Exercise 11.1 (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

Ex 11.1 Question 1.

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

Answer.

Here $\alpha=90^{\circ}, \beta=135^{\circ}$ and $\gamma=45^{\circ}$
Since direction cosines of a line making angles $\alpha, \beta, \gamma$ with the $x, y$ and $z$-axes respectively are $\cos \alpha, \cos \beta, \cos \gamma$.

Therefore, the direction cosines of the required line are:
$
\begin{aligned}
& \cos 90^{\circ}=0 ; \cos 135^{\circ}=\frac{-1}{\sqrt{2}} ; \cos 45^{\circ}=\frac{1}{\sqrt{2}} \\
& {\left[\because \cos 135^{\circ}=\cos \left(180^{\circ}-45^{\circ}\right)=-\cos 45^{\circ}=\frac{-1}{\sqrt{2}}\right]}
\end{aligned}
$

Ex 11.1 Question 2.

Find the direction cosines of a line which makes equal angles with the co-ordinate axes.

Answer.

Let a line make equal angles $\alpha, \alpha, \alpha$ with the co-ordinate axes.
$\therefore$ Direction cosines of the line are $\cos \alpha, \cos \alpha, \cos \alpha$ $\qquad$
$
\begin{aligned}
& \therefore \cos ^2 \alpha+\cos ^2 \alpha+\cos ^2 \alpha=1\left[\because \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1\right] \\
& \Rightarrow 3 \cos ^2 \alpha=1 \\
& \Rightarrow \cos ^2 \alpha=\frac{1}{3} \\
& \Rightarrow \cos \alpha= \pm \frac{1}{\sqrt{3}}
\end{aligned}
$

Putting $\cos \alpha= \pm \frac{1}{\sqrt{3}}$ in eq. (i), direction cosines of the required line making equal angles with the co-ordinate axes are $\pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}$.

Direction cosines of a line making equal angles with the co-ordinate axes in the positive i.e., first octant are $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$.

Ex 11.1 Question 3.

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

Answer.

We know that if $a, b, c$ are direction ratios of a line, then direction cosines of the line are:
$
\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}
$

Here direction ratios of the line are $-18,12,-4$
Putting the values in eq. (i),
$
\begin{aligned}
& \frac{-18}{\sqrt{(-18)^2+(12)^2+(-4)^2}}, \frac{12}{\sqrt{(-18)^2+(12)^2+(-4)^2}}, \frac{-4}{\sqrt{(-18)^2+(12)^2+(-4)^2}} \\
& \Rightarrow \frac{-18}{\sqrt{324+144+16}}, \frac{12}{\sqrt{324+144+16}}, \frac{-4}{\sqrt{324+144+16}} \\
& \Rightarrow \frac{-18}{\sqrt{484}}, \frac{12}{\sqrt{484}}, \frac{-4}{\sqrt{484}} \\
& \Rightarrow \frac{-18}{22}, \frac{12}{22}, \frac{-4}{22} \\
& \Rightarrow \frac{-9}{11}, \frac{6}{11}, \frac{-2}{11}
\end{aligned}
$

$\text { Hence, direction cosines of required line are } \frac{-9}{11}, \frac{6}{11}, \frac{-2}{11} \text {. }$

Ex 11.1 Question 4

how that the points $(2,3,4),(-1,-2,1),(5,8,7)$ are collinear.

Answer.

The given points are $\mathrm{A}(2,3,4), \mathrm{B}(-1,-2,1)$ and $\mathrm{C}(5,8,7)$
$\therefore$ Direction ratios of the line joining $\mathrm{A}$ and $\mathrm{B}$ are
$
\begin{aligned}
& -1-2,-2-3,1-4\left[\because x_2-x_1, y_2-y_1, z_2-z_1\right] \\
& \Rightarrow-3,-5,-3=a_1, b_1, c_1 \text { (say) }
\end{aligned}
$

Again Direction ratios of the line joining $\mathrm{B}$ and $\mathrm{C}$ are
$
5-(-1), 8-(-2), 7-1=6,10,6=a_2, b_2, c_2 \text { (say) }
$

