RD Sharma Class 12 Exercise 26.1 Directions cosines and direction Ratios Solutions Maths

RD Sharma Class 12 Exercise 26.1 Directions Cosines and Direction Ratios Solutions Maths - Download PDF

Edited By Satyajeet Kumar | Updated on Jan 24, 2022 07:34 PM IST

The RD Sharma class 12 solution of Directions cosines and Directions ratios exercise 26.1 is one of the most interesting chapters of mathematics Class 12, and students once gain interest in understanding the chapter then they can effortlessly solve these chapters. The Class 12 RD Sharma chapter 26 exercise 26.1 solution will provide you with the best possible ways to solve the questions given and make you understand each and every concept with utmost focus and explanation that are not tough in language to understand.

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RD Sharma Class 12 Solutions Chapter26 Directions Cosines and Directions Ratios - Other Exercise
Direction Cosines and Direction Ratios Excercise: 26.1
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Also Read - RD Sharma Solutions For Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter26 Directions Cosines and Directions Ratios - Other Exercise

Direction Cosines and Direction Ratios Excercise: 26.1

Direction Cosines and Direction Ratios exercise 26.1 question 1

Answer: The direction cosines are $0,\frac{1}{2},\frac{\sqrt{3}}{2}$
Given: Angle made by line with x, y and z-axis are $90^{\circ},60^{\circ},30^{\circ}$
Hint: Find direction cosines of a line $\cos \alpha ,\cos \beta ,\cos \gamma$
Solution:
Let l, m and n are the direction cosines of the line and $\alpha ,\beta ,\gamma$ are the angles made with axes.
$\alpha=90 ^{\circ},\beta=60 ^{\circ},\gamma=30^{\circ}$
Now,
$\begin{aligned} &I=\cos \alpha \Rightarrow I=\cos 90^{\circ} \Rightarrow I=0 \\ &m=\cos \beta \Rightarrow m=\cos 60^{\circ} \Rightarrow m=\frac{1}{2} \\ &n=\cos \gamma \Rightarrow n=\cos 30^{\circ} \Rightarrow n=\frac{\sqrt{3}}{2} \end{aligned}$
Therefore, the direction cosines of the line are $0,\frac{1}{2},\frac{\sqrt{3}}{2}$

Direction Cosines and Direction Ratios exercise 26.1 question 2

Answer: $\frac{2}{3},\frac{-1}{3},\frac{-2}{3}$
Given: Direction ratios of line are (2,-1,-2)
Hint: Find direction cosines using $I=\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, m=\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}, n=\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}$
Solution:
Let l, m and n are the direction cosines of the line
Here (a, b, c) = (2, -1, -2) are the direction ratios of the line.
Direction cosine is related to direction ratios are
$\begin{aligned} &I=\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}} \Rightarrow I=\frac{2}{\sqrt{(2)^{2}+(-1)^{2}+(-2)^{2}}} \Rightarrow \mid=\frac{2}{\sqrt{9}} \\ &m=\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}} \Rightarrow m=\frac{c}{\sqrt{(2)^{2}+(-1)^{2}+(-2)^{2}}} \Rightarrow m=\frac{-1}{\sqrt{9}} \\ &n=\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}} \Rightarrow n=\frac{-2}{\sqrt{(2)^{2}+(-1)^{2}+(-2)^{2}}} \Rightarrow n=\frac{-2}{\sqrt{9}} \end{aligned}$
Therefore, direction cosines are $\frac{2}{3},\frac{-1}{3},\frac{-2}{3}$

Direction Cosines and Direction Ratios exercise 26.1 question 3

Answer: $\frac{3}{\sqrt{77}},\frac{-2}{\sqrt{77}},\frac{8}{\sqrt{77}}$
Hint: Direction ratios are proportional to the length of line
Given: Two points (-2, 4, -5) and (1, 2, 3)
Solution:
The direction ratios of the line joining the two points are
(a, b, c) = (1-(-2), 2-4, 3-(-5))
(a, b, c)=(3, -2, 8)
Now, if l, m and n are the direction cosine, So
$\begin{aligned} &(1, \mathrm{~m}, \mathrm{n})=\left(\frac{\mathrm{a}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}}}, \frac{\mathrm{b}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}}}, \frac{\mathrm{c}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}}}\right) \\ &(1, \mathrm{~m}, \mathrm{n})=\left(\frac{3}{\sqrt{3^{2}+(-2)^{2}+8^{2}}}, \frac{-2}{\sqrt{3^{2}+(-2)^{2}+8^{2}}}, \frac{8}{\sqrt{3^{2}+(-2)^{2}+8^{2}}}\right) \\ &(1, \mathrm{~m}, \mathrm{n})=\left(\frac{3}{\sqrt{77}}, \frac{-2}{\sqrt{77}}, \frac{8}{\sqrt{77}}\right) \end{aligned}$
Therefore, the direction cosines of the line are $\frac{3}{\sqrt{77}},\frac{-2}{\sqrt{77}},\frac{8}{\sqrt{77}}$

