RD Sharma Solutions Class 12 Mathematics Chapter 26 MCQ

# RD Sharma Solutions Class 12 Mathematics Chapter 26 MCQ

Edited By Satyajeet Kumar | Updated on Jan 24, 2022 07:35 PM IST

RD Sharma Solutions is a pretty popular name that specializes in creating NCERT solutions. The RD Sharma class 12th exercise MCQ book especially has garnered a lot of praise for being a lifesaver for students in India. Students who are in class 12 should definitely practice the RD Sharma class 12 chapter 26 exercise MCQ to prepare for exams and test their maths skills at home.
RD Sharma class 12 solutions Direction Cosines & Direction Ratios MCQ should be utilized by all students attending maths for their board exams. Chapter 26 of the NCERT is titled Direction Cosines & Direction Ratios. RD Sharma Solutions The main concept covered in this chapter is finding the directional ratio of lines, direction cosines of line, Coordinates of the foot on the y-axis and collinear points. Exercise MCQ of this chapter has a total of 18 questions. The RD Sharma class 12th exercise MCQ will help the students in finding the right options for the multiple-choice questions.

## Direction Cosines and Direction Ratios Excercise: MCQ

### Directions Cosines and Direction Ratio Exercise Multiple Choice Question, question 1

Given: P(x,y,z) lie on xy-plane
Solution:

z-coordinate of every point on xy plane is zero
Therefore option (c) z=0 is correct for point P(x,y,z) on xy plane

Directions Cosines and Direction Ratio Exercise Multiple Choice Question, question 2

$y=0,z=0,x\neq 0$
Given: P(x,y,z) lie on x-axis except origin
Solution:

Both y and z coordinate on each point of the x-axis are equal to zero. The x-coordinate at origin is also equal to zero.
Therefore, y and z coordinates on each point of the x-axis, except the origin are equal to zero, hence option
$y=0,z=0,x\neq 0$
is correct.

Directions Cosines and Direction Ratio Exercise Multiple Choice Question, question 3

Hint: Distance between plane passing through points is the length of edge
Given: Points (5,7,9) and (2,3,7)
Solution: In geometry, parallelopiped is a three dimension figure with six parallelogram

$\leftarrow b\rightarrow$

Let P and Q be the points. So, all the plane passes through these points. Now distance between planes joining (a,0,0) and (b,0,0) is |b-a|. Similarly, we can generate 3 pairs of plane from each coordinate from these 2 points.
For x-coordinate, value of a is
a=|5-2|=3
For y-coordinate, value of b is
b=|7-3|=4
For z-coordinate value of c is
c=|9-7|=2
Therefore, the edges of the parallelopoid are 3,4,2. Hence option (d) all of these is correct.

Directions Cosines and Direction Ratio Exercise Multiple Choice Question, question 4

Hint: Distance between planes passing through points is the length of edge
Given: Points (2,3,5) and (5,9,7)
Solution: In geometry parallelopiped is a three dimension figure with six parallelogram

z
Let P and Q be the points. So, all the plane passes through these points. Now distance between planes joining (a,0,0) and (b,0,0) is |b-a|. Similarly, we can generate 3 pairs of plane from each coordinate from these 2 points.
For x-coordinate, value of a is
a=|2-5|=3
For y-coordinate, value of b is
b=|3-9|=6
For z-coordinate value of c is
c=|5-7|=2
The edges of parallelopiped are 3,6,2
Length of diagonal=
\begin{aligned} &\sqrt{a^{2}+b^{2}+c^{2}} \\ &=\sqrt{3^{2}+6^{2}+2^{2}} \\ &=\sqrt{9+36+4} \\ &=\sqrt{49} \\ &=7 \end{aligned}
Therefore, the length of diagonal is 7. Hence option (a) 7 is correct.

