RD Sharma Class 12 Exercise 27.1 Straight Line in Space Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 27.1 Straight Line in Space Solutions Maths - Download PDF Free Online

Edited By Lovekush kumar saini | Updated on Jan 24, 2022 08:09 PM IST

RD Sharma books are one of the best materials available for CBSE students for maths. CBSE schools use them all over the country, and most students prefer them as they are very informative. As maths is a vast subject that deals with many concepts, it is essential for students to refer to detailed material. This is where RD Sharma's class 12th exercise 27.1 comes in to help.

RD Sharma solutions of Straight line in space exercise 27.1 is an informative material that helps students get a better insight on the concepts. RD Sharma class 12th exercise 27.1 consists of 18 questions that are of Level-1 difficulty and cover most of the fundamentals of the chapter. Students can complete this exercise fairly quickly as their difficulty is low.

## Straight Line in Space Excercise: 27.1

Straight Line in Space exercise 27.1 question 1

\begin{aligned} &\vec{r}=5 \hat{\imath}+2 \hat{\jmath}-4 \hat{k}+\lambda(3 \hat{\imath}+2 \hat{\jmath}-8 \hat{k}): \\ &\frac{x-5}{3}=\frac{y-2}{2}=\frac{z+4}{-8} \end{aligned}
Hint :-
$\overrightarrow{r}=\overrightarrow{a}+\lambda\overrightarrow{b}$
Given :-
Line is through point (5 , 2 , -4) and parallel to vector $3\hat{i}+2\hat{j}-8\hat{k}$
Solution :-
Position vector of (5 , 2 , -4) is
\begin{aligned} &\vec{a}=5 \hat{\imath}+2 \hat{\jmath}-4 \hat{k} \\ &\text { and } \vec{b}=3 \hat{\imath}+2 \hat{\jmath}-8 \hat{k} \end{aligned}
Direction Ratio = (3 , 2 , -8 )
Vector equation of line which passes through (5 , 2 , -4 ) and parallel to vector $3\hat{i}+2\hat{j}-8\hat{k}$ is \begin{aligned} &\vec{a}=5 \hat{\imath}+2 \hat{\jmath}-4 \hat{k} \\ &\text { and } \vec{b}=3 \hat{\imath}+2 \hat{\jmath}-8 \hat{k} \end{aligned} , where ? is constant parameter
$=(5 \hat{\imath}+2 \hat{\jmath}-4 \hat{k})+\lambda(3 \hat{\imath}+2 \hat{\jmath}-8 \hat{k})$
Cartesian Equation :-
$\begin{gathered} \frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c} \\ \Rightarrow \frac{x-5}{3}=\frac{y-2}{2}=\frac{z+4}{-8} \end{gathered}$

Straight Line in Space exercise 27.1 question 2

$\vec{r}=(-\hat{\imath}+2 \hat{k})+\lambda(4 \hat{\imath}+4 \hat{\jmath}+4 \hat{k})$
Hint :-
$\vec{r}=\overrightarrow{a}+\lambda(\overrightarrow{b}-\overrightarrow{a})$
Given :-
Points (-1 , 0 , 2) and (3 , 4 , 6)
Solution :-
Position Vector of the point ( -1 , 0 , 2)
$\vec{a}=-\hat{i}+2\hat{k}$
Position Vector of the point (3 , 4 , 6)
$\vec{b}=-3\hat{i}+4\hat{j}+6\hat{k}$
Now equation of line passing through the point where position vector are a and b
\begin{aligned} \vec{r} &=\vec{a}+\lambda(\vec{b}-\vec{a}) \\ \vec{r} &=(-\hat{\imath}+2 \hat{k})+\lambda(3 \hat{\imath}+4 \widehat{\jmath}+6 \hat{k}+\hat{\imath}-2 \hat{k}) \\ &=(-\hat{\imath}+2 \hat{k})+\lambda(4 \hat{\imath}+4 \hat{\jmath}+4 \hat{k}) \end{aligned}

