Vectors — Maths STD 12 Science — Question

$\begin{aligned} & \therefore \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1 \\ & \therefore \cos ^2 45^{\circ}+\cos ^2 60^{\circ}+\cos ^2 \gamma=1\end{aligned}$

$\begin{aligned} & \therefore\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{2}\right)+\cos ^2 \gamma=1 \\ & \therefore \cos ^2 \gamma=1-\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\end{aligned}$

$\therefore \cos \gamma= \pm \frac{1}{2}$

$\therefore \cos \gamma=\frac{1}{2}$ or $\cos \gamma=-\frac{1}{2}$

$\therefore \cos \gamma=\cos \frac{\pi}{3}$ or $\cos \gamma=-\cos \frac{\pi}{3}$

$=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \frac{2 \pi}{3}$

$\therefore \gamma=\frac{\pi}{3}$ or $\gamma=\frac{2 \pi}{3}$

Hence, the third direction angle is $\frac{\pi}{3}$ or $\frac{2 \pi}{3}$.