RD Sharma Class 12 Exercise MCQ Straight line in space Solutions Maths

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

Edited By Lovekush kumar saini | Updated on Jan 25, 2022 09:48 AM IST

Class 12 students are stuck between completing the homework and preparing for their board exams. In mathematics, there are even possibilities to score grace marks for their steps in solving sums. But this is not the case in MCQs. The Multiple Choice Questions are easy as there are options provided, and this is why most of the students get confused while choosing the answer. RD Sharma solutions To help them prepare well to face these MCQs, the RD Sharma Class 12th Chapter 27 MCQ lends a helping hand.

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RD Sharma Class 12 Solutions Chapter27MCQ Straight line in space - Other Exercise

Straight Line in Space Excercise: 27 MCQ

Straight Line in Space exercise Multiple choice question 1

Answer: Correct answer is D.
Hint: Use vector dot product
Given: $\frac{x^{2}+1}{2}=\frac{y-2}{5}=\frac{z+3}{4} \text { and } \frac{x-1}{1}=\frac{y+2}{2}=\frac{z-3}{-3}$

Solution: The direction ratios of the given lines are proportional to $2,5,4$ and $1,2,-3$ . The given lines are parallel to the vectors $\Rightarrow \overrightarrow{b_{1}}=2 \hat{\imath}+5 \hat{\jmath}+4 \hat{k} \text { and } \overrightarrow{b_{2}}=\hat{\imath}+2 \hat{\jmath}-3 \hat{k}$
Let θ be the angle between the given lines.
Now,
$\begin{aligned} &\operatorname{Cos} \theta=\frac{\overrightarrow{b_{1}} \cdot \overrightarrow{b_{2}}}{\left|\overrightarrow{b_{1}}\right|\left|\overrightarrow{b_{2}}\right|} \\\\ &\operatorname{Cos} \theta=\frac{(2 \hat{\imath}+5 \hat{\jmath}+4 \hat{k}) \cdot(\hat{\imath}+2 \hat{\jmath}-3 \hat{k})}{\sqrt{2^{2}+5^{2}+4^{2}} \sqrt{1^{2}+2^{2}+(-3)^{2}}} \end{aligned}$
$\operatorname{Cos} \theta=\frac{2+10-12}{\sqrt{45} \sqrt{14}}$
$\operatorname{Cos} \theta=$
Hence, $\theta=\frac{\pi}{2}$
$\theta=90^{\circ}$
So, the correct option is (d) 90°

Straight Line in Space exercise Multiple choice question 2

Answer: Correct answer is A.
Hint: Use vector cross product.

Given: $: \frac{x}{1}=\frac{y}{2}=\frac{z}{3} \text { and } \frac{x-1}{-2}=\frac{y-2}{-4}=\frac{z-3}{-6}$

Solution: The equation of the given lines are
$\begin{aligned} &\frac{x}{1}=\frac{y}{2}=\frac{z}{3} \rightarrow(1) \\\\ &\text { and } \frac{x-1}{-2}=\frac{y-2}{-4}=\frac{z-3}{-6} \\\\ &\Rightarrow \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3} \rightarrow(2) \end{aligned}$
Thus the two lines are parallel to the vector $\vec{b}=\hat{\imath}+2 \hat{\jmath}+3 \hat{k}$ and pass through the points $(0,0,0)$ and $(1,2,3).$
Now,
$\begin{aligned} &\left(\overrightarrow{a_{1}}-\overrightarrow{a_{2}}\right) \times \vec{b} \\\\ &=(\hat{\imath}+2 \hat{\jmath}+3 \hat{k}) \times(\hat{\imath}+2 \hat{\jmath}+3 \hat{k}) \\\\ &=\overrightarrow{0} \\\\ &\Rightarrow(\vec{a} \times \vec{a}=\overrightarrow{0}) \end{aligned}$
Since the distance between the two parallel lines is 0, the given lines are coincident.
So, the correct option is (a) coincident.

Straight Line in Space exercise Multiple choice question 3

Answer: Correct Answer is A.
Hint: Use vector cross product.

