What is the direction cosines of a line in a 3D plane?

A (directed) line's direction cosines are the cosines of the angles formed by the line with the positive directions of the coordinate axes.

What is the direction angle?

The angles formed by a line intersecting the positive directions of the X, Y, and Z axes are known as directional angles.

Can I score good marks after studying from Rd Sharma Class 12th Exercise FBQ solutions?

Our RD Sharma Solutions for Class 12 are created in accordance with the most recent CBSE syllabus. Solving RD Sharma exercise questions with the assistance of RD Sharma Solutions will assist students in achieving high marks in the Class 12 board examination. It also includes key concepts and formulas to help you understand the chapter better.

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Is RD Sharma's book appropriate for CBSE Class 12?

The NCERT textbook is the most important study material or reference book for Class 12. After you have finished solving the NCERT textbook, you will need a lot of practice. Solving the Rd Sharma Class 12th Exercise FBQ questions will give you a lot of practice, which is important for CBSE Class 12 students.

RD Sharma Solutions Class 12 Mathematics Chapter 26 FBQ

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RD Sharma Solutions Class 12 Mathematics Chapter 26 FBQ

RD Sharma Solutions Class 12 Mathematics Chapter 26 FBQ

Updated on 24 Jan 2022, 07:35 PM IST

RD Sharma books have always set parameters for board exams for a long time. This book is very important from an exam point of view because teachers assign question papers from this book most of the time. Rd Sharma Class 12th Exercise FBQ solutions that you find here are not the work of a single person but a team of subject experts who always thrive for student’s success. Class 12th RD Sharma Chapter 26 Exercise 26 FBQ Solutions will help a student to revise a bit faster and save a lot of time. This particular exercise has 18 questions or fill in the blanks.

This Story also Contains

RD Sharma Class 12 Solutions Chapter26 Directions Cosines and Directions Ratios - Other Exercise
Direction Cosines and Direction Ratios Excercise: FBQ
RD Sharma Chapter wise Solutions

The concepts that are discussed in Rd Sharma Class 12th Exercise FBQ are Direction Cosines which include an easy way to identify the direction of a line in three-dimensional space, Direction ratios, directional angles, Projection of line segment, unit vector are also discussed in this exercise. RD Sharma Solutions All the topics are wonderfully covered for a quick revision for the student.

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

Direction Cosines and Direction Ratios Excercise: FBQ

Direction Cosines and Direction Ratios exercise Fill in the blanks question 1

Answer : $\sqrt{a^{2}+c^{2}}$

Hint:
Use Distance Formula

Given:
Point $P(a,b,c)$

To Find:
Distance of point P from y -axis

Solution:
The distance of the point $(a,b,c)$ from y-axis will be the perpendicular distance from point $(a,b,c)$ to y-axis whose co-ordinates are $(0,b,0)$
Distance formula is
$\begin{aligned} &d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}} \\\\ &d=\sqrt{(0-a)^{2}+(b-b)^{2}+(0-c)^{2}} \end{aligned}$
where $\left(x_{1}, y_{1}, z_{1}\right)=(a, b, c) \text { and }\left(x_{2}, y_{2}, z_{2}\right)=(0, b, 0)$
$d=\sqrt{a^{2}+c^{2}}$
Therefore distance from y axis is $\sqrt{a^{2}+c^{2}}$

Direction Cosines and Direction Ratios exercise Fill in the blanks question 2

Answer : $\sqrt{a^{2}+b^{2}}$

Hint:
Use Distance Formula

Given:
Point $P(a,b,c)$
To Find:
Distance of point P from z -axis
Solution:
The distance of the point $(a,b,c)$ from z-axis will be the perpendicular distance from point $(a,b,c)$ to z-axis whose co-ordinates are $(0,0,c)$
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}} \\\\ &d=\sqrt{(0-a)^{2}+(0-b)^{2}+(c-c)^{2}} \end{aligned}$$\begin{aligned} &d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}} \\\\ &d=\sqrt{(0-a)^{2}+(0-b)^{2}+(c-c)^{2}} \end{aligned}$$\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{(0-a)^{2}+(0-b)^{2}+(c-c)^{2}} \end{aligned}$
where $\left(x_{1}, y_{1}, z_{1}\right)=(a, b, c) \text { and }\left(x_{2}, y_{2}, z_{2}\right)=(0,0, c)$
$d=\sqrt{a^{2}+b^{2}}$
Therefore distance of the point $(a,b,c)$ from z axis is $\sqrt{a^{2}+b^{2}}$.

