Question

Find the direction ratio of a line perpendicular to the two lines whose direction ratios are
$-2,1,-1$ and $-3,-4,1$

Answer 1

Given direction ratios are :$-2,1,-1$ and $-3,-4,1$

Let $a,b$ and $c$ be the direction ratios of the line perpendicular to the given lines.

Thus, we have,

$-2a+b-c=0$

$-3a-4b+c=0$

Cross multiplying, we get

$\frac{a}{1\times 1-(-4)\times (-1)}=\frac{b}{(-3)\times (-1)-(-2)\times 1}=\frac{c}{-2\times -4-(-3)\times 1}$

$\Rightarrow \frac{a}{1-4}=\frac{b}{3+2}=\frac{c}{8+3}$

$\Rightarrow \frac{a}{-3}=\frac{b}{5}=\frac{c}{11}$

Let us find $\sqrt{a^2+b^2+c^2}=\sqrt{{(-3)}^2+5^2+{11}^2}=\sqrt{9+25+121}=\sqrt{155}$

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

The direction cosines are $:\frac{-3}{\sqrt{155}},\frac{5}{\sqrt{155}},\frac{11}{\sqrt{155}}$