Direction Cosines and Direction Ratios exercise 26.1 question 4

Answer: The points are collinear
Hint: Direction ratios of parallel line are proportional.
Given: Points A(2, 3, -4), B(1, -2, 3) and C(3, 8, -11)
Solution:
Direction ratio of AB are = (1-2, -2-3, 3-(-4))
= (-1, -5, -7)
Direction ratio of BC are = (3-1, 8-(-2), -11-3)
= (2, 10, -14)
Comparing direction ratio of AB and BC
$\frac{-1}{2}=\frac{-5}{10},\frac{7}{-14}$
Hence, both are proportional and B is the common point in parallel lines.
Therefore, A, B and C are collinear points.

Direction Cosines and Direction Ratios exercise 26.1 question 5

Answer: The direction cosine of the side of triangle ABC are
$\left(\frac{-2}{\sqrt{17}}, \frac{-2}{\sqrt{17}}, \frac{3}{\sqrt{17}}\right),\left(\frac{-2}{\sqrt{17}}, \frac{-3}{\sqrt{17}}, \frac{-2}{\sqrt{17}}\right),\left(\frac{-4}{\sqrt{42}}, \frac{-5}{\sqrt{42}}, \frac{1}{\sqrt{42}}\right)$
Hint: Direction ratios are proportional to the length of side.
Given: A(3, 5, -4), B(-1, 1, 2) and C(-5, -5, 2). Find direction cosine of AB, BC and AC.

Solution: A(3, 5, -4)

B(-1, 1, 2) C(-5, -5, 2)
Let us consider an $\sqcup ABC$
Direction ratio of AB = (-1-3, 1-5, 2+4)
= (-4, -4, 6)
Direction cosines of AB are
$\begin{aligned} &=\left(\frac{-4}{\sqrt{(4)^{2}+(4)^{2}+(-6)^{2}}}, \frac{-4}{\sqrt{(4)^{2}+(4)^{2}+(-6)^{2}}}, \frac{6}{\sqrt{(4)^{2}+(4)^{2}+(-6)^{2}}}\right) \\ &=\left(\frac{-4}{\sqrt{68}}, \frac{-4}{\sqrt{68}}, \frac{6}{\sqrt{68}}\right) \\ &=\left(\frac{-2}{\sqrt{17}}, \frac{-2}{\sqrt{17}}, \frac{3}{\sqrt{17}}\right) \end{aligned}$ ……………….. (1)
Direction ratio of BC = (-5+1, -5-1, -2-2)
= (-4, -6, -4)
Direction cosines of BC are
$\begin{aligned} &=\left(\frac{-4}{\sqrt{(-4)^{2}+(-6)^{2}+(-4)^{2}}}, \frac{-6}{\sqrt{(-4)^{2}+(-6)^{2}+(-4)^{2}}}, \frac{-4}{\sqrt{(-4)^{2}+(-6)^{2}+(-4)^{2}}}\right) \\ &=\left(\frac{-4}{\sqrt{68}}, \frac{-6}{\sqrt{68}}, \frac{-4}{\sqrt{68}}\right) \\ &=\left(\frac{-2}{\sqrt{17}}, \frac{-3}{\sqrt{17}}, \frac{-2}{\sqrt{17}}\right) \end{aligned}$ ………………. (2)
Direction ratio of AC = (-5-3, -5-5, -2+4)
= (-8, -10, 2)
Direction cosines of AC are
$\begin{aligned} &=\left(\frac{-8}{\sqrt{(-8)^{2}+(-10)^{2}+(2)^{2}}}, \frac{-10}{\sqrt{(-8)^{2}+(-10)^{2}+(2)^{2}}}, \frac{2}{\sqrt{(-8)^{2}+(-10)^{2}+(2)^{2}}}\right) \\ &=\left(\frac{-8}{\sqrt{168}}, \frac{-10}{\sqrt{168}}, \frac{2}{\sqrt{168}}\right) \\ &=\left(\frac{-4}{\sqrt{42}}, \frac{-5}{\sqrt{42}}, \frac{1}{\sqrt{42}}\right) \end{aligned}$ ………………… (3)
By (1), (2) and (3) we get direction cosine of side AB, BC and AC are
$\left(\frac{-2}{\sqrt{17}}, \frac{-2}{\sqrt{17}}, \frac{3}{\sqrt{17}}\right),\left(\frac{-2}{\sqrt{17}}, \frac{-3}{\sqrt{17}}, \frac{-2}{\sqrt{17}}\right),\left(\frac{-4}{\sqrt{42}}, \frac{-5}{\sqrt{42}}, \frac{1}{\sqrt{42}}\right)$