Directions Cosines and Direction Ratio Exercise Multiple Choice Question, question 5

Answer: (b) Externally in ratio 2:3
Hint: Use section formula
Given: Point(-1,3,4) and (2,-5,6)
To Find: Ratio in which xy plane divide the points
Solution:

Suppose xy plane divide the line segment joining the points O(-1,3,4) and Q(2,-5,6) in ratio k:1
Using section formula coordinates of the point of intersection are
$(x, y, z)=\left(\frac{k(2)+(-1)}{k+1}, \frac{k(-5)+3}{k+1}, \frac{k(6)+4}{k+1}\right)$
The z-coordinate of any point of the xy-plane is zero.
\begin{aligned} &\Rightarrow \frac{k(6)+4}{k+1}=0 \\ &\Rightarrow 6 k+4=0 \\ &\Rightarrow k=-\frac{4}{6} \end{aligned}
$k:1=-2:3$
Therefore, the xy plane divide the given points in 2:3 externally. Hence option(b) externally in the ratio 2:3 is correct.

Directions Cosines and Direction Ratio Exercise Multiple Choice Question, question 6

Hint: Use section formula
Given: x-coordinate of point P(x,y,z) on the join of Q(2,2,1) and R(5,1,-2) is 4
To Find: z-coordinate of point P(x,y,z)
Solution:

Suppose point P divide the line segment joining the points Q(2,2,1) and R(5,1,-2) in ratio t:1
Using section formula, coordinates of point P
$(x, y, z)=\left[\frac{t(5)+2}{t+1}, \frac{t(1)+2}{t+1}, \frac{t(-2)+1}{t+1}\right]$
x-coordinate of point P is 4
\begin{aligned} &\Rightarrow \frac{5 t+2}{t+1}=4 \\ &\Rightarrow 5 t+2=4(t+1) \\ &\Rightarrow t=2 \end{aligned}
Now z-coordinate will be
\begin{aligned} &z=\frac{t(-2)+1}{t+1} \\ &z=\frac{2(-2)+1}{t+1} \\ &z=\frac{-4+1}{2+1} \\ &z=\frac{-3}{3} \\ &z=-1 \end{aligned}
Therefore z coordinate of P is -1. Hence option (c) -1 is correct.

Directions Cosines and Direction Ratio Exercise Multiple Choice Question, question 7

$(a)\sqrt{b^2+c^2}$
Hint: Use distance formula
Given: Point P(a,b,c)
To Find: Distance of point P from x-axis
Solution:
The distance of the point (a,b,c) from x-axis will be perpendicular distance from point (a,b,c) to x-axis whose coordinates are (a,0,0)
Distance formula is given by
\begin{aligned} &d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}} \\ &d=\sqrt{(a-a)^{2}+(0-b)^{2}+(0-c)^{2}} \end{aligned}
where
\begin{aligned} &\left(x_{1}, y_{1}, z_{1}\right)=(a, b, c) \&\left(x_{2}, y_{2}, z_{2}\right)=(a, 0,0) \\ &d=\sqrt{b^{2}+c^{2}} \end{aligned}
Therefore, option
$(a)\sqrt{b^2+c^2}$ is correct.

Directions Cosines and Direction Ratio Exercise Multiple Choice Question, question 8

Hint: Use of section formula
Given: Points (1,2,3) and (4,2,1)
To find: Ratio in which xy plane divides the point
Solution: Let the xy plane divides the given point in the ratio t:1

Using section formula the point of intersection are:
$(x, y, z)=\left[\frac{t(4)+1}{t+1}, \frac{t(2)+2}{t+1}, \frac{t(1)+3}{t+1}\right]$
The z-coordinate of any point of xy plane is zero.
\begin{aligned} &\Rightarrow \frac{t(1)+3}{t+1}=0 \\ &\Rightarrow t+3=0 \\ &\Rightarrow t=-3 \\ &\Rightarrow t: 1=-3: 1 \end{aligned}
Therefore, the xy plane divide the given points in 3:1 externally. Hence option (b) 3:1 externally is correct.

Directions Cosines and Direction Ratio Exercise Multiple Choice Question, question 9

Hint: Use section formula
Given: Points P(3,2,-4), Q(5,4,-6) and R(9,8,-10)
To Find: R divides PQ in what ratio
Solution:

Let R divides the line PQ in the ratio k:1
Using section formula
\begin{aligned} &(x, y, z)=\left[\Rightarrow \frac{k\left(x_{2}\right)+x_{1}}{k+1}, \frac{k\left(y_{2}\right)+y_{1}}{k+1}, \frac{k\left(z_{2}\right)+z_{1}}{k+1}\right] \\ &(9,8,-10)=\left[\frac{k(5)+3}{k+1}, \frac{k(4)+2}{k+1}, \frac{k(-6)+(-4)}{k+1}\right] \\ &(9,8,-10)=\left[\frac{5 k+3}{k+1}, \frac{4 k+2}{k+1}, \frac{-6 k-4}{k+1}\right] \end{aligned}
comparing x-coordinate
\begin{aligned} &9=\frac{5 k+3}{k+1} \\ &9(k+1)=5 k+3 \\ &4 k=-6 \\ &k=\frac{-3}{2} \end{aligned}
Therefore R divides the line PQ in ratio 3:2 externally. Hence option (b) is correct.

