Question

Find the direction cosines of the line which is perpendicular to the lines with direction cosines proportional to 1, -2, -2 and 0, 2, 1.

• A. $\frac{1}{3},\frac{2}{3},-\frac{1}{3}$
• B. $\frac{2}{3},-\frac{1}{3},\frac{2}{3}$
• C. $1,-\frac{1}{3},\frac{2}{3}$
• D. None of these

Answer 1

The correct option is B $\frac{2}{3},-\frac{1}{3},\frac{2}{3}$

Let $l,m,n$ be the direction cosines of the required line. Since it is perpendicular to the lines whose direction cosines are proportional to $1,-2,-2$ and $0,2,1$ respectively.

Thus,

$l-2m-2n=0$

$0l+2m+n=0$

On solving these equations, we get,

$\frac{l}{2}=\frac{m}{-1}=\frac{n}{2}$

Thus, the direction ratios of the required line are proportional to $2,-1,2$. Hence, its direction cosines are :

$\frac{2}{\sqrt{2^2+(-1{)}^2+2^2}},\frac{-1}{\sqrt{2^2+(-1{)}^2+2^2}},\frac{2}{\sqrt{2^2+(-1{)}^2+2^2}}=\frac{2}{3},-\frac{1}{3},\frac{2}{3}$.