Direction Cosines and Direction Ratios exercise 26.1 question 6

Answer: Angle between two vectors is $\frac{\pi }{2}$
Hint: Use dot product formula
Given: $\left ( a_{1},b_{1},c_{1} \right )=\left ( 1,-2,1 \right )$ and $\left ( a_{2},b_{2},c_{2} \right )=\left ( 4,3,2 \right )$ . Find angle between two vectors.
Solution:
Let $\theta$ be the angle between two vectors with direction ratios $\left ( a_{1},b_{1},c_{1} \right )$ & $\left ( a_{2},b_{2},c_{2} \right )$
Now,
$\begin{aligned} &\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \\ &\cos \theta=\frac{(1)(4)+(-2)(3)+(1)(2)}{\sqrt{1^{2}+(-2)^{2}+(1)^{2}} \sqrt{4^{2}+3^{2}+2^{2}}} \\ &\cos \theta=\frac{4-6+2}{\sqrt{6} \sqrt{29}} \\ &\cos \theta=0 \\ &\theta=\cos ^{-1}(0)=\frac{\pi}{2} \end{aligned}$
Therefore, angle between two vectors is $\frac{\pi}{2}$

Direction Cosines and Direction Ratios exercise 26.1 question 7

Answer: Angle between two vectors is $\cos ^{-1}\left(\frac{-18 \sqrt{2}}{35}\right)$
Hint: Direction cosine is proportional to direction ratios
Given: $\left(1, m_{1}, n_{1}\right)=(2,3,-6) \text { and }\left(I_{2}, m_{2}, n_{2}\right)=(3,-4,5)$ . Find Angle between two vectors.
Solution:
Let $\theta$ be the angle between two vectors with direction ratios $\left(a_{1}, b_{1}, c_{1}\right) \&\left(a_{2}, b_{2}, c_{2}\right)$
As direction ratios are proportional to direction cosines
$\begin{aligned} &\left(a_{1}, b_{1}, c_{1}\right)=\left(I_{1}, m_{1}, n_{1}\right)=(2,3,-6) \\ &\left(a_{2}, b_{2}, c_{2}\right)=\left(I_{2}, m_{2}, n_{2}\right)=(3,-4,5) \end{aligned}$
Now,
$\begin{aligned} &\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \\ &\cos \theta=\frac{(2)(3)+(3)(-4)+(-6)(5)}{\sqrt{2^{2}+(3)^{2}+(-6)^{2}} \sqrt{3^{2}+(-4)^{2}+5^{2}}} \\ &\cos \theta=\frac{6-12-30}{\sqrt{49} \sqrt{50}} \\ &\cos \theta=\frac{-36 \times \sqrt{2}}{7 \times 5 \times \sqrt{2} \times \sqrt{2}} \\ &\cos \theta=\frac{-18 \sqrt{2}}{35} \\ \end{aligned}$
$\theta=\cos ^{-1}\left(\frac{-18 \sqrt{2}}{35}\right)$
Therefore, the angle between two vectors is $\cos ^{-1}\left(\frac{-18 \sqrt{2}}{35}\right)$

