Question

Find the direction cosines of the vector which is perpendicular to the vectors with direction ratios $-1,2,2$ and $0,2,1$.

Answer 1

We have direction ratio are

$-1,2,2$ and $0,2,1$

Let $a,b,c$ be the direction ratio of the line.

$-1a+2b+2c=0-------------\left(1\right)$

$3a-b+c=0---------------\left(2\right)$

By using cross multiplication method.

$\frac{a}{2+2}=\frac{b}{6+1}=\frac{c}{1-6}=k$ (say)

$\Rightarrow \frac{a}{4}=\frac{b}{7}=\frac{c}{-5}=k$ (say)

Now, $a=4k,b=7k,c=-5k$

Then,

$D^{\prime }$ ration are $4k,7k,-5k$ or $4,7,-5\Rightarrow (l,m,n)$

Now, we prove that

$D^{\prime }$ crossing are $\frac{l}{\sqrt{l^2+m^2+n^2}},\frac{m}{\sqrt{l^2+m^2+n^2}},\frac{n}{\sqrt{l^2+m^2+n^2}}$

$\Rightarrow \frac{4}{\sqrt{90}},\frac{7}{\sqrt{90}},\frac{-5}{\sqrt{90}}$

Hence, this is the answer.