Given: $\frac{x-7}{2}=\frac{y+17}{-3}=\frac{z-6}{-1} \text { and } \frac{x+5}{1}=\frac{y+3}{2}=\frac{z-4}{-2}$

Solution: The direction ratios of the given lines are proportional to $2,-3, 1$ and $1,2,-2.$
The given lines are parallel to the vectors
$\Rightarrow \overrightarrow{b_{1}}=2 \hat{\imath}-3 \hat{\jmath}+\hat{k} \operatorname{and} \overrightarrow{b_{2}}=\hat{\imath}+2 \hat{\jmath}-2 \hat{k}$
Vector perpendicular to the given two lines is
$\begin{aligned} &\Rightarrow \vec{b}=\overrightarrow{b_{1}} \times \overrightarrow{b_{2}} \\\\ &\Rightarrow\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & -3 & 1 \\ 1 & 2 & -2 \end{array}\right| \\\\ &\Rightarrow 4 \hat{\imath}+5 \hat{\jmath}+7 \hat{k} \end{aligned}$
Hence, the direction ratios of the line perpendicular to the given two lines are proportional to $4,5,7.$

Straight Line in Space exercise Multiple choice question 4

Answer: Correct answer is C.
Hint: Use vector dot product.

Given: $: \frac{x-1}{1}=\frac{y-1}{1}=\frac{z-1}{2} \text { and } \frac{x-1}{-\sqrt{3}-1}=\frac{y-1}{\sqrt{3}-1}=\frac{z-1}{4}$

Solution: The direction ratios of the given lines are proportional to $1 ,1, 2$ and - $-\sqrt{3}-1, \sqrt{3}-1,4$
The given lines are parallel to the vectors
$\Rightarrow \overrightarrow{b_{1}}=\hat{\imath}+\hat{\jmath}+2 \widehat{k} \text { and } \overrightarrow{b_{2}}=(-\sqrt{3}-1) \hat{\imath}+(\sqrt{3}-1) \hat{\jmath}+4 \hat{k}$
Let θ be the angle between the given lines.
Now,
$\begin{aligned} &\operatorname{Cos} \theta=\frac{\overrightarrow{b_{1}} \cdot \overrightarrow{b_{2}}}{\left|\overrightarrow{b_{1}}\right|\left|\overrightarrow{b_{2}}\right|} \\\\ &\operatorname{Cos} \theta=\frac{(\hat{\imath}+\hat{\jmath}+2 \hat{k}) \cdot\{(-\sqrt{3}-1) \hat{\imath}+(\sqrt{3}-1) \hat{\jmath}+4 \hat{k}\}}{\sqrt{1^{2}+1^{2}+2^{2}} \sqrt{(-\sqrt{3}-1)^{2}+(\sqrt{3}-1)^{2}+(4)^{2}}} \end{aligned}$
$\begin{aligned} &\operatorname{Cos} \theta=\frac{-\sqrt{3}-1+\sqrt{3}-1+8}{\sqrt{3} \sqrt{24}} \\\\ &\operatorname{Cos} \theta=\frac{6}{6 \sqrt{2}} \end{aligned}$
$\begin{aligned} &\operatorname{Cos} \theta=\frac{1}{\sqrt{2}} \\\\ &\theta=\frac{\pi}{3} \end{aligned}$
So, the correct option is $\text { (c) } \frac{\pi}{3}$

Straight Line in Space exercise Multiple choice question 5

Answer: Correct answer is A.
Hint: Solve Simultaneously.

Given: $x-y+z-5=0$ and $x-3y-6=0$

Solution: $x-y+z-5=0$ and $x-3y-6=0$
$\begin{gathered} \Rightarrow x-y+z-5=0 \\\\ x-3 y-6=0 \end{gathered}$
$\begin{aligned} &\Rightarrow x-y+z-5=0 \quad\dots(1)\\ &x=3 y+6\quad\dots(2) \end{aligned}$
From (1) and (2) we get,