Directions Cosines and Direction Ratios Exercise Multiple Choice Question, question 10.

(a) $\left [\frac{19}{8},\frac{57}{16},\frac{17}{16} \right ]$
Hint: Bisection divides triangle in ratios
$\frac{AB}{AC}=\frac{BD}{DC}$
Given: A(3,2,0), B(5,3,2) and C(-9,6,-3) are vertices of triangle ABC
To find: Point D coordinates which meets BC at bisector of $\angle BAC$
Solution:

As bisector AD meets BC at D, therefore
$\Rightarrow \frac{AB}{AC}=\frac{BD}{DC}$
Using distance formula
\begin{aligned} &A B=\sqrt{(5-3)^{2}+(3-2)^{2}+(2-0)^{2}} \\ &=\sqrt{4+1+4} \\ &=\sqrt{9} \\ &=3 \end{aligned}
and,
\begin{aligned} &A C=\sqrt{(-9-3)^{2}+(6-2)^{2}+(-3-0)^{2}} \\ &=\sqrt{(-12)^{2}+4^{2}+3^{2}} \\ &=\sqrt{144+16+9} \\ &=\sqrt{169} \\ &=13 \end{aligned}
Therefore,
$\frac{BD}{DC}=\frac{13}{8}$
D divides BC in the ratio 3:13 internally
Using section formula coordinates of D are
\begin{aligned} &(x, y, z)=\left[\Rightarrow \frac{-9 \times 3+5 \times 13}{3+13}, \frac{6 \times 3+13 \times 3}{3+13}, \frac{-3 \times 3+13 \times 2}{3+13}\right] \\ &(x, y, z)=\left[\frac{19}{8}, \frac{57}{16}, \frac{17}{16}\right] \end{aligned}
Hence coordinate of D are
$\left [\frac{19}{8},\frac{57}{16},\frac{17}{16} \right ]$
Therefore option (a) is correct

Directions Cosines and Direction Ratios Exercise Multiple Choice Question, question 11

Hint: Use direction to find direction cosine
Given:
$|\overrightarrow{O P}|=3 \&(a, b, c)=(-1,2,-2)$
To Find: Point P
Solution: Direction cosine is related to direction ratio as
\begin{aligned} &l=\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, m=\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}, n=\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}} \\ &l=\frac{-1}{\sqrt{(-1)^{2}+2^{2}+(-2)^{2}}}, m=\frac{2}{\sqrt{(-1)^{2}+2^{2}+(-2)^{2}}}, n=\frac{-2}{\sqrt{(-1)^{2}+2^{2}+(-2)^{2}}} \\ &l=-\frac{1}{3}, m=\frac{2}{3}, n=-\frac{2}{3} \end{aligned}
Now,
$\overrightarrow{OP}$
=Position of P- Position of O
\begin{aligned} &\overrightarrow{O P}=(x \hat{i}+y \hat{j}+z \hat{k})-(0 \hat{i}+0 \hat{j}+0 \hat{k}) \\ &\overrightarrow{O P}=(x \hat{i}+y \hat{j}+z \hat{k}) \end{aligned}
where (x,y,z) are the coordinates of P
Now,
\begin{aligned} &\overrightarrow{O P}=|\overrightarrow{O P}|\left(l^{\Lambda}+m \hat{j}+n \hat{k}\right) \\ &(x \hat{i}+y \hat{j}+z \stackrel{\Lambda}{k})=3\left(\frac{-1}{3} \hat{i}+\frac{2}{3} j-\frac{2}{3} \hat{k}\right) \\ &\left(x \stackrel{\Lambda}{i}+y \hat{j}+z^{\Lambda}\right)=(-\hat{i}+2 \hat{j}-2 \hat{k}) \end{aligned}
Comparing both,
(x,y,z)=(-1,2,-2)
Therefore coordinates of point P are (-1,2,-2). Hence option (a) is correct

Directions Cosines and Direction Ratios Exercise Multiple Choice Question, question 12.