straight line in space exercise 27.1 question 3

\begin{aligned} &\vec{r}=5 \hat{\imath}-\hat{2} \hat{\jmath}+4 \hat{k}+\lambda(2 \hat{\imath}-\hat{\jmath}+3 \hat{k}) \\ &\frac{x-5}{2}=\frac{y+2}{-1}=\frac{z-4}{3} \end{aligned}
Hint :-
$\vec{r}=\vec{a}+\lambda \vec{b}$
Given :-
Line passes through (5 , -2 , 4) and parallel to $2 \hat{\imath}-\hat{\jmath}+3 \hat{k}$
Solution :-
Position Vector of the point (5 , -2 , 4) is
$\vec{a}=5 \hat{\imath}-\hat{2} \hat{\jmath}+4 \hat{k}$ and
parallel to $\overrightarrow{b}=2 \hat{\imath}-\hat{\jmath}+3 \hat{k}$
Direction ratio = (2 , -1 , 3)
Vector equation of the line passes through the point (5 , -2 , 4) and parallel to $5 \hat{\imath}-\hat{2} \hat{\jmath}+4 \hat{k}$
$\vec{r}=\vec{a}+\lambda \vec{b}$
$=5 \hat{\imath}-\hat{2} \hat{\jmath}+4 \hat{k}+\lambda(2 \hat{\imath}-\hat{\jmath}+3 \hat{k}) \\$, where ? is constant
And Cartisian form
\begin{aligned} &\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c} \\ &\frac{x-5}{2}=\frac{y+2}{-1}=\frac{z-4}{3} \end{aligned}

Straight Line in Space exercise 27.1 question 4

\begin{aligned} &\vec{r}=(2 \hat{\imath}-3 \hat{\jmath}+4 \hat{k})+\lambda(3 \hat{\imath}+4 \hat{\jmath}-5 \hat{k}) \\ &\frac{x-2}{3}=\frac{y+3}{4}=\frac{z-4}{-5} \end{aligned}
Hint :-
$\vec{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$
Given :-
Position vector of the point $=2 \hat{\imath}-3 \hat{\jmath}+4 \hat{k}$ and in the direction of $3 \hat{\imath}+4 \hat{\jmath}-5 \hat{k}$
Solution :-
Position vector of point
$a=2 \hat{\imath}-3 \hat{\jmath}+4 \hat{k}$ , point = ( 2 , -3 , 4)
And position vector of parallel line(in the direction)
$b=3 \hat{\imath}+4 \hat{\jmath}-5 \hat{k}$ , Direction ratio = (3 , 4 , -5 )
Vector Equation :-
$\vec{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$ , where ? is constant parameter
\begin{aligned} &=(2 \hat{\imath}-3 \hat{\jmath}+4 \hat{k})+\lambda(3 \hat{\imath}+4 \hat{\jmath}-5 \hat{k}) \\ \end{aligned}
Cartesian Equation :-
\begin{aligned} &\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c} \\ &\frac{x-2}{3}=\frac{y+3}{4}=\frac{z-4}{-5} \end{aligned}

