Question

The direction ratios of two lines are $1,-2,-2$ and $0,2,1$. The direction cosines of the line perpendicular to the above lines are

• A. $\frac{2}{3},\frac{-1}{3},\frac{2}{3}$
• B. $\frac{-1}{3},\frac{2}{3},\frac{2}{3}$
• C. $\frac{1}{4},\frac{3}{4},\frac{1}{2}$
• D. None of these

Answer 1

Answer: (A)

$\frac{2}{3},\frac{-1}{3},\frac{2}{3}$

Let $a,b,c$ be the direction ratios of the line whose direction cosines are required.
Then as this line is perpendicular to the given lines so we have
$a\left(1\right)+b(-2)+c(-2)=0$ and $a\left(0\right)+b\left(2\right)+c\left(1\right)=0$
Solving these simultaneously, we get
$\frac{a}{(-2)\left(1\right)-(-2)\left(2\right)}=\frac{b}{(-2)\left(0\right)-\left(1\right)\left(1\right)}=\frac{c}{\left(1\right)\left(2\right)-\left(0\right)(-2)}$
$\Rightarrow \frac{a}{2}=\frac{b}{-1}=\frac{c}{2}\Rightarrow a:b:c=2:-1:2$
Therefore the required direction cosines are
$\frac{2}{\sqrt{2^2+1^2+2^2}},\frac{-1}{\sqrt{2^2+1^2+2^2}},\frac{2}{\sqrt{2^2+1^2+2^2}}\Rightarrow \frac{2}{3},\frac{-1}{3},\frac{2}{3}$