From eq. (i) and (ii),
$
\begin{aligned}
& \frac{-3}{6}=\frac{-1}{2}, \frac{-5}{10}=\frac{-1}{2}, \frac{-3}{6}=\frac{-1}{2} \\
& \Rightarrow \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}
\end{aligned}
$

Therefore, $\mathrm{AB}$ is parallel to $\mathrm{BC}$. But point $\mathrm{B}$ is common to both $\mathrm{AB}$ and $\mathrm{BC}$. Hence points $\mathrm{A}, \mathrm{B}$, C are collinear.

Ex 11.1 Question 5.

Find the direction cosines of the sides of the triangle whose vertices are $(3,5,-4),(-1,1,2)$ and $(-5,-5,-2)$.

Answer.

Direction ratios of the line joining $\mathrm{A}$ and $\mathrm{B}$ are $-1-3,1-5,2-(-4)$
$
\Rightarrow-4,-4,6\left[\because x_2-x_1, y_2-y_1, z_2-z_1\right]
$
$\therefore$ Direction cosines of line $\mathrm{AB}$ are

$
\begin{aligned}
& \frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}} \\
& \Rightarrow \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}} \\
& \Rightarrow \frac{-4}{\sqrt{16+16+36}}, \frac{-4}{\sqrt{16+16+36}}, \frac{6}{\sqrt{16+16+36}} \\
& \Rightarrow \frac{-4}{\sqrt{68}}, \frac{-4}{\sqrt{68}}, \frac{6}{\sqrt{68}} \\
& \Rightarrow \frac{-4}{2 \sqrt{17}}, \frac{-4}{2 \sqrt{17}}, \frac{6}{2 \sqrt{17}} \\
& \Rightarrow \frac{-2}{\sqrt{17}}, \frac{-2}{\sqrt{17}}, \frac{3}{\sqrt{17}}
\end{aligned}
$

Now Direction ratios of the line joining $\mathrm{B}$ and $\mathrm{C}$ are $-5-(-1),-5-1,-2-2=-4,-6,-4$
$\therefore$ Direction cosines of line $\mathrm{BC}$ are

$\begin{aligned}
& \frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}} \\
& \Rightarrow \frac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}}, \frac{-6}{\sqrt{(-4)^2+(-6)^2+(-4)^2}}, \frac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}} \\
& \Rightarrow \frac{-4}{\sqrt{16+36+16}}, \frac{-6}{\sqrt{16+36+16}}, \frac{-4}{\sqrt{16+36+16}} \\
& \Rightarrow \frac{-4}{\sqrt{68}}, \frac{-6}{\sqrt{68}}, \frac{-4}{\sqrt{68}} \\
& \Rightarrow \frac{-4}{2 \sqrt{17}}, \frac{-6}{2 \sqrt{17}}, \frac{-4}{2 \sqrt{17}}
\end{aligned}$

$
\Rightarrow \frac{-2}{\sqrt{17}}, \frac{-3}{\sqrt{17}}, \frac{-2}{\sqrt{17}}
$

Direction ratios of the line joining $\mathrm{C}$ and $\mathrm{A}$ are $3-(-5), 5-(-5),-4-(-2)=8,10,-2$
$\therefore$ Direction cosines of line $\mathrm{CA}$ are
$
\begin{aligned}
& \frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}} \\
& \Rightarrow \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}} \\
& \Rightarrow \frac{8}{\sqrt{64+100+4}}, \frac{10}{\sqrt{64+100+4}}, \frac{-2}{\sqrt{64+100+4}} \\
& \Rightarrow \frac{8}{\sqrt{168}}, \frac{10}{\sqrt{168}}, \frac{-2}{\sqrt{168}} \\
& \Rightarrow \frac{8}{2 \sqrt{42}}, \frac{10}{2 \sqrt{42}}, \frac{-2}{2 \sqrt{42}} \\
& \Rightarrow \frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, \frac{-1}{\sqrt{42}}
\end{aligned}
$