Direction Cosines and Direction Ratios exercise Fill in the blanks question 3

Answer : $(l, m, n)=\left(0,-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$

Hint:
Direction cosine is cosine of angle made with the axes

Given:
$\alpha=\frac{\pi}{2}, \beta=\frac{3 \pi}{4} \text { and } \gamma=\frac{\pi}{4}$ are the angles made with $x,y,z$ axes.
To Find:
Direction cosine of the line
Solution:
$l=\cos \alpha, \quad m=\cos \beta, \quad n=\cos \gamma$
$\begin{aligned} &l=\cos \frac{\pi}{2}, \quad m=\cos \frac{3 \pi}{4}, \quad n=\cos \frac{\pi}{4} \\ &l=0, \quad m=-\frac{1}{\sqrt{2}}, \quad n=\frac{1}{\sqrt{2}} \end{aligned}$
Therefore, the direction cosine of the line are $(l, m, x)=\left(0,-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$

Direction Cosines and Direction Ratios exercise Fill in the blanks question 4

Answer : $-1$

Hint:
$l^{2}+m^{2}+n^{2}=1$
Given:
Line makes angles $\alpha, \beta, \gamma \text { with } x, y, z \text { axes }$
To Find:
$\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma$
Solution:
Direction cosine of a line making angles $\alpha ,\beta ,\gamma$ with co-ordinate axes are
$\begin{aligned} &l=\cos \alpha \\ &m=\cos \beta \text { and } \\ &n=\cos \gamma \end{aligned} and l^{2}+m^{2}+n^{2}=1$
$\begin{aligned} &\text { and } l^{2}+m^{2}+n^{2}=1 \\ &\Rightarrow \cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1 \\ &\Rightarrow 2 \cos ^{2} \alpha+2 \cos ^{2} \beta+2 \cos ^{2} \gamma=2 \end{aligned}$
Now, using the formula: $\left[1+\cos 2 x=2 \cos ^{2} x\right]$
$\begin{aligned} &\Rightarrow 1+\cos 2 \alpha+1+\cos 2 \beta+1+\cos 2 \gamma=2 \\ &\Rightarrow \cos 2 \alpha+\cos 2 \beta+\cos 2 y=2-3 \\ &\Rightarrow \cos 2 \alpha+\cos 2 \beta+\cos 2 y=-1 \end{aligned}$
Therefore, value of $(\cos 2 \alpha+\cos 2 \beta+\cos 2 y) \text { is }-1$

Direction Cosines and Direction Ratios exercise Fill in the blanks question 5

Answer : $2$

Hint:
$l^{2}+m^{2}+n^{2}=1$
Given:
Line makes angles α,β, γ with x,y,z axes
To Find:
$\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma$
Solution:
Direction cosine of a line making angles α,β,y are
$\begin{aligned} &l=\cos \alpha \\ &m=\cos \beta \text { and } \end{aligned}$
$\begin{aligned} &n=\cos \gamma \text { and } \\ &l^{2}+m^{2}+n^{2}=1 \\ &\therefore \cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1 \end{aligned}$ ......(i)
Now as$\left[\cos ^{2} x=1-\sin ^{2} x\right]$ ....... (ii)
∴ Using above eq(ii) in (i), we get
$\begin{aligned} &\left(1-\sin ^{2} \alpha\right)+\left(1-\sin ^{2} \beta\right)+\left(1-\sin ^{2} x\right)=1 \\\\ &\Rightarrow 3-\left(\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma\right)=1 \\\\ &\Rightarrow \sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=3-1 \\\\ &\Rightarrow \sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=2 \end{aligned}$
Therefore, the value of $\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma \text { is } 2$.

Direction Cosines and Direction Ratios exercise Fill in the blanks question 6
Answer : $\frac{\pi }{2}$

Hint:
Using property of direction cosine $l^{2}+m^{2}+n^{2}=1$
Given:
Line makes angles with y and z axes is $\frac{\pi }{4}$
Hence $\beta=\gamma=\frac{\pi}{4}$
To Find:
Angle made with x-axes?
Solution:
Direction cosine of a line making angle α,β,γ are
$l=\cos \alpha, m=\cos \beta \text { and } n=\cos \gamma\\\\ and \; \; l^{2}+m^{2}+n^{2}=1\\\\ \Rightarrow \cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1$
$\Rightarrow \cos ^{2} \alpha+\cos ^{2}\left(\frac{\pi}{4}\right)+\cos ^{2}\left(\frac{\pi}{4}\right)=1 \quad\left[\cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\right]$
$\begin{aligned} &\Rightarrow \cos ^{2} \alpha+\frac{1}{2}+\frac{1}{2}=1 \\\\ &\Rightarrow \cos ^{2} \alpha=1-1 \\\\ &\Rightarrow \cos ^{2} \alpha=0 \\\\ &\Rightarrow \cos \alpha=0 \end{aligned}$
$\Rightarrow \alpha=\frac{\pi}{2}$
Therefore, the angle made by line with x -axis is $\frac{\pi }{2}$.