Direction Cosines and Direction Ratios exercise 26.1 question 8

Answer: Angle between two vectors is $\cos ^{-1}\left(\frac{20}{21}\right)$
Hint: Use dot product formula.
Given: $\left(a_{1}, b_{1}, c_{1}\right)=(2,3,6) \&\left(a_{2}, b_{2}, c_{2}\right)=(1,2,2)$ . Find angle between two lines
Solution: here we have $\left(a_{1}, b_{1}, c_{1}\right)=(2,3,6) \&\left(a_{2}, b_{2}, c_{2}\right)=(1,2,2)$
Let $\theta$ be the angle between two lines whose direction ratios are $\left(a_{1}, b_{1}, c_{1}\right)$ & $\left(a_{2}, b_{2}, c_{2}\right)$
$\begin{aligned} &\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \\ &\cos \theta=\frac{(2)(1)+(3)(2)+(6)(2)}{\sqrt{2^{2}+(3)^{2}+(6)^{2}} \sqrt{1^{2}+(2)^{2}+2^{2}}} \\ &\cos \theta=\frac{2+6+12}{\sqrt{49} \sqrt{9}} \\ &\cos \theta=\frac{20}{21} \\ &\theta=\cos ^{-1}\left(\frac{20}{21}\right) \end{aligned}$
Therefore the angle between lines is $\cos ^{-1}\left(\frac{20}{21}\right)$

Direction Cosines and Direction Ratios exercise 26.1 question 9

Answer: The points are collinear.
Hint: Direction ratios of parallel lines are proportional.
Given: Points A (2, 3, 4), B (-1, -2, -1), C (5, 8, 7). Prove A, B, C are collinear
Solution: we have Points A (2, 3, 4), B (-1, -2, -1), C (5, 8, 7)
Direction ratio of AB are $=\left ( -1-2,-2-3,1-4 \right )$
$=\left ( -3,-5,-3\right )$
Direction ratio of BC are $=\left ( 5-\left ( -1 \right ),8-\left ( -2 \right ),7-1\right )$
$=\left ( 6,10,6 \right )$
Comparing direction ratio of AB and BC
$\frac{-3}{6}=\frac{-5}{10}=\frac{-3}{6}$
Hence, both are proportional and B is the common point in parallel line.
Therefore, A, B, C are collinear points.

Direction Cosines and Direction Ratios exercise 26.1 question 10

Answer: Both lines are parallel to each other
Hint: Direction ratio of parallel lines are proportional
Given: $\mathrm{A}(4,7,8), \mathrm{B}(2,3,4), \mathrm{C}(-1,-2,1) \& \mathrm{D}(1,2,5)$ . Show that AB and CD are parallel.
Solution: we have $\mathrm{A}(4,7,8), \mathrm{B}(2,3,4), \mathrm{C}(-1,-2,1) \& \mathrm{D}(1,2,5)$
Direction ratio of AB are $=\left ( 2-4,3-7,4-8 \right )$
$=\left (-2,-4,-4 \right )$
Direction ratio of BC are $=\left ( 1-\left ( -1 \right ),2-\left ( -2 \right ),5-1 \right )$
$=\left ( 2,4,4 \right )$
Comparing direction ratio of AB and BC
$\frac{-2}{2}=\frac{-4}{4}=\frac{-4}{4}$
Hence, both are proportional to each other
Therefore, AB and CD are parallel lines

Direction Cosines and Direction Ratios exercise 26.1 question 11

Answer: Both the lines are perpendicular to each other
Hint: Angle between perpendicular line is $90^{\circ}$
Given: $\mathrm{A}(1,-1,2), \mathrm{B}(3,4,-2), \mathrm{C}(0,3,2) \& \mathrm{D}(3,5,6)$ . Show AB is perpendicular to CD
Solution: we have $\mathrm{A}(2,3,6), \mathrm{B}(1,2,2), \mathrm{C}(0,3,2) \& \mathrm{D}(3,5,6)$
Let $\theta$ be the angle between two lines whose direction cosines are $\left ( a_{1},b_{1},c_{1} \right )$ & $\left ( a_{2},b_{2},c_{2} \right )$
Direction ratio of $AB=\left ( 3-1,4-\left ( -1 \right ),-2-2 \right )$
$\left ( a_{1},b_{1},c_{1} \right )=\left ( 2,5,-4 \right )$
Direction ratio of $CD=\left ( 3-0,5-3,6-2 \right )$
$\left ( a_{2},b_{2},c_{2} \right )=\left ( 3,2,4 \right )$
Now,
$\begin{aligned} &\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \\ &\cos \theta=\frac{(2)(3)+(5)(2)+(-4)(4)}{\sqrt{2^{2}+(5)^{2}+(-4)^{2}} \sqrt{3^{2}+(2)^{2}+4^{2}}} \\ &\cos \theta=\frac{6+10-16}{\sqrt{45} \sqrt{29}} \\ &\cos \theta=0 \\ &\theta=\cos ^{-1}(0) \\ &\theta=90^{\circ} \end{aligned}$
Therefore, AB and CD are perpendicular to each other.