Straight Line in Space exercise 27.1 question 5

\begin{aligned} &\vec{r}=(2 \hat{\imath}-3 \hat{\jmath}+4 \hat{k})+\lambda(\hat{\imath}-13 \hat{\jmath}+17 \hat{k}) \\ &\frac{x-2}{1}=\frac{y+3}{13}=\frac{z-4}{17} \end{aligned}
Hint: Position vector of the mid point
Given : ABCD is a Parallelogram
Solution:
We know that the position vector of the mid-point of \begin{aligned} &\vec{a} \end{aligned} and is \begin{aligned} &\vec{b} \end{aligned}is $\frac{\vec{a}+\vec{b}}{2}$
Let the position vector of D be $x\hat{i}+y\hat{j}+z\hat{k}$
Position vector of mid-point of A and C = Position vector of mid-point of B and D
\begin{aligned} &\frac{(4 \hat{\mathrm{i}}+5 \hat{\mathbf{j}}-10 \hat{\mathrm{k}})+(-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})}{2}=\frac{(2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})+(x\hat{\mathrm{i}}+ y\hat{\mathbf{j}}+z\hat{\mathrm{k}})}{2} \\ &\Rightarrow \frac{3}{2} \hat{\mathrm{i}}+\frac{7}{2} \hat{\mathrm{j}}-\frac{9}{2} \hat{\mathrm{k}}=\left(\frac{\mathrm{x}+2}{2}\right) \hat{\mathrm{i}}+\left(\frac{-3+\mathrm{y}}{2}\right) \hat{\mathrm{j}}+\left(\frac{4+\mathrm{z}}{2}\right) \hat{\mathrm{k}} \end{aligned}
Comparing the coefficients of $\hat{i},\hat{j}$ & $\hat{k}$ we get
\begin{aligned} &\left(\frac{x+2}{2}\right)=\frac{3}{2} \\ &\Rightarrow x=1 \\ &\frac{-3+y}{2}=\frac{7}{2} \\ &\Rightarrow y=10 \\ &\frac{4+z}{2}=-\frac{9}{2} \\ &\Rightarrow z=-13 \end{aligned}
Position vector of point $D=\hat{i}+10\hat{j}-13\hat{k}$. The vector equation of line BD passing through the points with position vectors
$\overrightarrow{\mathrm{a}}(\mathrm{B}) and \overrightarrow{\mathrm{b}}(\mathrm{D})$
$\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}})$
Here,
\begin{aligned} &\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathrm{k}} \\ &\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+10 \hat{\mathrm{j}}-13 \hat{\mathrm{k}} \end{aligned}
Vector equation of the required line is
\begin{aligned} &\overrightarrow{\mathrm{r}}=(2 \hat{\mathrm{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathrm{k}})+\lambda\{(\hat{\mathrm{i}}+10 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})-(2 \hat{\mathrm{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathrm{k}})\} \\ &\Rightarrow \overrightarrow{\mathrm{r}}=(2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})+\lambda(-\hat{\mathrm{i}}+13 \hat{\mathbf{j}}-17 \hat{\mathrm{k}}) \end{aligned} ………….. (1)
Here, $\lambda$is a parameter.
Reducing (1) to Cartesian form, we get
\begin{aligned} &x \hat{\mathbf{i}}+y \hat{\mathrm{j}}+\mathrm{z} \hat{\mathbf{k}}=(2 \hat{\mathrm{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathrm{k}})+\lambda(-\hat{\mathrm{i}}+13 \hat{\mathrm{j}}-17 \hat{\mathrm{k}})[\text { Putting } \overrightarrow{\mathrm{r}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}} \text { in }(1)] \\ &\Rightarrow \mathrm{x} \hat{\mathbf{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}=(2-\lambda) \hat{\mathrm{i}}+(-3+13 \lambda) \hat{\mathrm{j}}+(4-17 \lambda) \hat{\mathrm{k}} \end{aligned}
Comparing the coefficients of $\hat{i},\hat{j}$ & $\hat{k}$ we get
\begin{aligned} &x=2-\lambda, y=-3+13 \lambda, \mathrm{z}=4-17 \lambda \\ &\Rightarrow \frac{x-2}{-1}=\lambda, \frac{y+3}{13}=\lambda, \frac{z-4}{-17}=\lambda \\ &\Rightarrow \frac{x-2}{-1}=\frac{y+3}{13}=\frac{z-4}{-17}=\lambda \\ &\Rightarrow \frac{x-2}{1}=\frac{y+3}{-13}=\frac{z-4}{17}=-\lambda \end{aligned}
Hence, the Cartesian form of (1) is
$\frac{x-2}{1}=\frac{y+3}{-13}=\frac{z-4}{17}$