(d) $\cos ^{-1}\frac{1}{3}$
Given: A cube
To Find: Angle between its diagonals
Solution: Let us consider a cube OABCDEFG with vertices as shown

O(0,0,0), A(a,0,0), B(a,a,0), C(0,a,0), D(0,a,a), E(0,0,a), F(a,0,a), G(a,a,a)
There are four diagonals OG, CF, AD, and BE
Let us consider OG and AD
$\overrightarrow{OG}$ = Position vector of G- Position vector of O
\begin{aligned} &\overrightarrow{O G}=(a-0) \hat{i}+(a-0) \hat{j}+(a-0) \hat{k} \\ &\overrightarrow{O G}=a \hat{i}+a j+a \hat{k} \end{aligned}
Also,
$\overrightarrow{AD}$ =Position vector of D - Position vector of A
\begin{aligned} &\overrightarrow{A D}=(0-a) \hat{i}+(a-0) \hat{j}+(a-0) \hat{k} \\ &\overrightarrow{A D}=-a i+a \hat{j}+a \hat{k} \end{aligned}
Now using dot product
\begin{aligned} &\overrightarrow{O G} \cdot \overrightarrow{A D}=|\overrightarrow{O G} \| \overrightarrow{A D}| \cos \theta \\ &\cos \theta=\frac{\overrightarrow{O G} \cdot \overrightarrow{A D}}{|\overrightarrow{O G}||\overrightarrow{A D}|} \\ &\cos \theta=\frac{(a \hat{i}+a \hat{j}+a \hat{k}) \cdot(-a \hat{i}+a \hat{j}+a k)}{\sqrt{a^{2}+a^{2}+a^{2}} \sqrt{a^{2}+a^{2}+a^{2}}} \end{aligned}
\begin{aligned} &\cos \theta=\frac{-a^{2}+a^{2}+a^{2}}{a \sqrt{3} \times a \sqrt{3}} \\ &\cos \theta=\frac{a^{2}}{a^{2} \times 3} \\ &\theta=\cos ^{-1}\left(\frac{1}{3}\right) \end{aligned}

Therefore angle between diagonals of cube is
$\cos ^{-1}\frac{1}{3}$
Hence option (d) is correct.

Directions Cosines and Direction Ratios Exercise Multiple Choice Question, question 13.

(c) $\frac{4}{3}$
Hint: Use direction cosines
Given: Line makes angles
$\alpha ,\beta ,\gamma ,\delta$
with diagonals of cube.
To Find:
$\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma+\cos ^{2} \delta$
Solution: Let us consider a cube as shown

Direction ratio of OP are (a-0,a-0,a-0)
=(a,a,a)
Direction cosine of OP are
\begin{aligned} &\left(\frac{a}{\sqrt{a^{2}+a^{2}+a^{2}}}, \frac{a}{\sqrt{a^{2}+a^{2}+a^{2}}}, \frac{a}{\sqrt{a^{2}+a^{2}+a^{2}}}\right) \\ &\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \end{aligned}
Similarly, direction cosine of all diagonals
\begin{aligned} &O P=\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \\ &B N=\left(\frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \\ &A M=\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\right) \\ &C L=\left(\frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \end{aligned}
Let (l,m,n) are the direction cosine of line which is inclined at angle
$\alpha ,\beta ,\gamma ,\delta$ with diagonals. Now, using dot product
\begin{aligned} &\cos \alpha=\frac{l+m+n}{\sqrt{3}} \\ &\cos \beta=\frac{-l+m+n}{\sqrt{3}} \\ &\cos \gamma=\frac{l+m-n}{\sqrt{3}} \\ &\cos \delta=\frac{l-m+n}{\sqrt{3}} \end{aligned}
Squaring and adding above equations, we get
$\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma+\cos ^{2} \delta=\frac{1}{3}\left[(l+m+n)^{2}+(-l+m+n)^{2}+(l+m-n)^{2}+(l-m+n)^{2}\right]$\
\begin{aligned} &=\frac{1}{3}\left[4\left(l^{2}+m^{2}+n^{2}\right)\right] \\ &=\frac{4}{3} \end{aligned}
Therefore, value of
$\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma+\cos ^{2} \delta=\frac{4}{3}$
Hence option(c) is correct.

