Question

The coordinates of a point P are $(3,12,4)$ w.r.t origin O, then the direction cosines of $OP$ are

• A. $3,12,4$
• B. $\frac{1}{4},\frac{1}{3},\frac{1}{2}$
• C. $\frac{3}{\sqrt{13}},\frac{1}{\sqrt{13}},\frac{2}{\sqrt{13}}$
• D. $\frac{3}{13},\frac{12}{13},\frac{4}{13}$

Answer 1

The correct option is D

$\frac{3}{13},\frac{12}{13},\frac{4}{13}$

Given points are $O(0,0,0)=(x_1,y_1,z_1)$ and $P(3,12,4)=(x_2,y_2,z_2)$

Direction cosines of $OP$ is given by

$\cos \alpha =\frac{x_2-x_1}{\sqrt{x_2^2+y_2^2+z_2^2}}$

$\cos \beta =\frac{y_2-y_1}{\sqrt{x_2^2+y_2^2+z_2^2}}$

$\cos \gamma =\frac{z_2-z_1}{\sqrt{x_2^2+y_2^2+z_2^2}}$

$⟹$ $\cos \alpha =\frac{3-0}{\sqrt{3^2+{12}^2+4^2}}=\frac{3}{\sqrt{169}}=\frac{3}{13}$

$⟹$ $\cos \beta =\frac{12-0}{\sqrt{3^2+{12}^2+4^2}}=\frac{12}{\sqrt{169}}=\frac{12}{13}$

$⟹$ $\cos \gamma =\frac{4-0}{\sqrt{3^2+{12}^2+4^2}}=\frac{4}{\sqrt{169}}=\frac{4}{13}$

Hence, the direction cosines are $\frac{3}{13},\frac{12}{13},\frac{4}{13}$.