Direction Cosines and Direction Ratios exercise 26.1 question 12

Answer: Both the lines are perpendicular to each other
Hint: Angle between perpendicular line is $90^{\circ}$
Given: $A\left ( 0,0,0 \right ),B\left ( 2,1,1 \right ),C\left ( 3,5,-1 \right )$ & $B\left ( 4,3,-1 \right )$
. Show AB is perpendicular to CD
Solution: we have $A\left ( 0,0,0 \right ),B\left ( 2,1,1 \right ),C\left ( 3,5,-1 \right )$ & $B\left ( 4,3,-1 \right )$
Let $\theta$ be the angle between two lines whose direction cosines are $\left ( a_{1}, b_{1}, c_{1} \right )$ & $\left ( a_{2}, b_{2}, c_{2} \right )$
Direction ratio of $AB=\left ( 2-0,1-0,1-0 \right )$
$\left ( a_{1}, b_{1}, c_{1} \right )$ $=\left ( 2,1,1 \right )$
Direction ratio of $CD=\left ( 4-3,3-5,-1-\left ( -1 \right ) \right )$
$\left ( a_{2}, b_{2}, c_{2} \right )$ = $=\left ( 1,-2,0 \right )$
Now,
$\begin{aligned} &\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \\ &\cos \theta=\frac{(2)(1)+(1)(-2)+(1)(0)}{\sqrt{2^{2}+(1)^{2}+(1)^{2}} \sqrt{1^{2}+(-2)^{2}+0^{2}}} \\ &\cos \theta=\frac{2-2}{\sqrt{6} \sqrt{5}} \\ &\cos \theta=0 \\ &\theta=\cos ^{-1}(0) \\ &\theta=90^{\circ} \end{aligned}$
Therefore, AB and CD are perpendicular to each other.

Direction Cosines and Direction Ratios exercise 26.1 question 13

Answer: Angle between lines is $\frac{\pi }{2}$
Hint: Use dot product formula
Given: $\left(a_{1}, b_{1}, c_{1}\right)=(a, b, c) \&\left(a_{2}, b_{2}, c_{2}\right)=(b-c, c-a, a-b)$ . Find angle between two lines.
Solution: we have $\left(a_{1}, b_{1}, c_{1}\right)=(a, b, c) \&\left(a_{2}, b_{2}, c_{2}\right)=(b-c, c-a, a-b)$
Now,
$\begin{aligned} &\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \\ &\cos \theta=\frac{(a)(b-c)+(b)(c-a)+(c)(a-b)}{\sqrt{a^{2}+b^{2}+c^{2}} \sqrt{(b-c)^{2}+(c-a)^{2}+(a-b)^{2}}} \\ &\cos \theta=\frac{(a b-a c)+(b c-b a)+(a c-b c)}{\sqrt{a^{2}+b^{2}+c^{2}} \sqrt{(b-c)^{2}+(c-a)^{2}+(a-b)^{2}}} \\ &\cos \theta=0 \\ &\theta=\cos ^{-1}(0) \\ &\theta=\frac{\pi}{2} \end{aligned}$
Therefore, the angle between line is $\theta=\frac{\pi}{2}$

Direction Cosines and Direction Ratios exercise 26.1 question 14

Answer: Angle between AB and CD is $0^{\circ}$

Given: $A(1,2,3), B(4,5,7), C(-4,3,-6) \& D(2,9,2)$ . Find angle between AB and CD
Solution: we have $A(1,2,3), B(4,5,7), C(-4,3,-6) \& D(2,9,2)$
Direction ratio of $AB=\left ( 4-1,5-2,7-3 \right )$
$\left ( a_{1},b_{1},c_{1} \right )=\left ( 3,3,4 \right )$
Direction ratio of $CD=\left ( 2-\left ( -4 \right ),9-3,2-\left ( -6 \right ) \right )$
$\left ( a_{2},b_{2},c_{2} \right )=\left ( 6,6,8 \right )$
Let $\theta$ be the angle between two lines whose direction cosines are $\left ( a_{1},b_{1},c_{1} \right )$ & $\left ( a_{2},b_{2},c_{2} \right )$
$\begin{aligned} &\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \\ &\cos \theta=\frac{(3)(6)+(3)(6)+(4)(8)}{\sqrt{3^{2}+(3)^{2}+(4)^{2}} \sqrt{6^{2}+6^{2}+8^{2}}} \\ &\cos \theta=\frac{18+18+32}{\sqrt{9+9+16} \sqrt{36+36+64}} \\ &\cos \theta=\frac{68}{\sqrt{34} \sqrt{136}} \\ &\cos \theta=\frac{68}{34 \times 2} \\ \end{aligned}$
$\theta=\cos ^{-1}(1) \\$
$\theta=0^{\circ}$
Therefore, angle between AB and CD is $\theta=0^{\circ}$