Directions Cosines and Direction Ratios Exercise Multiple Choice Question, question 14.

Given: Point (2,5,7)
To Find: Foot of perpendicular on x-axis
Solution: As the foot of perpendicular of the point (2,5,7) lie on x-axis, the x-coordinate at the foot is 2, while y and z coordinates on the x axis are always zero. So, the coordinates at foot of perpendicular are (2,0,0)
Therefore, opton (a) (2,0,0) is correct.

Directions Cosines and Direction Ratios Exercise Multiple Choice Question, question 15.

Hint: Use section formula
Given: Points (3,2,-1) and (6,2,-2) and P(5,y,z)
To Find: y-coordinate of P(5,y,z)
Solution: Let P divide the line joining (3,2,-1)and (6,2,-2) in t:1

Using section formula
\begin{aligned} &(5, y, z)=\left[\frac{t(6)+3}{t+1}, \frac{t(2)+2}{t+1}, \frac{t(-2)+(-1)}{t+1}\right] \\ &(5, y, z)=\left[\frac{6 t+3}{t+1}, \frac{2 t+2}{t+1}, \frac{-2 t-1)}{t+1}\right] \end{aligned}
Comparing x-coordinate
\begin{aligned} &5=\frac{6 t+3}{t+1} \\ &5(t+1)=6 t+3 \\ &t=2 \end{aligned}
Now comparing y-coordinate
\begin{aligned} &y=\frac{2 t+2}{t+1} \\ &y=\frac{2(2)+2}{2+1} \\ &y=2 \end{aligned}
Therefore, y-coordinate of P(5,y,z) is 2. hence option(a) is correct.

Directions Cosines and Direction Ratios Exercise Multiple Choice Question, question 16.

(d) $\sqrt{\alpha^2+\gamma ^2}$
Hint: Use distance formula
Given:
Point P $(\alpha ,\beta ,\gamma )$
To Find: Distance of P from y-axis
Solution: The distance of the point
$(\alpha ,\beta ,\gamma )$ from y-axis whose coordinates are
$(0 ,\beta ,0 )$
Distance formula is
\begin{aligned} &d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}} \\ &=\sqrt{(0-\alpha)^{2}+(\beta-\beta)^{2}+(0-\gamma)^{2}} \end{aligned}
where
\begin{aligned} &\left(x_{1}, y_{1}, z_{1}\right)=(\alpha, \beta, \gamma) \&\left(x_{2}, y_{2}, z_{2}\right)=(0, \beta, 0) \\ &d=\sqrt{\alpha^{2}+\gamma^{2}} \end{aligned}
Therefore, option(d) is correct.

Directions Cosines and Direction Ratios Exercise Multiple Choice Question, question 17.

(d) $k =\frac{1}{\sqrt{3}} \text{or}\frac{-1}{\sqrt{3}}$
Hint: Use property of direction cosines
$l^2+m^2+n^2=1$
Given: Direction cosine of line are k,k,k
To Find: value of k
Solution: Since direction cosine of line are k,k,k
$\therefore l=k,m=k \; \; \&\; \; n=k$
We know that
\begin{aligned} &l^{2}+m^{2}+n^{2}=1 \\ &k^{2}+k^{2}+k^{2}=1 \\ &3 k^{2}=1 \\ &k=\pm \frac{1}{\sqrt{3}} \end{aligned}
Therefore, option (d) is correct
i.e. $k=\pm \frac{1}{\sqrt{3}}$

Directions Cosines and Direction Ratios Exercise Multiple Choice Question, question 18.

Given: Point (2,-3,4)
To Find: Foot of the perpendicular on y-axis
Solution: As the foot of the perpendicular of point (2,-3,4) lie on y-axis, the y-coordinate at the foot is -3, while x and z coordinate are zero on y-axis. So, the coordinate of foot of perpendicular are (0,-3,0)
Therefore, option (c) is correct.

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