# NCERT Class 12 Maths Three Dimensional Geometry

## Chapter 11 Ex.11.1 Question 1

If a line makes angles $$90^\circ ,135^\circ ,45^\circ$$ with $$x,y$$ and $$z -$$axes respectively, find its direction cosines.

### Solution

Let direction cosines of the line be $$l,m$$ and $$n$$.

Hence,

\begin{align}l &= \cos 90^\circ = 0\\m &= \cos 135^\circ = \cos \left( {90^\circ + 45^\circ } \right)\\&= - \sin 45^\circ = - \frac{1}{{\sqrt 2 }}\\n& = \cos 45^\circ = \frac{1}{{\sqrt 2 }}\end{align}

Thus, the direction cosines of the line are $$0, - \frac{1}{{\sqrt 2 }}$$ and $$\frac{1}{{\sqrt 2 }}$$

## Chapter 11 Ex.11.1 Question 2

Find the direction cosines $$l,m$$ and $$n$$ of a line which makes equal angles with the coordinates axes.

### Solution

Let the direction cosines of the line make an angle $$\alpha$$ with each of the coordinates axes.

Hence,

\begin{align}l &= \cos \alpha \\m &= \cos \alpha \\n &= \cos \alpha\end{align}

Since, $${l^2} + {m^2} + {n^2} = 1$$

Hence,

\begin{align} &\Rightarrow \; {\cos ^2}\alpha + {\cos ^2}\alpha + {\cos ^2}\alpha = 1\\& \Rightarrow \;3{\cos ^2}\alpha = 1\\& \Rightarrow \;{\cos ^2}\alpha = \frac{1}{3}\\ &\Rightarrow \;\cos \alpha = \pm \frac{1}{{\sqrt 3 }}\end{align}

Thus, the direction cosines of the line, which is equally inclined to the coordinate axes, are $$\pm \frac{1}{{\sqrt 3 }}, \pm \frac{1}{{\sqrt 3 }}$$ and $$\pm \frac{1}{{\sqrt 3 }}$$.

## Chapter 11 Ex.11.1 Question 3

If a line has the direction ratios $$-18,12, - 4$$, then what are its direction cosines?

### Solution

If a line has direction ratios $$-18,12, - 4$$, then its direction cosines are

\begin{align}l &= \frac{{ - 18}}{{\sqrt {{{\left( { - 18} \right)}^2} + {{\left( {12} \right)}^2} + {{\left( { - 4} \right)}^2}} }} = \frac{{ - 18}}{{\sqrt {484} }} = \frac{{ - 18}}{{22}} = \frac{{ - 9}}{{11}}\\m &= \frac{{12}}{{\sqrt {{{\left( { - 18} \right)}^2} + {{\left( {12} \right)}^2} + {{\left( { - 4} \right)}^2}} }} = \frac{{12}}{{\sqrt {484} }} = \frac{{12}}{{22}} = \frac{6}{{11}}\\n &= \frac{{ - 4}}{{\sqrt {{{\left( { - 18} \right)}^2} + {{\left( {12} \right)}^2} + {{\left( { - 4} \right)}^2}} }} = \frac{{ - 4}}{{\sqrt {484} }} = \frac{{ - 4}}{{22}} = \frac{{ - 2}}{{11}}\end{align}

Hence, the direction cosines are $$\frac{{ - 9}}{{11}},\frac{6}{{11}}$$ and $$\frac{{ - 2}}{{11}}$$.

## Chapter 11 Ex.11.1 Question 4

Show that the points $$\left( {{\rm{2}},{\rm{3}},{\rm{4}}} \right),\left( { - {\rm{1}}, - {\rm{2}},{\rm{1}}} \right),\left( {{\rm{5}},{\rm{8}},{\rm{7}}} \right)$$ are collinear.

### Solution

Given points are $$A\left( {{\rm{2}},{\rm{3}},{\rm{4}}} \right),B\left( { - {\rm{1}}, - {\rm{2}},{\rm{1}}} \right)$$ and $$C\left( {{\rm{5}},{\rm{8}},{\rm{7}}} \right)$$.

As we know that the direction cosines of points,

$$\left( {{x_1},{y_1},{z_1}} \right)$$ and $$\left( {{x_2},{y_2},{z_2}} \right)$$ are given by $$\left( {{x_2} - {x_1}} \right),\left( {{y_2} - {y_1}} \right)$$ and $$\left( {{z_2} - {z_1}} \right)$$ .

Therefore, the direction ratios of $$AB$$ are

$$\left( { - {\rm{1}} - {\rm{2}}} \right),\left( { - {\rm{2}} - {\rm{3}}} \right)$$ and $$\left( {{\rm{1}} - {\rm{4}}} \right)$$

$$\Rightarrow \;- {\rm{3}}, - {\rm{5 }}$$ and $$- {\rm{3}}$$

The direction ratios of $$BC$$ are

$$\left[ {{\rm{5}}-\left( { - {\rm{1}}} \right)} \right]{\rm{,}}\left[ {{\rm{8}}-\left( { - {\rm{2}}} \right)} \right]$$ and $$\left( {{\rm{7}}-{\rm{1}}} \right)$$

$$\Rightarrow\; 6,\,10$$ and $$6$$

It can be seen that the direction ratios of $$BC$$ are $$-2$$ times that $$AB$$ i.e., they are proportional.

