Question

Find the direction ratios of a vector perpendicular to the lines whose direction ratios are $-2,1,1$ and $-3,-4,1$.

Answer 1

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

Thus, $l(-2)+m\left(1\right)+n(-1)=0$

$-2l+m-n=0$ ....... (i)

and $l(-3)+m(-4)+n\left(1\right)=0$

$-3l-4m+n=0$ ...... (ii)

By cross-multiplication between (i) and (ii), we have

$\frac{l}{1-4}=\frac{m}{3+2}=\frac{n}{8+3}$

$\Rightarrow \frac{l}{-3}=\frac{m}{5}=\frac{n}{11}$

Let us find $\sqrt{l^2+m^2+n^2}$

$\sqrt{l^2+m^2+n^2}=\sqrt{(-3{)}^2+(5{)}^2+(11{)}^2}$

$=\sqrt{9+25+121}$

$=\sqrt{155}$

Thus, the direction ratios of the line are, $-3,5,11$

The direction cosines are

$l=\frac{-3}{\sqrt{155}}$

$m=\frac{5}{\sqrt{155}}$

$n=\frac{11}{\sqrt{155}}$