Browse Questions

# Write the direction cosines of a line parallel to the line $\large\frac{3-x}{3} = \frac{y+2}{-2} = \frac{z+2}{6}.$

Toolbox:
• If two lines are parallel then the ratio of their direction ratios are proportional.
• $l,m,n$ are the direction ratios, then its direction cosines are $\large\frac{l}{\sqrt {l^2+m^2+n^2}}$$,\large\frac{m}{\sqrt {l^2+m^2+n^2}}$$\large\frac{n}{\sqrt {l^2+m^2+n^2}}$
Step 1:
Equation of the given line $AB$ is $\large\frac{3-x}{3}=\frac{y+2}{-2}=\frac{z+2}{6}$
$\Rightarrow \large\frac{x-3}{-3}=\frac{y+2}{-2}=\frac{z+2}{6}$
If two lines are parallel then the ratio of their direction ratios are proportional.
Hence the direction ratios of the line parallel to AB is
$-3k,-2k,6k$
dircetion ration are $(-3,-2,6)$
Step 2:
If $l,m,n$ are the direction ratios, then its direction cosines are
$\large\frac{l}{\sqrt {l^2+m^2+n^2}}$$,\large\frac{m}{\sqrt {l^2+m^2+n^2}}$$\large\frac{n}{\sqrt {l^2+m^2+n^2}}$
Here $l=-3,m=-2\; and\; n=6$
Therfore $\sqrt {l^2+m^2+n^2}=\sqrt {(-3)^2+(-2)^2+(6)^2}$
$\qquad\qquad\qquad\qquad\;\;\;=\sqrt {(9)+(4)+(36)}$
$\qquad\qquad\qquad\qquad\;\;\;=\sqrt {49}$
$\qquad\qquad\qquad\qquad\;\;\;=7$
Step 3:
Therefore The direction cosines are $\bigg( \large\frac{-3}{7},\large \frac{-2}{7}, \large\frac{6}{7} \bigg)$