Direction Cosines and Direction Ratios exercise Fill in the blanks question 7

Answer : $(l, m, n)=\left(\frac{2}{3}, \frac{2}{3}, \frac{-1}{3}\right)$

Hint:
$(a,b,c)$ of a vector $\left(a i^{{\wedge}}+b j^{\wedge}+c k^{\wedge}\right)$ are direction ratio of that vector.
Given:
Vector?$\left(2 i^{\wedge}+2 \hat{\jmath}-k^{\wedge}\right)$
To Find:
Direction cosine of vector.
Solution:
If $l,m,n$ Direction cosine and $a,b,c$ ane direction ratios, then
$l=\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, \quad m=\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}, \quad n=\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}$
$\therefore l=\frac{2}{\sqrt{2^{2}+2^{2}+(-1)^{2}}}, \quad m=\frac{2}{\sqrt{2^{2}+2^{2}+(-1)^{2}}}, \quad n=\frac{-1}{\sqrt{2^{2}+2^{2}+(-1)^{2}}}$
$\therefore l=\frac{2}{\sqrt{9}}, \quad m=\frac{2}{\sqrt{9}}, \quad n=\frac{-1}{\sqrt{9}}$
Therefore $(l, m, n)=\left(\frac{2}{3}, \frac{2}{3},-\frac{1}{3}\right)$.

Direction Cosines and Direction Ratios exercise Fill in the blanks question 8

Answer : $\gamma=\frac{\pi}{3}$

Hint:
Use property of direction cosine $l^{2}+m^{2}+n^{2}=1$
Given:
$\alpha=\frac{\pi}{4}, \beta=\frac{\pi}{3}$
To Find:
$\gamma$ = angle with z-axis
Solution:
Since line makes angle $\frac{\pi}{4} \text { and } \frac{\pi}{3}$ with x and y axes, there direction cosines are
$l=\cos \alpha, m=\cos \beta \text { and } n=\cos \gamma$
$\begin{aligned} &\text { As } l^{2}+m^{2}+n^{2}=1 \\\\ &\therefore \cos ^{2} \frac{\pi}{4}+\cos ^{2} \frac{\pi}{3}+\cos ^{2} \gamma=1 \end{aligned}$ $\left[\cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}, \cos \left(\frac{\pi}{3}\right)=\frac{1}{2}\right]$
$\begin{aligned} &\frac{1}{2}+\frac{1}{4}+\cos ^{2} \gamma=1 \\\\ &\cos ^{2} \gamma=1-\frac{3}{4} \\\\ &\cos ^{2} \gamma=\frac{1}{4} \end{aligned}$
$\begin{aligned} &\cos \gamma=\frac{1}{2} \\\\ &\cos \gamma=\cos \frac{\pi}{3} \quad\left[\cos \frac{\pi}{3}=\frac{1}{2}\right] \end{aligned}$
$\gamma=\frac{\pi}{3}$
Therefore, the angle made with z -axls is $\frac{\pi }{3}$.

Direction Cosines and Direction Ratios exercise Fill in the blanks question 9

Answer: $5\sqrt{2}$

Hint:
Length of projection = Length × Direction cosine
Given:
Projections on coordinate axes are 3,4,5
To Find:
Length of line segment
Solution:
Let Line segment PQ makes angle α,β,γ with x,y and z axes, such that length of projection in coordinate axes are:
$\begin{aligned} &P Q \cos \alpha=3 \Rightarrow \cos \alpha=\frac{3}{P Q} \\ &P Q \cos \beta=4 \Rightarrow \cos \beta=\frac{4}{P Q} \\ &P Q \cos \gamma=5 \Rightarrow \cos \gamma=\frac{5}{P Q} \end{aligned}$
Now by properties of direction cosine
$\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1$
$\begin{aligned} &\therefore\left(\frac{3}{P Q}\right)^{2}+\left(\frac{4}{P Q}\right)^{2}+\left(\frac{5}{P Q}\right)^{2}=1 \\\\ &\frac{9}{P Q^{2}}+\frac{16}{P Q^{2}}+\frac{25}{P Q^{2}}=1 \\\\ &\frac{50}{P Q^{2}}=1 \end{aligned}$
$\begin{aligned} &P Q^{2}=50 \\\\ &P Q=\sqrt{50} \\\\ &P Q=5 \sqrt{2} \end{aligned}$
Therefore, the total length of line segment is $5\sqrt{2}$.