Hence, $$AB$$ is parallel to $$BC$$. Since point $$B$$ is common to both $$AB$$ and $$BC,$$ points $$A, \,B,$$ and $$C$$ are collinear.

## Chapter 11 Ex.11.1 Question 5

Find the direction cosines of the sides of the triangle whose vertices are $$\left( {{\rm{3}},{\rm{5}}, - {\rm{4}}} \right),\left( { - {\rm{1}},{\rm{1}},{\rm{2}}} \right)$$ and $$\left( { - {\rm{5}}, - {\rm{5}}, - {\rm{2}}} \right)$$.

### Solution

Vertices of the triangle are $$A\left( {{\rm{3}},{\rm{5}}, - {\rm{4}}} \right),B\left( { - {\rm{1}},{\rm{1}},{\rm{2}}} \right)$$ and $$C\left( { - {\rm{5}}, - {\rm{5}}, - {\rm{2}}} \right)$$

The direction ratios of the side $$AB$$ are

$$\left( { - {\rm{1}} - {\rm{3}}} \right),\left( {{\rm{1}} - {\rm{5}}} \right)$$ and $$\left[ {{\rm{2}}-\left( { - {\rm{4}}} \right)} \right]$$

$$\Rightarrow - 4, - 4$$ and $$6$$

Hence, the direction cosines of $$AB$$ are

\begin{align}{l_1} &= \frac{{ - 4}}{{\sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 6 \right)}^2}} }} = \frac{{ - 4}}{{\sqrt {68} }} = \frac{{ - 4}}{{2\sqrt {17} }} = \frac{{ - 2}}{{\sqrt {17} }}\\{m_1} &= \frac{{ - 4}}{{\sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 6 \right)}^2}} }} = \frac{{ - 4}}{{\sqrt {68} }} = \frac{{ - 4}}{{2\sqrt {17} }} = \frac{{ - 2}}{{\sqrt {17} }}\\{n_1} &= \frac{6}{{\sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 6 \right)}^2}} }} = \frac{6}{{\sqrt {68} }} = \frac{6}{{2\sqrt {17} }} = \frac{3}{{\sqrt {17} }}\end{align}

The direction ratios of $$BC$$ are

$$\left[ { - 5\left( { - 1} \right)} \right],\left( { - 5 - 1} \right)$$ and $$\left( { - 2 - 2} \right)$$

$$\Rightarrow - 4, - 6$$ and $$- 4$$

Hence, the direction cosines of $$BC$$ are

\begin{align}{l_2} &= \frac{{ - 4}}{{\sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 6} \right)}^2} + {{\left( { - 4} \right)}^2}} }} = \frac{{ - 4}}{{\sqrt {68} }} = \frac{{ - 4}}{{2\sqrt {17} }} = \frac{{ - 2}}{{\sqrt {17} }}\\{m_2}& = \frac{{ - 6}}{{\sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 6} \right)}^2} + {{\left( { - 4} \right)}^2}} }} = \frac{{ - 6}}{{\sqrt {68} }} = \frac{{ - 6}}{{2\sqrt {17} }} = \frac{{ - 3}}{{\sqrt {17} }}\\{n_2}& = \frac{{ - 4}}{{\sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 6} \right)}^2} + {{\left( { - 4} \right)}^2}} }} = \frac{{ - 4}}{{\sqrt {68} }} = \frac{{ - 4}}{{2\sqrt {17} }} = \frac{{ - 2}}{{\sqrt {17} }}\end{align}

The direction ratios of $$CA$$ are

$$\left( { - 5 - 3} \right),\left( { - 5 - 5} \right)$$ and $$\left[ { - {\rm{2}}-\left( { - {\rm{4}}} \right)} \right]$$

$$\Rightarrow - {\rm{8}}, - {\rm{1}}0$$ and $${\rm{2}}$$

Hence, the direction cosines of $$AC$$ are

\begin{align}{l_3} &= \frac{{ - 8}}{{\sqrt {{{\left( { - 8} \right)}^2} + {{\left( {10} \right)}^2} + {{\left( 2 \right)}^2}} }} = \frac{{ - 8}}{{\sqrt {168} }} = \frac{{ - 8}}{{2\sqrt {42} }} = \frac{{ - 4}}{{\sqrt {42} }}\\{m_3}& = \frac{{ - 10}}{{\sqrt {{{( - 8)}^2} + {{(10)}^2} + {{(2)}^2}} }} = \frac{{ - 10}}{{\sqrt {168} }} = \frac{{ - 10}}{{2\sqrt {42} }} = \frac{{ - 5}}{{\sqrt {42} }}\\{n_3}& = \frac{2}{{\sqrt {{{( - 8)}^2} + {{(10)}^2} + {{(2)}^2}} }} = \frac{2}{{\sqrt {168} }} = \frac{2}{{2\sqrt {42} }} = \frac{1}{{\sqrt {42} }}\end{align}

Thus, the direction cosines of the sides of the triangle are

$$\left( {\frac{{ - 2}}{{\sqrt {17} }},\frac{{ - 2}}{{\sqrt {17} }},\frac{3}{{\sqrt {17} }}} \right),\left( {\frac{{ - 2}}{{\sqrt {17} }},\frac{{ - 3}}{{\sqrt {17} }},\frac{{ - 2}}{{\sqrt {17} }}} \right)$$ and $$\left( {\frac{{ - 4}}{{\sqrt {42} }},\frac{{ - 5}}{{\sqrt {42} }},\frac{1}{{\sqrt {42} }}} \right)$$