Straight Line in Space exercise 27.1 question 6

\begin{aligned} &\vec{r}=(\hat{\imath}+2 \hat{\jmath}-\hat{k})+\lambda(\hat{\imath}-\hat{\jmath}+2 \hat{k}) \\ &\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z+1}{2} \end{aligned}
Hint :-
\begin{aligned} &\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a}) \\ &\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}} \end{aligned}
Given :-
Two points A (1 , 2 , -1) and B ( 2 , 1 , 1)
Solution :-
Position vector of A (1 , 2 , -1)
$\vec{a}=\hat{\imath}-\hat{\jmath}+2 \hat{k}$
Position vector of B (2 , 1 , 1)
\begin{aligned} &\vec{b}=2 \hat{\imath}+\hat{\jmath}+\hat{k} \\ &\vec{b}-\vec{a}=\hat{\imath}-\hat{\jmath}+2 \hat{k} \end{aligned}
Vector Equation passes through A (1 , 2 , -1) and B ( 2 , 1 , 1)
\begin{aligned} &\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a}) \\ \end{aligned} , where ? is parameter
$=\left ( \hat{\imath}-2\hat{\jmath}+ \hat{k} \right )+\lambda \left ( \hat{\imath}-\hat{\jmath}+2 \hat{k} \right )$
Cartisian Equation passes through A (1 , 2 , -1) and B ( 2 , 1 , 1)
\begin{aligned} &\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}} \\ &\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z+1}{2} \end{aligned}

Straight Line in Space exercise 27.1 question 7

\begin{aligned} &\vec{r}=(\hat{\imath}+2 \hat{\jmath}+3 \hat{k})+\lambda(\hat{\imath}-2 \hat{\jmath}+3 \hat{k}) \\ &\frac{x-1}{1}=\frac{y-2}{-2}=\frac{x-3}{3} \end{aligned}
Hint :-
\begin{aligned} &\vec{r}=\overrightarrow{a}+\lambda \overrightarrow{b} \end{aligned}
Given :-
The line passes through (1 , 2 , 3) and parallel to the vector $\hat{\imath}+2 \hat{\jmath}+3 \hat{k}$
Solution :-
Position vector of (1 , 2 , 3)$\Rightarrow \overrightarrow{a}=\left ( \hat{\imath}+2 \hat{\jmath}+3 \hat{k} \right )$
$\overrightarrow{b}=\hat{\imath}-2 \hat{\jmath}+3 \hat{k}$
Direction Ratio (1 , -2 , 3)
Vector equation of the line passes through (1 , 2 , 3) and
parallel to $\hat{\imath}-2 \hat{\jmath}+3 \hat{k}$
\begin{aligned} &\vec{r}=\overrightarrow{a}+\lambda \overrightarrow{b} \end{aligned} , where ? is parameter
$=\left ( \hat{\imath}+2 \hat{\jmath}+3 \hat{k} \right )+\lambda \left ( \hat{\imath}-2 \hat{\jmath}+3 \hat{k} \right )$
Cartesian Equation
\begin{aligned} &\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c} \\ &\frac{x-1}{1}=\frac{y-2}{-2}=\frac{x-3}{3} \end{aligned}

Straight Line in Space exercise 27.1 question 8

$(2 \hat{\imath}-\hat{\jmath}+\hat{k})+\lambda(2 \hat{\imath}+7 \hat{\jmath}-3 \hat{k})$
Hint :-
$\overrightarrow{a}+\lambda \overrightarrow{b}$
Given :-
The line passes through (2 , -1 , 1) and parallel to equation $\frac{x-3}{2}=\frac{y+1}{7}=\frac{z-2}{-3}$
Solution :-
P.V of (2 , -1 , 1) $\Rightarrow \overrightarrow{a}=2 \hat{\imath}-\hat{\jmath}+\hat{k}$
D.R of $\frac{x-3}{2}=\frac{y+1}{7}=\frac{z-2}{-3}$ is (2 , 7 , -3)
$\therefore \overrightarrow{b}=2\hat{i}+7\hat{j}-3\hat{k}$
Vector Equation
$\overrightarrow{r}=$$\overrightarrow{a}+\lambda \overrightarrow{b}$ , where ? is parameter
$=$ $(2 \hat{\imath}-\hat{\jmath}+\hat{k})+\lambda(2 \hat{\imath}+7 \hat{\jmath}-3 \hat{k})$

