Examples (Revised) - Chapter 11 - Three Dimensional Geometry - Ncert Solutions class 12 - Maths

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Ncert Solutions Class 12 Maths: Chapter 11 - Three Dimensional Geometry

Example 1

If a line makes angle $90^{\circ}, 60^{\circ}$ and $30^{\circ}$ with the positive direction of $x, y$ and $z$-axis respectively, find its direction cosines.
Solution

Let the $d . c$.' $s$ of the lines be $l, m, n$. Then $l=\cos 90^{\circ}=0, m=\cos 60^{\circ}=\frac{1}{2}$,
$
n=\cos 30^{\circ}=\frac{\sqrt{3}}{2} \text {. }
$

Example 2

If a line has direction ratios $2,-1,-2$, determine its direction cosines.
Solution

Direction cosines are
$
\frac{2}{\sqrt{2^2+(-1)^2+(-2)^2}}, \frac{-1}{\sqrt{2^2+(-1)^2+(-2)^2}}, \frac{-2}{\sqrt{2^2+(-1)^2+(-2)^2}}
$
or $\quad \frac{2}{3}, \frac{-1}{3}, \frac{-2}{3}$

Example 3

Find the direction cosines of the line passing through the two points $(-2,4,-5)$ and $(1,2,3)$.
Solution

We know the direction cosines of the line passing through two points $\mathrm{P}\left(x_1, y_1, z_1\right)$ and $\mathrm{Q}\left(x_2, y_2, z_2\right)$ are given by
$
\begin{aligned}
& \frac{x_2-x_1}{\mathrm{PQ}}, \frac{y_2-y_1}{\mathrm{PQ}}, \frac{z_2-z_1}{\mathrm{PQ}} \\
& \mathrm{PQ}=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}
\end{aligned}
$
where
$
\mathrm{PQ}=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}
$

Here $\mathrm{P}$ is $(-2,4,-5)$ and $\mathrm{Q}$ is $(1,2,3)$.
So
$
\mathrm{PQ}=\sqrt{(1-(-2))^2+(2-4)^2+(3-(-5))^2}=\sqrt{77}
$

Thus, the direction cosines of the line joining two points is
$
\frac{3}{\sqrt{77}}, \frac{-2}{\sqrt{77}}, \frac{8}{\sqrt{77}}
$

Example 4

Find the direction cosines of $x, y$ and $z$-axis.
Solution

The $x$-axis makes angles $0^{\circ}, 90^{\circ}$ and $90^{\circ}$ respectively with $x, y$ and $z$-axis. Therefore, the direction cosines of $x$-axis are $\cos 0^{\circ}, \cos 90^{\circ}, \cos 90^{\circ}$ i.e., $1,0,0$. Similarly, direction cosines of $y$-axis and $z$-axis are $0,1,0$ and $0,0,1$ respectively.

Example 5

Show that the points A $(2,3,-4)$, B $(1,-2,3)$ and C $(3,8,-11)$ are collinear.
Solution

Direction ratios of line joining $A$ and $B$ are
$
1-2,-2-3,3+4 \text { i.e., }-1,-5,7 \text {. }
$

The direction ratios of line joining $B$ and $C$ are
$
3-1,8+2,-11-3 \text {, i.e., } 2,10,-14 \text {. }
$

It is clear that direction ratios of $\mathrm{AB}$ and $\mathrm{BC}$ are proportional, hence, $\mathrm{AB}$ is parallel to $B C$. But point $B$ is common to both $A B$ and $B C$. Therefore, $A, B, C$ are collinear points.

Example 6

Find the vector and the Cartesian equations of the line through the point $(5,2,-4)$ and which is parallel to the vector $3 \hat{i}+2 \hat{j}-8 \hat{k}$.
Solution

We have
$
\vec{a}=5 \hat{i}+2 \hat{j}-4 \hat{k} \text { and } \vec{b}=3 \hat{i}+2 \hat{j}-8 \hat{k}
$

Therefore, the vector equation of the line is
$
\vec{r}=5 \hat{i}+2 \hat{j}-4 \hat{k}+\lambda(3 \hat{i}+2 \hat{j}-8 \hat{k})
$

Now, $\vec{r}$ is the position vector of any point $\mathrm{P}(x, y, z)$ on the line.
Therefore,
$
\begin{aligned}
x \hat{i}+y \hat{j}+z \hat{k} & =5 \hat{i}+2 \hat{j}-4 \hat{k}+\lambda(3 \hat{i}+2 \hat{j}-8 \hat{k}) \\
& =(5+3 \lambda) \hat{i}+(2+2 \lambda) \hat{j}+(-4-8 \lambda) \hat{k}
\end{aligned}
$

Eliminating $\lambda$, we get
$
\frac{x-5}{3}=\frac{y-2}{2}=\frac{z+4}{-8}
$
which is the equation of the line in Cartesian form.