Direction Cosines and Direction Ratios exercise 26.1 question 15

Answer: $\left(\pm \frac{1}{\sqrt{6}}, \pm \frac{1}{\sqrt{6}}, \pm \frac{2}{\sqrt{6}}\right) \text { and }\left(\pm \frac{1}{\sqrt{6}}, \pm \frac{2}{\sqrt{6}}, \pm \frac{1}{\sqrt{6}}\right)$
Given: $I+m+n=0$ and $2Im+2In-mn=0$ . Find direction cosine of line
Solution: we have
$I+m+n=0$ ………………. (1)
$I=-m-n$ ………………. (2)
$\begin{aligned} &\Rightarrow 2 \mid m+2 \ln -m n=0 \quad[1=-m-x] \\ &\Rightarrow 2(-m-n) m+2(-m-n) n-m n=0 \\ &\Rightarrow-2 m^{2}-2 m n-2 m n-2 n^{2}-m n=0 \\ &\Rightarrow 2 m^{2}+2 n^{2}+5 m n=0 \\ &\Rightarrow(m+2 n)(2 m+n)=0 \end{aligned}$
Given $m+2 n=0 \Rightarrow m=-2 n$
Or
$2 \mathrm{~m}+\mathrm{n}=0 \Rightarrow \mathrm{m}=\frac{-\mathrm{n}}{2}$
$If m=-2 n, then from (1), \mid=n$
$If m=\frac{-n}{2}, then from (1), 1=\frac{-n}{2}$
Thus, direction ratios are proportional to $(n,-2 n, n) \text { and }\left(\frac{-n}{2}, \frac{-n}{2}, n\right)$
$\Rightarrow(1,-2,1) \text { and }\left(-\frac{1}{2},-\frac{1}{2}, 1\right)$
Now direction cosines are
$\left(\pm \frac{1}{\sqrt{6}}, \pm \frac{1}{\sqrt{6}}, \pm \frac{2}{\sqrt{6}}\right) \text { and }\left(\pm \frac{1}{\sqrt{6}}, \pm \frac{2}{\sqrt{6}}, \pm \frac{1}{\sqrt{6}}\right)$

Direction Cosines and Direction Ratios exercise 26.1 question 16 (i)

Answer: Angle between lines is $\frac{\pi }{3}$
Given:
$\begin{aligned} &1+m+n=0 \\ &1^{2}+m^{2}+n^{2}=0 \end{aligned}$ . Find angle between the lines
Hint: Use $\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
Solution: we have
$I+m+n=0$ ………………. (1)
$m=-I-n$ ………………. (2)
$\begin{aligned} &\left.\Rightarrow\right|^{2}+m^{2}+n^{2}=0 \quad[m=-1-n] \\ &\left.\Rightarrow\right|^{2}+(-1-n)^{2}-n^{2}=0 \\ &\left.\Rightarrow\right|^{2}+\left.\right|^{2}+n^{2}+2 \mid n-n^{2}=0 \\ &\left.\Rightarrow 2\right|^{2}+2 \ln =0 \\ &\Rightarrow 2 l(1+n)=0 \end{aligned}$
Either $\mid=0 or \mid=-n$
$If \mid=0, then equ (2), m=-n$
$If \mid=-n$, then equ $(2), m=0$
Thus, direction ratio of two lines are proportional to
$\begin{aligned} &(0,-n, n) \text { and }(-n, 0, n) \\ &\Rightarrow(0,-1,1) \text { and }(-1,0,1) \\ &\Rightarrow\left(a_{1}, b_{1}, c_{1}\right)=(0,-1,1) \\ &\Rightarrow\left(a_{2}, b_{2}, c_{2}\right)=(-1,0,1) \end{aligned}$
Angle between two lines is