Straight Line in Space exercise 27.1 question 9

$(5 \hat{\imath}-4 \hat{\jmath}+6 \hat{k})+\lambda(3 \hat{\imath}+7 \hat{\jmath}+2 \hat{k})$
Hint :-
$\overrightarrow{r}=x \hat{\imath}-y \hat{\jmath}+z \hat{k}$
Given :-
\begin{aligned} &\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2} \end{aligned}
Solution :-
\begin{aligned} &\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}=\lambda(\text { say }) \\ &\therefore x=3 \lambda+5, \quad y=7 \lambda-4, \quad z=2 \lambda+6 \end{aligned}
We know that
\begin{aligned} \vec{r} &=x \hat{\imath}-y \hat{\jmath}+z \hat{k} \\ &=(3 \lambda+5) \hat{\imath}+(7 \lambda-4) \hat{\jmath}+(2 \lambda+6) \hat{k} \\ &=(5 \hat{\imath}-4 \hat{\jmath}+6 \hat{k})+\lambda(3 \hat{\imath}+7 \hat{\jmath}+2 \hat{k}) \end{aligned}….where ? is constant parameter.

Straight Line in Space exercise 27.1 question 10

\begin{aligned} &\frac{x-1}{1}=\frac{y+1}{2}=\frac{z-2}{-2} \\ &\vec{r}=(\hat{\imath}-\hat{\jmath}+2 \hat{k})+\lambda(\hat{\imath}+2 \hat{\jmath}-2 \hat{k}) \end{aligned}\\
Hint :-
$\vec{r}=a+\lambda b$
Given :-
The line passing through (1 , -1 , 2) and parallel to the line whose equation $\frac{x-3}{1}=\frac{y-1}{2}=\frac{z+1}{-2}$
Solution :-
Cartesian Equation of a line passing through a point (1,-1,2) and parallel to line
$\frac{x-3}{1}=\frac{y-1}{2}=\frac{z+1}{-2} is$
$\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}$
$\left ( x_{1} ,y_{1},z_{1}\right )$ is the point and a,b,c are the direction ratios of the parallel line.)
$\frac{x-1}{1}=\frac{y+1}{2}=\frac{z-2}{-2}$is the required cartesian equation of the line.
Let, $\frac{x-1}{1}=\frac{y+1}{2}=\frac{z-2}{-2}=\lambda \text { (say) }$
We know that
\begin{aligned} \vec{r} &=x \hat{\imath}-y \hat{\jmath}+z \hat{k} \\ &=(\lambda+1) \hat{\imath}+(2 \lambda-1) \hat{\jmath}+(-2 \lambda+2) \hat{k} \\ &=(\hat{\imath}-\hat{\jmath}+2 \hat{k})+\lambda(\hat{\imath}+2 \hat{\jmath}-2 \hat{k}) \end{aligned}is the required vector equation.

Straight Line in Space exercise 27.1 question 11

$\frac{-2}{7}, \frac{6}{7},-\frac{3}{7} ; \vec{r}=(4 \hat{\imath}+\hat{k})+\lambda(-2 \hat{\imath}+6 \hat{\jmath}-3 \hat{k})$
Hint :-
$\vec{r}=x \hat{\imath}-y \hat{\jmath}+z \hat{k}$
Given :-
$\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$
Solution :-
\begin{aligned} & \frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3} \\ \Rightarrow & \frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3} \end{aligned}
Direction ratios are (-2 , 6 , -3)
So, its directional cosines will be
$\frac{-2}{\sqrt{(-2)^{2}+6^{2}+(-3)^{2}}}, \frac{6}{\sqrt{(-2)^{2}+6^{2}+(-3)^{2}}}, \frac{-3}{\sqrt{(-2)^{2}+6^{2}+(-3)^{2}}} \\$
$=\frac{-2}{7}, \frac{6}{7},-\frac{3}{7}$
$Let \frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3}=\lambda (say)$
$\Rightarrow x=-2 \lambda+4 \quad, \quad y=6 \lambda, \quad z=-3 \lambda+1$
We know that
$\vec{r}=x \hat{\imath}+y \hat{\jmath}+z \hat{k} \\$
$=(-2 \lambda+4) \hat{\imath}+(6 \lambda) \hat{\jmath}+(-3 \lambda+1) \hat{k} \\$
$=(4 \hat{\imath}+\hat{k})+\lambda(-2 \hat{\imath}+6 \hat{\jmath}-3 \hat{k})$
Where ? is constant parameter.

