Question

If $\hat{i} + \hat{j}, \hat{j} + \hat{k}, \hat{i} + \hat{k}$ are the position vectors of the vertices of a triangle $ABC$ taken in order, then $\angle A$ is equal to

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{5}$
• C. $\frac{\pi}{6}$
• D. $\frac{\pi}{3}$

Answer 1

Answer: (D)

$\frac{\pi}{3}$

Answer 2

Now , $\vec{AC} = \hat{k} - \hat{j}$ and $\vec{AB} = \hat{k} - \hat{i}$
Let $\theta$ be the angle between $\vec{AC}$ and $\vec{AB}$ .
$\cos\theta = \frac{1-0-0+0}{\sqrt{2} \sqrt{2}} = \frac{1}{2}$
$\Rightarrow \theta = 60^{\circ} = \frac{\pi}{3}$