Direction Cosines and Direction Ratios exercise Fill in the blanks question 10

Answer: $6 i^{\wedge}-9j^{\wedge}+18 k^{\wedge}$

Hint:
Use direction ratio to find direction cosine.
Given:
$|\vec{r}|=21 \text { and }(a, b, c)=(2,-3,6)$
To Find:
vector $\vec{r}$
Solution:
Direction cosine is related to Direction ratio as
$l=\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, m=\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}, n=\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}$
$l=\frac{2}{\sqrt{2^{2}+(-3)^{2}+6^{2}}}, m=\frac{-3}{\sqrt{2^{2}+(-3)^{2}+6^{2}}}, n=\frac{6}{\sqrt{2^{2}+(-3)^{2}+6^{2}}}$
$\begin{aligned} &l=\frac{2}{7}, m=\frac{-3}{7}, n=\frac{6}{7} \\\\ \end{aligned}$
$\vec{r}=|\vec{r}|\left(l i^{\wedge}+m_{j}^{\wedge}+n k^{\wedge}\right)$
$\begin{aligned} &\text { Now } \vec{r}=21\left(\frac{2}{7} i^{\wedge}+\frac{(-3)}{7} \hat{\jmath}+\frac{6 k^{\wedge}}{7}\right) \\\\ &\vec{r}=3\left(2 i^{\wedge}-3 \hat{\jmath}+6 k^{\wedge}\right) \\\\ &\vec{r}=6 i^{\wedge}-9 \hat{\jmath}+18 k^{\wedge} \end{aligned}$
Therefore, the vector $\vec{r} \text { is }\left(6 i^{\wedge}-9 j^{\wedge}+18 k^{\wedge}\right)$

Direction Cosines and Direction Ratios exercise Fill in the blanks question 11

Answer: -$\left(\frac{-6}{7}, \frac{-2}{7}, \frac{-3}{7}\right) \text { or }\left(\frac{6}{7}, \frac{2}{7}, \frac{3}{7}\right)$
Hint:
Use direction ratio of the line
Given:
Points $(4, 3,-5)$ and $(-2, 1,-8)$
To Find:
Direction cosine of line joining point
Solution:
Let point $P(4, 3,-5)$ and $Q (-2, 1,-8)$ are joined to form line $PQ$
Direction Ratio of line $PQ=$ Position vector of $P-$ Position vector of $Q$
Direction ratio of line $PQ=[4-(-2), 3-1, -5-(-8)]$
$(a,b,c)=(6, 2, 3)$
Direction Cosine of line $PQ$ are:
$l=\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, m=\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}, n=\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}$
$l=\frac{6}{\sqrt{6^{2}+2^{2}+3^{2}}}, m=\frac{2}{\sqrt{6^{2}+2^{2}+3^{2}}}, n=\frac{3}{\sqrt{6^{2}+2^{2}+3^{2}}}$
$\begin{aligned} &l=\frac{6}{\sqrt{49}}, m=\frac{2}{\sqrt{49}}, n=\frac{3}{\sqrt{49}} \\\\ &l=\pm \frac{6}{7}, m=\pm \frac{2}{7}, n=\pm \frac{3}{7} \end{aligned}$
$\begin{aligned} &(l, m, n)=\left(\frac{-6}{7},-\frac{2}{7}, \frac{-3}{7}\right) \\\\ &\text { or }\left(\frac{6}{7}, \frac{2}{7}, \frac{3}{7}\right) \end{aligned}$
Therefore the direction cosine of the line segment are $\left(\frac{-6}{7}, \frac{-2}{7}, \frac{-3}{7}\right) \text { or }\left(\frac{6}{7}, \frac{2}{7}, \frac{3}{7}\right)$.