Straight Line in Space exercise 27.1 question 12

\begin{aligned} &\frac{x-b}{a}=\frac{y-0}{1}=\frac{z-d}{c} \\ &\vec{r}=(b \hat{\imath}+d \hat{k})+\lambda(a \hat{\imath}+\hat{\jmath}+c \hat{k}) \end{aligned}
(a , 1 , c)
Hint :-
Take $x=ay+b$ and $z=cy+d$ and find y
Given :-
$x=ay+b$ and $z=cy+d$
Solution :-
$x=ay+b$ and $z=cy+d$
? $x-b=ay$ $cy=z-d$
? $y=\frac{x-b}{a}$ ----- (1) ? $y=\frac{z-d}{c}$ ----- (2)
By (1) and (2)
\begin{aligned} &\frac{x-b}{a}=y=\frac{z-d}{c} \\ &\frac{x-b}{a}=\frac{y-0}{1}=\frac{z-d}{c} \end{aligned}
D.R = ( a , 1 , c)
Let $\frac{x-b}{a}=y=\frac{z-d}{c}=\lambda \left ( say \right )$
$\Rightarrow x=a \lambda+b \quad, \quad y=\lambda \quad, \quad z=c \lambda+d$
We know that
$\vec{r} =x \hat{\imath}+y \hat{\jmath}+z \hat{k} \\$
$=(a \lambda+b) \hat{\imath}+\lambda \hat{\jmath}+(c \lambda+d) \hat{k}$
$=(b \hat{\imath}+d \hat{k})+\lambda(a \hat{\imath}+\hat{\jmath}+c \hat{k})$

Straight Line in Space exercise 27.1 question 13

\begin{aligned} &\frac{x-1}{1}=\frac{y+2}{2}=\frac{z+3}{-2} \\ &\vec{r}=(\hat{\imath}-2 \hat{\jmath}-3 \hat{k})+\lambda(\hat{\imath}+2 \hat{\jmath}-2 \hat{k}) \end{aligned}
Hint :-
$\vec{r}=x\hat{\imath}-y \hat{\jmath}+z \hat{k}$
Given :-
The line passes through the point $\hat{\imath}-2 \hat{\jmath}-3 \hat{k}$ and parallel to the line joining with position vectors $\hat{\imath}- \hat{\jmath}+4 \hat{k}$ and $2\hat{\imath}+ \hat{\jmath}+2 \hat{k}$
Solution :-
The points of position vector $\hat{\imath}- \hat{\jmath}+4 \hat{k}$ and $2\hat{\imath}+ \hat{\jmath}+2 \hat{k}$ are P(1 , -1 , 4) and Q(2 , 1 , 2)
D.R of PQ =Q-P= (1 , 2 , -2) , $\overrightarrow{b}=\hat{\imath}+2 \hat{\jmath}-2 \hat{k}$
Points of P.V $\overrightarrow{b}=\hat{\imath}-2 \hat{\jmath}-3 \hat{k}$ are (1 , -2 , -3)
The Cartesian equation of the line having the point (1 , -2 , -3) and
D.R (1 , 2 , -2) are
$\frac{x-1}{1}=\frac{y+2}{2}=\frac{z+3}{-2}$
And vector equation are
\begin{aligned} \vec{r} &=\vec{a}+\lambda \vec{b} \\ &=(\hat{\imath}-2 \hat{\jmath}-3 \hat{k})+\lambda(\hat{\imath}+2 \hat{\jmath}-2 \hat{k}) \end{aligned}
Where $\lambda$ is constant parameter.

