Vectors — Maths STD 12 Science — Question

Then $l=\frac{a}{\sqrt{a^2+b^2+c^2}}=\frac{4}{\sqrt{4^2+(-12)^2+(18)^2}}$

$=\frac{4}{\sqrt{16+144+324}}=\frac{4}{22}=\frac{2}{11}$

$m=\frac{b}{\sqrt{a^2+b^2+c^2}}=\frac{-12}{\sqrt{4^2+(-12)^2+(18)^2}}$

$=\frac{-12}{\sqrt{16+144+324}}=\frac{-12}{22}=\frac{-6}{11}$

and $n=\frac{c}{\sqrt{a^2+b^2+c^2}}=\frac{18}{\sqrt{4^2+(-12)^2+(18)^2}}$

$=\frac{18}{\sqrt{16+144+324}}=\frac{18}{22}=\frac{9}{11}$

Hence, the direction cosines of the line are $\frac{2}{11}, \frac{-6}{11}, \frac{9}{11}$.