$\begin{aligned} &3 y+6-y+z-5=0 \\\\ &2 y+z+1=0 \\\\ &y=\frac{-z-1}{2} \end{aligned}$
Also,
$\begin{aligned} &y=\frac{x-6}{3} \rightarrow \text { from }(2) \\\\ &\therefore \frac{x-6}{3}=y=\frac{-z-1}{2} \end{aligned}$

So, the given equation can be re-written as
$\frac{x-6}{3}=\frac{y}{1}=\frac{-z-1}{2}$
Hence the direction ratios of the given line are proportional to $3, 1, -2.$
So, the correct option is (a) $3, 1, -2.$

Straight Line in Space exercise Multiple choice question 6

Answer: Correct answer is A.
Hint: Use vector cross product.
Given: $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$
Solution: Since, We have $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$
Let point $(1,2,3)$ be $P$ and the point through which the line passes be $Q (6,7,7)\Rightarrow$ (Given)
Also, the line is parallel to the vector $\Rightarrow$
$\begin{aligned} &\vec{b}=3 \hat{\imath}+2 \hat{j}-2 \hat{k} \Rightarrow(\text { Given }) \\\\ &\overrightarrow{P Q}=5 \hat{\imath}+5 \hat{j}+4 \hat{k} \end{aligned}$
Now,
$\vec{b} \times \overrightarrow{P Q}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 3 & 2 & -2 \\ 5 & 5 & 4 \end{array}\right|=18 \hat{\imath}-22 \hat{\jmath}+5 \hat{k}$
$\begin{aligned} &\Rightarrow|\vec{b} \times \overrightarrow{P Q}|=\sqrt{18^{2}+(-22)^{2}+5^{2}} \\\\ &=\sqrt{324+484+25} \\\\ &=\sqrt{833} \end{aligned}$
$\begin{aligned} &\therefore d=\frac{|\vec{b} \times \overrightarrow{P Q}|}{|\vec{b}|} \\\\ &=\frac{\sqrt{833}}{\sqrt{17}} \\\\ &=\sqrt{49} \\\\ &=7 \end{aligned}$
So, the correct option is $(a) 7$

Straight Line in Space exercise Multiple choice question 7

Answer: Correct answer is C.
Hint: Use formula of vector addition.
Given: $a_{1} \hat{\imath}+a_{2} \hat{\jmath}+a_{3} \hat{k} \text { and } b_{1} i+b_{2} \hat{\jmath}+b_{3} \hat{k}$
Solution: Line between two given points $a_{1} \hat{\imath}+a_{2} \hat{\jmath}+a_{3} \hat{k} \text { and } b_{1} i+b_{2} \hat{\jmath}+b_{3} \hat{k} \text { is: }$

$\begin{aligned} &\Rightarrow a_{1} \hat{\imath}+a_{2} \hat{\jmath}+a_{3} \hat{k}+t\left(b_{1}-a_{1}\right) \hat{\imath}+\left(b_{2}-a_{2}\right) \hat{\jmath}+\left(b_{3}-3\right) \hat{k} \\\\ &\Rightarrow a_{1}(1-t) \hat{\imath}+a_{2}(1-t) \hat{\jmath}+a_{3}(1-t) \hat{k}+t\left(b_{1} \hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k}\right) \end{aligned}$
So, the correct option is (c)

Straight Line in Space exercise Multiple choice question 8

Answer: Correct answer is B.
Hint: Use trigonometric formulae
Given: $\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma$
Solution: $\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma=2\left(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma\right)-3$
using $\left(\cos 2 \theta=2 \cos ^{2} \theta-1\right)$
And as the question states: A line makes angle $\alpha ,\beta, \gamma$ with the axes respectively
$\begin{aligned} &\text { So, } \cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma=2\left(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma\right)-3 \\\\ &=2-3 \\\\ &=-1 \end{aligned}$
So, the correct option is $(b)-1$