Straight Line in Space exercise 27.1 question 14

$\left ( x,y,z \right )=\left ( -2,-1,3 \right ) or \left ( 4,3,7 \right )$
Hint :-
Distance Formula : $\sqrt{\left(x_{1-} x_{2}\right)^{2}+\left(y_{1-} y_{2}\right)^{2}+\left(z_{1-} z_{2}\right)^{2}}$
Given :-
Point are P (1 , 3 , 3)
And line are $\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}$
Solution :-
$\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}$ -------- (1)
Let the point be $x_{1},y_{1},z_{1}$
$\therefore \frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}=\lambda \left ( say \right )$
$x_{1}=3\lambda -2$ , $y_{1}=2\lambda -1$ , $z_{1}=2\lambda +3$
The distance of $\left ( x_{1},y_{1},z_{1} \right )$ and P (1 , 3 , 3) = 5
$\sqrt{(3 \lambda-2-1)^{2}+(2 \lambda-1-3)^{2}+(2 \lambda+3-3)^{2}}=5$
Squaring both sides
\begin{aligned} &(3 \lambda-3)^{2}+(2 \lambda-4)^{2}+(2 \lambda)^{2}=25 \\ &\Rightarrow 9 \lambda^{2}-18 \lambda+9+4 \lambda^{2}-16 \lambda+16+4 \lambda^{2}=25 \\ &\Rightarrow 17 \lambda^{2}-34 \lambda=0 \\ &\Rightarrow 17 \lambda(\lambda-2)=0 \\ &\lambda=0, \lambda=2 \end{aligned}
∴ when $\lambda =0$;$x=-2$ , $y=-1$ , $z=3$
& when $\lambda =2$;$x=4$ , $y=3$ , $z=7$
$\left ( x,y,z \right )=\left ( -2,-1,3 \right ) or \left ( 4,3,7 \right )$

Straight Line in Space exercise 27.1 question 15

Answer :- The points are collinear
Hint :-
$\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}$
Given :-
P.V are $2 \hat{\imath}+3 \hat{\jmath}, \hat{\imath}+2 \widehat{\jmath}+3 \hat{k}, 7 \hat{\imath}+9 \hat{k}$
Solution :-
The points of P.V $2 \hat{\imath}+3 \hat{\jmath}, \hat{\imath}+2 \widehat{\jmath}+3 \hat{k}, 7 \hat{\imath}+9 \hat{k}$ are (-2 , 3 , 0) , (1 , 2 , 3) , (7 , 0 , 9)
The equation of the line passing through the point (-2,3,0) and (1 , 2 , 3) are
\begin{aligned} &\frac{x+2}{1+2}=\frac{y-3}{2-3}=\frac{z-0}{3-0} \\ &\Rightarrow \frac{x+2}{3}=\frac{y-3}{-1}=\frac{z-0}{3} \end{aligned} ----- (1)
Putting (7 , 0 , 9) in (1) , we get
\begin{aligned} &\frac{7+2}{3}=\frac{0-3}{-1}=\frac{9}{3} \\ \end{aligned}
$\Rightarrow 3,3,3$
Since ( 7 , 0 , 9) satisfying equation (1)
Hence it is collinear.

