Solution

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

$l=\cos 90^{\circ}=0$

$m=\cos 135^{\circ}=-\frac{1}{\sqrt{2}}$

$n=\cos 45^{\circ}=\frac{1}{\sqrt{2}}$

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

Solution

Let the direction cosines of the line make an angle $a$ with each of the coordinate axes.

$\therefore l=\cos a, m=\cos a, n=\cos a$

$ \begin{aligned} & l^{2}+m^{2}+n^{2}=1 \\ & \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{aligned} $

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 $\pm \frac{1}{\sqrt{3}}$.

Solution

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

$\frac{-18}{\sqrt{(-18)^{2}+(12)^{2}+(-4)^{2}}}, \frac{12}{\sqrt{(-18)^{2}+(12)^{2}+(-4)^{2}}}, \frac{-4}{\sqrt{(-18)^{2}+(12)^{2}+(-4)^{2}}}$

i.e., $\frac{-18}{22}, \frac{12}{22}, \frac{-4}{22}$

$\frac{-9}{11}, \frac{6}{11}, \frac{-2}{11}$

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

Solution

The given points are $A(2,3,4), B(-1,-2,1)$, and $C(5,8,7)$.

It is known that the direction ratios of line joining the points, $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, are given by, $x_2-x_1, y_2-y_1$, and $z_2-z_1$.

The direction ratios of $A B$ are $(-1-2),(-2-3)$, and $(1-4)$ i.e., $-3,-5$, and -3 . The direction ratios of $BC$ are $(5-(-1)),(8-(-2))$, and $(7-1)$ i.e., 6,10 , and 6 . It can be seen that the direction ratios of $B C$ are -2 times that of $A B$ i.e., they are proportional.

Therefore, $A B$ is parallel to $B C$. Since point $B$ is common to both $A B$ and $B C$, points $A, B$, and $C$ are collinear.

Solution

The vertices of $\triangle A B C$ are $A(3,5,-4), B(-1,1,2)$, and $C(-5,-5,-2)$.

The direction ratios of side $A B$ are $(-1-3),(1-5)$, and $(2-(-4))$ i.e., $-4,-4$, and 6 .

Then, $\sqrt{(-4)^{2}+(-4)^{2}+(6)^{2}}=\sqrt{16+16+36}$

$ \begin{aligned} & =\sqrt{68} \\ & =2 \sqrt{17} \end{aligned} $

Therefore, the direction cosines of $A B$ are

$ \begin{aligned} & \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}}} \\ & \frac{-4}{2 \sqrt{17}},-\frac{4}{2 \sqrt{17}}, \frac{6}{2 \sqrt{17}} \\ & \frac{-2}{\sqrt{17}}, \frac{-2}{\sqrt{17}}, \frac{3}{\sqrt{17}} \end{aligned} $

The direction ratios of $BC$ are $(-5-(-1)),(-5-1)$, and $(-2-2)$ i.e., $-4,-6$, and -4 . Therefore, the direction cosines of $B C$ are

$ \begin{aligned} & \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}}} \\ & \text{ i.e., } \frac{-4}{2 \sqrt{17}}, \frac{-6}{2 \sqrt{17}}, \frac{-4}{2 \sqrt{17}} \end{aligned} $

The direction ratios of CA are $(-5-3),(-5-5)$, and $(-2-(-4))$ i.e., $-8,-10$, and 2 . Therefore, the direction cosines of $A C$ are

$ \begin{aligned} & \frac{-8}{\sqrt{(-8)^{2}+(10)^{2}+(2)^{2}}}, \frac{-5}{\sqrt{(-8)^{2}+(10)^{2}+(2)^{2}}}, \frac{2}{\sqrt{(-8)^{2}+(10)^{2}+(2)^{2}}} \\ & \text{ i.e., } \frac{-8}{2 \sqrt{42}}, \frac{-10}{2 \sqrt{42}}, \frac{2}{2 \sqrt{42}} \end{aligned} $