Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

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

Since the line is perpendicular to the given lines, its direction ratios satisfy:
$2a-3b+c=0$
$a+2b-2c=0$

Solving by cross-multiplication:
$\frac{a}{(-3)(-2)-(1)(2)}=\frac{b}{(1)(1)-(2)(-2)}=\frac{c}{(2)(2)-(-3)(1)}$

$\frac{a}{6-2}=\frac{b}{1+4}=\frac{c}{4+3}$

$\frac{a}{4}=\frac{b}{5}=\frac{c}{7}$

The direction ratios are proportional to $4,5,7$.

The magnitude is $\sqrt{4^2+5^2+7^2}=\sqrt{16+25+49}=\sqrt{90}=3\sqrt{10}$.

Therefore, the direction cosines are $\frac{\pm 4}{3\sqrt{10}},\frac{\pm 5}{3\sqrt{10}},\frac{\pm 7}{3\sqrt{10}}$.