Direction Cosines and Direction Ratios exercise Fill in the blanks question 12

Answer:$\pm \sqrt{3}$
Hint:
Use property of direction cosine.
Given:
Direction cosine of line $(l, m, n)=\left(\frac{1}{c}, \frac{1}{c}, \frac{1}{c}\right)$
To Find:
Value of c
Solution:
$(l, m, n)=\left(\frac{1}{c}, \frac{1}{c}, \frac{1}{c}\right)\\\\ Now \; \; as\; \; l^{2}+m^{2}+n^{2}=1\\\\ \left(\frac{1}{c}\right)^{2}+\left(\frac{1}{c}\right)^{2}+\left(\frac{1}{c}\right)^{2}=1$
$\begin{aligned} &\frac{3}{c^{2}}=1 \\\\ &c^{2}=3 \\\\ &c=\pm \sqrt{3} \end{aligned}$
Therefore the value of c is $\pm \sqrt{3}$.

Direction Cosines and Direction Ratios exercise Fill in the blanks question 13

Answer: Point $P=(-2, 4,-4)$

Hint:
Use direction ratio to find direction cosine
Given:
$\overrightarrow{\mid O P} \mid=6\; and\\\\ (a, b, c)=(-1,2,-2)$
To Find:
Point P
Solution:
Direction cosine is related to direction ratio as :
$l=\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, \frac{m}{\sqrt{a^{2}+b^{2}+c^{2}}} n=\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}$
$l=\frac{-1}{\sqrt{(-1)^{2}+2^{2}+(-2)^{2}}}, m=\frac{2}{\sqrt{(-1)^{2}+2^{2}+(-2)^{2}}}, n=\frac{-2}{\sqrt{(-1)^{2}+2^{2}+(-2)^{2}}}$
$l=\frac{-1}{3}, m=\frac{2}{3}, n=\frac{-2}{3}$
Now, $\overrightarrow{O P}=$ Position vector of $P -$ Position vector of $O$
$\Rightarrow \overrightarrow{\mathrm{1}}=x i^{\wedge}+y \mathrm{l}^{\wedge}+z k^{\wedge}$
where x,y,z are the coordinate of P
Now $\overrightarrow{O P}=|\overrightarrow{O P}|\left(l \hat{\imath}+m \hat{j}+n k^{n}\right)$
$\begin{aligned} &\left(x i^{\wedge}+y i^{\wedge}+2 k^{\wedge}\right)=6\left(-\frac{1}{3} i^{\wedge}+\frac{2}{3} \hat{\jmath}-\frac{2}{3} k^{\wedge}\right) \\\\ &\left.\left(x i^{\wedge}+y\right)^{\wedge}+2 k^{\wedge}\right)=\left(-2 \hat{1}+4 \hat{\jmath}-4 k^{\wedge}\right) \end{aligned}$
Comparing both
$(x, y, z)=(-2,4,-4)$
Therefore, the coordinate of point P are $(-2,4,-4).$

Direction Cosines and Direction Ratios exercise Fill in the blanks question 14

Answer:$\frac{\pi }{3}$
Hint:
Dot product of two direction cosine two vectors
Given:
$\begin{aligned} &\left(a_{1}, b_{1}, c_{1}\right)=(1,1,2) \text { and } \\ &\left(a_{2}, b_{2}, c_{2}\right)=(\sqrt{3}-1,-\sqrt{3}-1,4) \end{aligned}$
To Find:
Angle between two vectors
Solution:
Angle between two vectors with direction ratios $\left(a, b, c_{1}\right) \text { and }\left(a_{2}, b_{2}, c_{2}\right)$ is
given by
$\cos \theta=\left[\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \cdot \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\right]$
$\text { Now, } \sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}=\sqrt{(1)^{2}+(1)^{2}+(2)^{2}}=\sqrt{1+1+4}=\sqrt{6}$
$\text { Also } \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}=\sqrt{(\sqrt{3}-1)^{2}+(-\sqrt{3}-1)^{2}+4^{2}}$
$=\sqrt{3+1-2 \sqrt{3}+3+1+2 \sqrt{3}+16}=\sqrt{24}$
$\text { Hence } \cos \theta=\left[\frac{1(\sqrt{3}-1)+1 \cdot(-\sqrt{3}-1)+2 \cdot 4}{\sqrt{6} \cdot \sqrt{24}}\right]$
$\begin{aligned} &\cos \theta=\left[\frac{\sqrt{3}-1-\sqrt{3}-1+8}{\sqrt{6 \times 24}}\right] \\ &\cos \theta=\frac{6}{12} \end{aligned}$
$\begin{aligned} &\theta=\cos ^{-1}\left[\frac{1}{2}\right] \\ &\theta=\frac{\pi}{3} \end{aligned}$
Therefore, the angle between two vectors is $\frac{\pi }{3}$.