Straight Line in Space exercise 27.1 question 16

\begin{aligned} &\frac{x-1}{-2}=\frac{y-2}{14}=\frac{z-3}{3} \\ &(\hat{\imath}+2 \hat{\jmath}+3 \hat{k})+\lambda(\widehat{-2} \imath+14 \hat{\jmath}+3 \hat{k}) \end{aligned}
Hint :-
$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$
Given :-
The line passes through the point ( 1 , 2 , 3) and parallel to the line
\begin{aligned} &\frac{-x-2}{1}=\frac{y+3}{7}=\frac{2z-6}{3} \\ \end{aligned}
Solution :-
The P.V of the point (1 , 2 , 3) will be $\overrightarrow{r}=\hat{i}+2\hat{j}+3\hat{k}$
The line \begin{aligned} &\frac{-x-2}{1}=\frac{y+3}{7}=\frac{2z-6}{3} \\ \end{aligned} can be written as \begin{aligned} &\frac{x+2}{-1}=\frac{y+3}{7}=\frac{z-3}{\frac{3}{2}} \\ \end{aligned} ---------- (1)
So the direction ratios can be $-1,7,\frac{3}{2} or -2,14,3$ $PV=\widehat{-2}i+14\hat{j}+3\hat{k}$
Cartesian equation of the line passing through ( 1 , 2 , 3) and parallel to (1)
\begin{aligned} &\frac{x-1}{-2}=\frac{y-2}{14}=\frac{z-3}{3} \\ \end{aligned}
Vector Equation
$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$
=\begin{aligned} &(\hat{\imath}+2 \hat{\jmath}+3 \hat{k})+\lambda(\widehat{-2} \imath+14 \hat{\jmath}+3 \hat{k}) \end{aligned}

Straight Line in Space exercise 27.1 question 17

$\frac{x+\frac{1}{3}}{2}=\frac{y-\frac{1}{3}}{1}=\frac{z-1}{-6}$…cartesian equation of line
$\left(-\frac{1}{3} \hat{\imath}+\frac{1}{3} \hat{\jmath}+\hat{k}\right)+\lambda(2 \hat{\imath}+\hat{\jmath}-\widehat{6 k})$…vector equation of line
$\left(\frac{-1}{3}, \frac{1}{3}, 1\right)$…the fixed point through which the line passes ;
$(2,1,-6)$….direction ratios of the line
Hint :-
$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$
Given :-
$3x+1=6y-2=1-z$
Solution :-
The given line can be written as $\frac{x+\frac{1}{3}}{\frac{1}{3}}=\frac{y-\frac{2}{6}}{\frac{1}{6}}=\frac{z-1}{-1}$
$\Rightarrow \frac{x+\frac{1}{3}}{2}=\frac{y-\frac{1}{3}}{1}=\frac{z-1}{-6}$
Thus the line passes through the point $\left(\frac{-1}{3}, \frac{1}{3}, 1\right)$
vector equation of the point will be $\left(\overrightarrow{a}=-\frac{1}{3} \hat{\imath}+\frac{1}{3} \hat{\jmath}+\hat{k}\right)$
The direction ratios are proportional to $(2,1,-6)$
and its vector equation will be $\overrightarrow{b}=2 \hat{\imath}+\hat{\jmath}-\widehat{6 k}$
Vector Equation of line
$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$
$=\left(-\frac{1}{3} \hat{\imath}+\frac{1}{3} \hat{\jmath}+\hat{k}\right)+\lambda(2 \hat{\imath}+\hat{\jmath}-\widehat{6 k})$

Straight Line in Space exercise 27.1 question 18

$(\hat{\imath}+2 \hat{\jmath}-\hat{k})+\lambda(7 \hat{\imath}-5 \hat{\jmath}+\hat{k})$
Hint :-
$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$
Given :-
$A(1,2,-1) \& 5 x-25=14-7 y=35 z$
Solution :-
$5 x-25=14-7 y=35 z \\ \Rightarrow \frac{x-5}{\frac{1}{5}}=\frac{y-2}{\frac{-1}{7}}=\frac{z}{\frac{1}{35}} \\ \Rightarrow \frac{x-5}{7}=\frac{y-2}{-5}=\frac{z}{1}$ -------------- (1)
Direction ratios of (1) will be ( 7 , -5 , 1 )
hence, $\vec{b}=7 \hat{\imath}-5 \hat{\jmath}+\hat{k}$
P.V of $A(1,2,-1)=>\vec{a}=\hat{\imath}+2 \hat{\jmath}-\hat{k} \\$
$\vec{r}=\vec{a}+\lambda \vec{b} \\ \quad$
$=(\hat{\imath}+2 \hat{\jmath}-\hat{k})+\lambda(7 \hat{\imath}-5 \hat{\jmath}+\hat{k})$

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RD Sharma Chapter wise Solutions

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