Straight Line in Space exercise Multiple choice question 9

Answer: Correct answer is A.
Hint: Use properties of vector cosines.
Given: Direction ratio is proportional to 1,-3, 2
Solution: If the direction ratios of a line are proportional to $1,-3, 2$ then its direction cosines are also proportional to $1,-3, 2.$
So let directional cosines be $x,-3x,2x$
Now, sum of the squares of directional cosines =1 (property)
$\text { So, } x^{2}+9 x^{2}+4 x^{2}=1$
$\begin{aligned} &\Rightarrow x^{2}=\frac{1}{14} \\\\ &\Rightarrow x=\pm \frac{1}{\sqrt{14}} \end{aligned}$
So, direction cosines can be $\pm\left(\frac{1}{\sqrt{14}} \hat{\imath}-\frac{3}{\sqrt{14}} \hat{j}+\frac{2}{\sqrt{14}} \hat{k}\right)$
So, the correct option is (A)

Straight Line in Space exercise Multiple choice question 10

Answer: Correct answer is B.
Hint: $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1$
Given: Lines make angle $\frac{\pi}{3} \text { and } \frac{\pi}{4} \text { with } x-\text { axis and } y-\text { axis }$
Solution: If a line makes angles α,β and γ with the axes then,
$\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1 \rightarrow(1)$
Hence,
$\alpha=\frac{\pi}{3} \text { and } \beta=\frac{\pi}{4}$
Now,
$\begin{aligned} &\Rightarrow \cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1 \\\\ &\Rightarrow \cos ^{2} \frac{\pi}{3}+\cos ^{2} \frac{\pi}{4}+\cos ^{2} \gamma=1 \end{aligned}$
$\begin{aligned} &\Rightarrow \frac{1}{4}+\frac{1}{2}+\cos ^{2} \gamma=1 \\\\ &\Rightarrow \frac{3}{4}+\cos ^{2} \gamma=1 \end{aligned}$
$\begin{aligned} &\Rightarrow \cos \gamma=\frac{1}{2} \\\\ &\Rightarrow \gamma=\frac{\pi}{3} \end{aligned}$
So, the correct option is $(B) \frac{\pi}{3}$

Straight Line in Space exercise Multiple choice question 11

Answer: Correct answer is A.
Hint: Use vector dot product.
Given: x,y and z are 12, 4 and 3
Solution: Let the line segment be represented in vector form as $\vec{a}=a_{1} \vec{\imath}+a_{2} \vec{\jmath}+a_{3} \vec{k}$ vector along co-ordinate axes are $\vec{i}, \vec{j}, \vec{k}$ vector along given that projection of $\vec{a} \text { on } x-\text { axis is } 12$ , that with $y-\text { axis is } 4$ and
that with $z-\text { axis is } 3$ .

$\begin{aligned} &\Rightarrow \vec{a} \cdot \vec{l}=12 \\\\ &\Rightarrow a_{1}=12 \end{aligned}$
Similarly $a_{2}=4, a_{3}=3$
$\therefore \vec{a}=12 \vec{\imath}+4 \vec{\jmath}+3 \vec{k}$
The line segment of the length $|\vec{a}|=\sqrt{144+16+9} \Rightarrow 13$ and the direction cosines of the line segment are $<13, \frac{12}{13}, \frac{4}{13}, \frac{3}{13}>$

Straight Line in Space exercise Multiple choice question 12

Answer: Correct answer is A.
Hint: Use vector cross product.
Given: $\frac{x}{1}=\frac{y}{2}=\frac{z}{3} \text { and } \frac{x-1}{-2}=\frac{y-2}{-4}=\frac{z-3}{-6}$
Solution: The equations of the given lines are
$\Rightarrow \frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ ..........(1)
Direction ratios are 1,2,3
$\begin{aligned} &\frac{x-1}{-2}=\frac{y-2}{-4}=\frac{z-3}{-6} \\\\ &\Rightarrow \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3} \end{aligned}$ ............(2)
Direction ratios are 1,2,3
Since, direction ratios of the two lines are equal so, the given lines are parallel.
So, the correct option is (a) parallel.

