hat{k})+\lambda(3 \hat{i}+2 \hat{j}+6 \hat{k}) \\ & \end{aligned}$
$\vec{r}=(7 \hat{i}-6 \hat{k})+\mu(\hat{i}+2 \hat{j}+2 k)$
Solution:
Let $\theta$ be the angle between the given lines
The given lines are parallel to the vectors $\vec{b}_{1}=3 \hat{i}+2 \hat{j}+6 \hat{k} \text { and } \vec{b}_{2}=\hat{i}+2 \hat{j}+2 \hat{k}$ respectively
So, the angle $\theta$ between the given lines are given by
$\begin{aligned} &\cos \theta=\frac{\vec{b}_{1} \cdot \vec{b}_{2}}{\left|\vec{b}_{1}\right|\left|\vec{b}_{2}\right|} \\ & \end{aligned}$
$=\frac{(3 \hat{i}+2 \hat{j}+6 \hat{k}) \cdot(\hat{i}+2 \hat{j}+2 \hat{k})}{\sqrt{3^{2}+2^{2}+6^{2}} \sqrt{1^{2}+2^{2}+2^{2}}}$
$\begin{aligned} &=\frac{19}{\sqrt{49} \sqrt{9}} \\ & \end{aligned}$
$=\frac{19}{21} \\$
$\theta=\cos ^{-1} \frac{19}{21}$

Straight Line in Space Exercise Very Short Answer Question 21

Answer:
Required angle is $\frac{\pi}{2}$
Hint:
Use equation of a line in space.
Given:
$2 x=3 y=-z \text { and } 6 x=-y=-4 z$
Solution:
The equation of the given lines can be rewritten as,
$\frac{x}{3}=\frac{y}{2}=\frac{z}{-6} \text { and } \frac{x}{2}=\frac{y}{-12}=\frac{z}{-3}$
We know that angle between the lines
$\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}} \text { and } \frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}$ is given by,
$\cos \theta=a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}$
Let,
$\theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
Let $\theta$ be the angle between the given lines
$\begin{aligned} &\cos \theta=\frac{3 \times 2+2 \times(-12)+(-6) \times(-3)}{\sqrt{3^{2}+2^{2}+(-6)^{2}} \sqrt{2^{2}+(-12)^{2}+(-3)^{2}}} \\ & \end{aligned}$
$=\frac{6-24+18}{\sqrt{49} \sqrt{157}}$
$\begin{aligned} &=0 \\ & \end{aligned}$
$\theta=\frac{\pi}{2}$
Thus, the angle between the given lines is $\frac{\pi}{2}$

Straight Line in Space Exercise Very Short Answer Question 22

Answer:
Required answer is $\vec{r}=(3 \hat{i}+4 \hat{j}+5 \hat{k})+\lambda(2 \hat{i}+2 \hat{j}-3 \hat{k})$
Hint:
Use equation of a line in space.
Given:
Point $\left ( 3,4,5 \right )$ and parallel vector $2 \hat{i}+2 \hat{j}-3 \hat{k}$
Solution:
The line is passing through the point $\left ( 3,4,5 \right )$ through the point having position vector is $\vec{a}=3 \hat{i}+4 \hat{j}+5 \hat{k}$ and parallel to the vector $\vec{b}=2 \hat{i}+2 \hat{j}-3 \hat{k}$
$\begin{aligned} &\vec{r}=\vec{a}+\lambda \vec{b} \\ & \end{aligned}$
$\vec{r}=(3 \hat{i}+4 \hat{j}+5 \hat{k})+\lambda(2 \hat{i}+2 \hat{j}-3 \hat{k})$

Also, see,

RD Sharma Solutions Class 12 mathematics chapter 27 exercise 27.1
RD Sharma Solutions Class 12 mathematics chapter 27 exercise 27.2
RD Sharma Solutions Class 12 mathematics chapter 27 exercise 27.3
RD Sharma Solutions Class 12 mathematics chapter 27 exercise 27.4
RD Sharma Solutions Class 12 mathematics chapter 27 exercise 27.5
RD Sharma Solutions Class 12 mathematics chapter 27 exercise MCQ
RD Sharma Solutions Class 12 mathematics chapter 27 exercise FBQ

The number of exercises in class 12, chapter 27, Straight Line in Space, is five. The concepts in the ex 27.1 to ex 27.5 are included in the Very Short Answers part, where there are 22 questions in total. The topics from which the questions are asked include finding the angle between the lines, symmetric and unsymmetric, vector equations, cartesian equations, equation of a plane, equation of a line, direction cosines, conditions for lines to intersect, direction cosine, etc. The RD Sharma Class 12 Chapter 27 VSA solution guide directs the students on how these sums can be solved easily.

Students cannot contact their teachers or tutors every time a doubt strikes them; simple and complex doubts will also vanish when a glance is made into the RD Sharma Class 12th Chapter 27 VSA reference book. The various practice questions in this book will motivate the students to work out more sums that eventually sharpen their knowledge. Soon, they would cross their benchmark scores effortlessly due to the high confidence level they gained from the practice.

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RD Sharma Chapter-wise Solutions

Chapter 1: Relations

Chapter 2: Functions

Chapter 3: Inverse Trigonometric Functions

Chapter 4: Algebra of Matrices

Chapter 5: Determinants

Chapter 6: Adjoint and Inverse of a Matrix

Chapter 7: Solution of Simultaneous Linear Equations

Chapter 23: Continuity

Chapter 9: Differentiability

Chapter 10: Differentiation

Chapter 11: Higher Order Derivatives