Thank you for reporting, we will resolve it shortly
Q.
If $\hat{ i }+\hat{ j }, \hat{ j }+\hat{ k }, \hat{ i }+\hat{ k }$ are the position vectors of the vertices of a $\Delta A B C$ taken in order, then $\angle A$ is equal to
Let position vector of the vertices are
$O A =\hat{ i }+\hat{ j }, O B =\hat{ j }+\hat{ k }$
and $ O C =\hat{ i }+\hat{ k }$
Now,$ A B =-\hat{ i }+\hat{ k } $
and $ A C =\hat{ k }-\hat{ j }$
$\therefore \cos \theta=\frac{( A B ) \cdot( A C )}{| A B || A C |}$
$=\frac{(-\hat{ i }+\hat{ k }) \cdot(\hat{ k }-\hat{ j })}{\sqrt{1^{2}+1^{2}} \sqrt{1^{2}+1^{2}}}$
$=\frac{(\hat{ k })^{2}}{\sqrt{2} \sqrt{2}}=\frac{1}{2}$
$\Rightarrow \theta=\frac{\pi}{3}$