$\begin{aligned} &\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \\ &\cos \theta=\frac{(0)(-1)+(-1)(0)+(1)(1)}{\sqrt{0^{2}+(-1)^{2}+(1)^{2}} \sqrt{(-1)^{2}+0^{2}+1^{2}}} \\ &\cos \theta=\frac{1}{\sqrt{2} \sqrt{2}} \\ &\cos \theta=\frac{1}{2} \\ &\theta=\cos ^{-1}\left(\frac{1}{2}\right) \\ \end{aligned}$
$\theta=\frac{\pi}{3}$
Therefore, angle between two lines is $\frac{\pi}{3}$ .

Direction Cosines and Direction Ratios exercise 26.1 question 16 (ii)

Answer: angle between two lines is $\frac{\pi }{2}$
Given:
$\begin{aligned} &2 l-m+2 n=0 \\ &m n+n l+\mid m=0 \end{aligned}$ , Find angle between the lines
Hint: Use $\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
Solution: we have
$2 l-m+2 n=0$ …………….. (1)
$m=2I+2 n$ …………….. (2)

$\Rightarrow m n+n l+\operatorname{lm}=0$ $\quad[m=2 \mid+2 n]$

$\Rightarrow(2 \mid+2 n) n+n l+l(2 l+2 n)=0$
$\Rightarrow\left(2\left|n+2 n^{2}+n\right|+\left.2\right|^{2}+2 n \mid\right)=0$
$\left.\Rightarrow 2\right|^{2}+51 n+2 n^{2}=0$
$\Rightarrow(1+2 n)(21+n)=0$
$\Rightarrow \mid=-2 \mathrm{n}$ or $\mathrm{I}=\frac{-\mathrm{n}}{2}$
$If I=-2 n$, then equ $(2), m=-2 n$
$If \mathrm{I}=\frac{-\mathrm{n}}{2}$, then equ $(2), \mathrm{m}=\mathrm{n}$
Thus, direction ratio of two lines are proportional to
$\begin{aligned} &(-2 n,-2 n, n) \text { and }\left(\frac{-n}{2}, n, n\right) \\ &\Rightarrow(-2,-2,1) \text { and }\left(-\frac{1}{2}, 1,1\right) \\ &\Rightarrow\left(a_{1}, b_{1}, c_{1}\right)=(-2,-2,1) \\ &\Rightarrow\left(a_{2}, b_{2}, c_{2}\right)=\left(-\frac{1}{2}, 1,1\right) \end{aligned}$
Angle between two lines is
$\begin{aligned} &\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \\ &\cos \theta=\frac{(-2)\left(-\frac{1}{2}\right)+(-2)(1)+(1)(1)}{\sqrt{(-2)^{2}+(-2)^{2}+(1)^{2}} \sqrt{\left(-\frac{1}{2}\right)^{2}+1^{2}+1^{2}}} \\ &\cos \theta=\frac{1-2+1}{\sqrt{4+4+1} \sqrt{\frac{1}{4}+1+1}} \\ &\cos \theta=0 \\ &\theta=\cos ^{-1}(0) \\ \end{aligned}$
$\theta=\frac{\pi}{2}$
Therefore, angle between two lines is $\frac{\pi}{2}$ .

The RD Sharma class 12 solutions chapter 26 exercise 26.1 consists of a total of 19, which covers the essential concepts of this chapter mentioned below-

Direction Cosines
Using direction ratio, show that points are collinear
To find the angle between the lines
Show the points are collinear

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Frequently Asked Questions (FAQs)

1. How do you find the direction ratio of direction cosines?

Any numbers that are proportional to the direction cosines are called direction ratios usually represented as a, b, c. So we can write, a=kl, b=km, c=kn where k is a constant.

2. What is the difference between direction cosine and direction ratios?

We can say that cosines of direction angles of a vector are the coefficients of the unit vectors and when the unit vector is resolved in terms of its rectangular components.

3. Is this solution helpful for the preparation of public exams?

Yes, the RD Sharma class 12 solution is helpful for the preparation of public exams as well as the board exams.

4. How do you find the direction of A ratio?

To find the components of a unit vector, divide the original three components of the vector by the magnitude of the vector. the three components of the unit vector are known as direction ratios because they represent the ratio of each coordinate to the total magnitude.

5. Do parallel lines have the same direction ratios?

Yes, parallel lines do have the same direction ratios.