Straight Line in Space exercise Multiple choice question 13

Answer: Correct answer is D
Hint: The given line is parallel to the vectors
Given: $\frac{x-3}{3}=\frac{y-2}{1}=\frac{z-1}{0}$
Solution: Given the equation is
$\frac{x-3}{3}=\frac{y-2}{1}=\frac{z-1}{0}$
Also the given line is parallel to the vector,
$\overrightarrow{b_{1}}=3 \hat{\imath}-\hat{\jmath}+0 \hat{k}$
Let $x \hat{\imath}+y \hat{\jmath}+z \hat{k}$ be perpendicular to the given line.
Now,
$3x+4y+Oz=0$
it is satisfied by the coordinates of z- axis, i.e, (0,0,1)
Hence, the given line is perpendicular to z – axis
So, the correct option is (d)

Straight Line in Space exercise Multiple choice question 14

Answer: Correct answer is D.
Hint: use vector Dot product

Given: $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1} \text { and } \frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$

Solution:
$\begin{aligned} &\vec{a}=3 \hat{\imath}+3 \hat{\jmath}+3 \hat{k} \quad, \vec{b}=3 \hat{\imath}-7 \hat{\jmath}+6 \hat{k} \quad, \vec{p}=3 \hat{\imath}-\hat{\jmath}+\hat{k} \\\\ &\vec{q}=-3 \hat{\imath}-2 \hat{\jmath}+4 \hat{k} \end{aligned}$
$\vec{p} \times \vec{q}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{array}\right|=-6 \hat{\imath}+15 \hat{\jmath}+3 \hat{k}$
$\begin{aligned} &\text { S. } D=\left|\frac{\mid \overrightarrow{(a)}-(\overrightarrow{b)}(\vec{p} x \vec{q})}{|\vec{p} x \vec{q}|}\right| \\\\ &=\left|\frac{(6 \hat{\imath}+15 \hat{\jmath}+3 \hat{k})}{\sqrt{36+225+9}}\right| \end{aligned}$
$\begin{aligned} &=\left|\frac{-36-225-9}{\sqrt{36+225+9}}\right| \\\\ &=\left|\frac{-270}{\sqrt{270}}\right| \\\\ &=\sqrt{270} \Rightarrow 3 \sqrt{30} \end{aligned}$

Straight Line in Space exercise Multiple choice question 15

Answer: Correct answer is A.
Hint: y-axis lies on xy plane and y²
Given: The equation of y-axis in space.
Solution: As -axis lies on xy plane and y²
So, its distance from xy and yz plane is 0.
∴ By basic definition of 3-D coordinate we can say that x-coordinate and z-coordinate =0
∴ Any point on y-axis has x and z-coordinate =0
∴x=0 and z=0 Can be considered as the equation of y-axis.
Hence, the correct option is (A)

Straight Line in Space exercise Multiple choice question 16

Answer: Correct answer is d.
Hint: Use Matrix Product

Given: $\frac{x-2}{1}=\frac{y-3}{1}=\frac{4-z}{k} \text { and } \frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{-2}$

Solution:
$\begin{aligned} &\Rightarrow \frac{x-2}{1}=\frac{y-3}{1}=\frac{4-z}{k} \ldots(i) \\ &\text { and } \frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{-2} \ldots(i i) \end{aligned}$
Obviously parallel vectors b₁ and b₂ of line (i) and respectively are:
$\begin{aligned} &\Rightarrow \overrightarrow{b_{1}}=-1 \hat{\imath}-2 k \hat{\jmath}+2 \hat{k} \\\\ &\Rightarrow \overrightarrow{b_{2}}=k \hat{\jmath}=2 \hat{\jmath}-2 \hat{k} \\\\ &\text { Lines }(i) \perp(i i) \Rightarrow \overrightarrow{b_{1}} \perp \overrightarrow{b_{2}} \end{aligned}$
$\begin{aligned} &\Rightarrow \overrightarrow{b_{1}} \cdot \overrightarrow{b_{2}}=0 \\\\ &\Rightarrow-3 k+2 k+10=0 \\\\ &\Rightarrow-5 k+10=0 \end{aligned}$
$\begin{aligned} &\Rightarrow k=\frac{-10}{-5} \\\\ &\Rightarrow=2(d) \end{aligned}$
So, the correct option is $(d)2$

