Question

If the direction cosines of a line are $(\frac{1}{c},\frac{1}{c},\frac{1}{c})$ then show that $c=\pm \sqrt{3}$.

Answer 1

Direction cosines

$\alpha =\frac{v_x}{\sqrt{v_x^2+v_y^2+v_z^2}}=\frac{1}{c}$

$\beta =\frac{v_x}{\sqrt{v_x^2+v_y^2+v_z^2}}=\frac{1}{c}$

$\gamma =\frac{v_x}{\sqrt{v_x^2+v_y^2+v_z^2}}=\frac{1}{c}$

$\Rightarrow v_x=v_y=v_z=1$

and

$\sqrt{v_x^2+v_y^2+v_z^2}=c$

$\sqrt{1^2+1^2+1^2}=c$

$\pm \sqrt{3}=c$