Direction Cosines and Direction Ratios exercise Fill in the blanks question 15

Answer: $\frac{\sqrt{23}}{6}$

Hint:
Use the property of direction cosine
Given:
Direction cosine $=\left(\frac{1}{2}, \frac{1}{3}, n\right)$
To Find:
Value of $n$
Solution:
By property of direction cosine, we get
$\begin{aligned} &l^{2}+m^{2}+n^{2}=1 \\\\ &\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{3}\right)^{2}+n^{2}=1 \\\\ &\frac{1}{4}+\frac{1}{9}+n^{2}=1 \end{aligned}$
$\begin{aligned} &n^{2}=1-\frac{1}{4}-\frac{1}{9} \\ &n^{2}=\frac{23}{36} \\\\ &n=\sqrt{\frac{23}{36}} \\\\ &n=\frac{\sqrt{23}}{6} \end{aligned}$

Direction Cosines and Direction Ratios exercise Fill in the blanks question 16

Answer:$\frac{\pi }{2}$
Hint:
Use property $\left[\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1\right]$
Given:
$\alpha+\beta=\frac{\pi}{2}$
To Find:
$\gamma$ (Angle between line and z axis)
Solution:
If a line makes α,β,γ, then
$\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1$ ......(i)
It is given that $\alpha+\beta=\frac{\pi}{2}$
$\alpha=\frac{\pi}{2}-\beta$
Taking cosine both side
$\begin{aligned} &\cos \alpha=\cos \left(\frac{\pi}{2}-\beta\right) \\\\ &\cos \alpha=\sin \beta \\\\ &\cos ^{2} \alpha=\sin ^{2} \beta \end{aligned}$ $\left[\sin ^{2} x+\cos ^{2} x=1\right]$
$\begin{aligned} &\cos ^{2} \alpha=1-\cos ^{2} \beta \\\\ &\cos ^{2} \alpha+\cos ^{2} \beta=1 \end{aligned}$ .......(ii)
Using the above equation (ii) in (i) we get
$\begin{aligned} &\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1 \\\\ &1+\cos ^{2} x=1 \\\\ &\cos ^{2} x=0 \end{aligned}$
$\gamma=\frac{\pi}{2}\\\\ or \; \; 90^{\circ}$
Therefore, the value of $\gamma$ is $\frac{\pi }{2}$

Direction Cosines and Direction Ratios exercise Fill in the blanks question 17

Answer: 4
Hint:
Use the property $\left[\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} y=1\right]$
Given:
$\alpha=\beta=\gamma$
To Find:
Number of equally inclined lines
Solution:
If α,β,γ are the angles made by line with axes, then
$\begin{array}{ll} \Rightarrow \cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1 & {[\alpha=\beta=\gamma]} \end{array}$
$\begin{aligned} &\Rightarrow 3 \cos ^{2} \alpha=1 \\ &\Rightarrow \cos ^{2} \alpha=\frac{1}{3} \\ &\Rightarrow \cos \alpha=\cos \beta=\cos \gamma=\pm \frac{1}{\sqrt{3}} \end{aligned}$
Possible direction cosine are $\left(\pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}\right)$
Different set of direction cosine are
$\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right),\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\right),\left(\frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \text { and }\left(\frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$

Therefore, 4 lines are equally inclined to axes.

Direction Cosines and Direction Ratios exercise Fill in the blanks question 18

Answer →$-\frac{3}{7}$

Hint :
In zx plane y-coordinate is zero and y coordinate of points divide line segment.
Given :
Points are $(2,3,1) \text { and }(6,7,1)$
To find:
Ratio in which line is divided
Solution :
Let zx-plane divide the line segment in ratio
$\left(m_{1}\right) \quad\quad\quad\quad\left(m_{2}\right)$
$p\quad \quad\quad\quad\quad\quad\quad1$
$(2,3,1)\quad\quad\quad\quad (x, 0, z)\quad\quad\quad\quad (6,7,1)$
$\left(x_{1}, y_{1}, z_{1}\right) \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\left(x_{2}, y_{2}, z_{2}\right)$
$\text { zx plane }$
Using section formula,
$\begin{aligned} &y=\frac{m_{1} y_{2}+m_{2} y_{2}}{m_{1}+m_{2}} \\\\ &0=\frac{7 p+3}{p+1} \end{aligned}$
$\begin{aligned} &7 p+3=0 \\\\ &p=\frac{-3}{7} \end{aligned}$
Therefore, the zx plane divide the line in ratio $\frac{-3}{7}$

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