Example 7

Find the angle between the pair of lines given by
$
\vec{r}=3 \hat{i}+2 \hat{j}-4 \hat{k}+\lambda(\hat{i}+2 \hat{j}+2 \hat{k})
$

and
$
\vec{r}=5 \hat{i}-2 \hat{j}+\mu(3 \hat{i}+2 \hat{j}+6 \hat{k})
$

Solution

Here $\vec{b}_1=\hat{i}+2 \hat{j}+2 \hat{k}$ and $\vec{b}_2=3 \hat{i}+2 \hat{j}+6 \hat{k}$
The angle $\theta$ between the two lines is given by
$
\begin{aligned}
\cos \theta & =\left|\frac{\vec{b}_1 \cdot \vec{b}_2}{\left|\vec{b}_1\right|\left|\vec{b}_2\right|}\right|=\left|\frac{(\hat{i}+2 \hat{j}+2 \hat{k}) \cdot(3 \hat{i}+2 \hat{j}+6 \hat{k})}{\sqrt{1+4+4} \sqrt{9+4+36}}\right| \\
& =\left|\frac{3+4+12}{3 \times 7}\right|=\frac{19}{21} \\
\text { Hence } \quad \theta & =\cos ^{-1}\left(\frac{19}{21}\right)
\end{aligned}
$

Example 8

Find the angle between the pair of lines
$
\begin{aligned}
& \frac{x+3}{3}=\frac{y-1}{5}=\frac{z+3}{4} \\
& \frac{x+1}{1}=\frac{y-4}{1}=\frac{z-5}{2}
\end{aligned}
$
Solution

The direction ratios of the first line are 3, 5, 4 and the direction ratios of the second line are $1,1,2$. If $\theta$ is the angle between them, then
$
\cos \theta=\left|\frac{3.1+5.1+4.2}{\sqrt{3^2+5^2+4^2} \sqrt{1^2+1^2+2^2}}\right|=\frac{16}{\sqrt{50} \sqrt{6}}=\frac{16}{5 \sqrt{2} \sqrt{6}}=\frac{8 \sqrt{3}}{15}
$

Hence, the required angle is $\cos ^{-1}\left(\frac{8 \sqrt{3}}{15}\right)$.

Example 9

Find the shortest distance between the lines $l_1$ and $l_2$ whose vector equations are
and
$
\begin{aligned}
\vec{r} & =\hat{i}+\hat{j}+\lambda(2 \hat{i}-\hat{j}+\hat{k}) \\
\vec{r} & =2 \hat{i}+\hat{j}-\hat{k}+\mu(3 \hat{i}-5 \hat{j}+2 \hat{k})
\end{aligned}
$

Solution

Comparing (1) and (2) with $\vec{r}=\vec{a}_1+\lambda \vec{b}_1$ and $\overrightarrow{\mathrm{r}}=\vec{a}_2+\mu \overrightarrow{\mathrm{b}}_2$ respectively, we get
$
\begin{aligned}
& \overrightarrow{\mathrm{a}}_1=\hat{i}+\hat{j}, \vec{b}_1=2 \hat{i}-\hat{j}+\hat{k} \\
& \overrightarrow{\mathrm{a}}_2=2 \hat{i}+\hat{j}-\hat{k} \text { and } \vec{b}_2=3 \hat{i}-5 \hat{j}+2 \hat{k}
\end{aligned}
$

Therefore
$
\vec{a}_2-\vec{a}_1=\hat{i}-\hat{k}
$
and
$
\begin{aligned}
\vec{b}_1 \times \vec{b}_2 & =(2 \hat{i}-\hat{j}+\hat{k}) \times(3 \hat{i}-5 \hat{j}+2 \hat{k}) \\
& =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & -1 & 1 \\
3 & -5 & 2
\end{array}\right|=3 \hat{i}-\hat{j}-7 \hat{k}
\end{aligned}
$

So
$
\left|\vec{b}_1 \times \vec{b}_2\right|=\sqrt{9+1+49}=\sqrt{59}
$

Hence, the shortest distance between the given lines is given by
$
d=\left|\frac{\left(\overrightarrow{\mathrm{b}}_1 \times \overrightarrow{\mathrm{b}}_2\right) \cdot\left(\overrightarrow{\mathrm{a}}_2-\overrightarrow{\mathrm{a}}_1\right)}{\left|\overrightarrow{\mathrm{b}}_1 \times \overrightarrow{\mathrm{b}}_2\right|}\right|=\frac{|3-0+7|}{\sqrt{59}}=\frac{10}{\sqrt{59}}
$

Example 10

Find the distance between the lines $l_1$ and $l_2$ given by
$
\text { and } \quad \begin{aligned}
\vec{r} & =\hat{i}+2 \hat{j}-4 \hat{k}+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k}) \\
\quad \vec{r} & =3 \hat{i}+3 \hat{j}-5 \hat{k}+\mu(2 \hat{i}+3 \hat{j}+6 \hat{k})
\end{aligned}
$

Solution

The two lines are parallel (Why? ) We have
$
\overrightarrow{\mathrm{a}}_1=\hat{i}+2 \hat{j}-4 \hat{k}, \overrightarrow{\mathrm{a}}_2=3 \hat{i}+3 \hat{j}-5 \hat{k} \text { and } \vec{b}=2 \hat{i}+3 \hat{j}+6 \hat{k}
$

Therefore, the distance between the lines is given by
$
\begin{aligned}
& \left.d=\left|\frac{\vec{b} \times\left(\vec{a}_2-\vec{a}_1\right)}{|\vec{b}|}\right|=\left|\frac{\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & 6 \\
2 & 1 & -1
\end{array}\right|}{\sqrt{4+9+36}}\right| \right\rvert\, \\
& =\frac{|-9 \hat{i}+14 \hat{j}-4 \hat{k}|}{\sqrt{49}}=\frac{\sqrt{293}}{\sqrt{49}}=\frac{\sqrt{293}}{7} \\
&
\end{aligned}
$