Straight Line in Space exercise Multiple choice question 17

Answer: Correct answer is b.
Hint:First find the co-ordinates
Given: $(-1,3,2)$ and $(5,0,6)$
Solution: Equation of line passing through $(-1,3,2)$ and $(5,0,6.)$
$\begin{aligned} &\Rightarrow \frac{x+1}{5-(-1)}=\frac{y-3}{0-3}=\frac{z-2}{6-2} \\\\ &\Rightarrow \frac{x+1}{6}=\frac{y-3}{-3}=\frac{z-2}{4}=k \end{aligned}$
Any point on it
$P(6 k-1,-3 k+3,4 k+2)$
x-coordinate:
$\begin{aligned} &\Rightarrow 2=6 k-1 \\\\ &\Rightarrow k=\frac{1}{2} \end{aligned}$
z-coordinate:
$\begin{aligned} &\Rightarrow 4 k+2 \\\\ &\Rightarrow 4\left(\frac{1}{2}\right)+2 \\\\ &\Rightarrow 2+2 \\\\ &\Rightarrow 4 \end{aligned}$
So, the correct option is (b) 4.

Straight Line in Space exercise Multiple choice question 18

Answer: Correct answer is d.
Hint: Use vector cross product.
Given: $L_{1}: x=5, \quad \frac{y}{3-2}=\frac{z}{-2}, \quad L_{2}: x=2, \quad \frac{y}{-1}=\frac{z}{2-d}$
Solution:
$\begin{aligned} &x=5, \frac{y}{3-2}=\frac{z}{-2} \ldots \ldots \ldots . . . \text { (i) }\\ \\ &x=2, \frac{y}{-1}=\frac{z}{2-2}\ldots \ldots \ldots . . . \text { (ii) } \end{aligned}$
∴ Parallel vectors b₁ and b₂ of line (i) and (ii) are:
$\overrightarrow{b_{1}}=0 \hat{\imath}+(3-2) \hat{\jmath}-2 \hat{k}, \overrightarrow{b_{2}}=0 \hat{\imath}-\hat{\jmath}+(2-2) \hat{k}$
$\begin{aligned} &\text { Lines (i) } \perp(i i) \Rightarrow \overrightarrow{b_{1}} \perp \overrightarrow{b_{2}} \\\\ &\Rightarrow \overrightarrow{b_{1}} \overrightarrow{b_{2}}=0 \\\\ &\Rightarrow(0 \times 0) \hat{\imath}+\{(3-2)(-1)\} \hat{j}+\{(-2)(2-2)\} \hat{k}=0 \end{aligned}$
$\begin{aligned} &\Rightarrow d 3+2 x-4=0 \\\\ &\Rightarrow 3 x-7=0 \\\\ &\Rightarrow d=7 / 3 \end{aligned}$
And, hence the correct option is (D).

Chapter 27 of class 12 mathematics contains five exercises, ex 27.1 to ex 27.5. In addition to the sums given in these exercises, 18 multiple choice questions are given in the textbook. The concepts like finding the shortest distance between the lines, finding the x, y, z coordinates, projections of a line segment, direction ratios of perpendicular lines, equation of a plane and line passing through the given points, etc. The answers to these questions can be found easily using the RD Sharma Class 12 Chapter 27 MCQ guide. The students are advised to try the answers by themselves and recheck with the ones given in the solution guide.

It is essential to know the tricks and shortcuts to find the answers for the MCQs in an instant. The RD Sharma Class 12th Chapter 27 MCQ reference book contains all the possible methods in which these questions can be solved; students can use the ways they find easy to adapt. In addition, there are various practice questions to test the conceptual knowledge before tests and exams. All these questions are updated according to the latest NCERT syllabus. This makes the Class 12 RD Sharma Chapter 27 MCQ Solution book popular among CBSE school students.

The more MCQs are practiced, the more marks will be scored by the students. They can refer to the answers given in the RD Sharma Class 12 Solutions Straight Line in Space Chapter 27 MCQ book while doing homework or assignments. These solution books help the students in acquiring more marks in their public and JEE mains exams. Most of the students have a copy of the RD Sharma Class 12th Chapter 27 MCQ book to keep in touch with the practice.

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RD Sharma Chapter-wise Solutions

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