Two adjacent sides of a parallelogram ABCD are given by $\vec{AB}=2\hat i+10\hat j + 11\hat k$ and $\vec {AD} = -\hat i+2\hat j+2\hat k$. The side AD is rotated by an acute angle $α$ in the plane of the parallelogram so that AD becomes AD'. If AD' makes a right angle with the side AB, then the cosine of the angle $α$ is given by

$\frac{\sqrt{17}}{9}$

Clearly, AD' is perpendicular to $\vec{AB}$ and lies in the plane of $\vec{AB}$ and $\vec{AD}$. Therefore,

$\vec{AD}'=\vec{AB}×(\vec{AB}×\vec{AD})$

$⇒\vec{AD}'=(\vec{AB}.\vec{AD})\vec{AB}-(\vec{AB}.\vec{AB})\vec{AD}$

$⇒\vec{AD}'=5(61\hat i-10\hat j-2\hat k)$

$∴\cos α=\frac{|\vec{AD}.\vec{AD}'|}{|\vec{AD}||\vec{AD}'|}$

$⇒\cos α=(-\hat i+2\hat j+2\hat k).5(61\hat i-10\hat j-2\hat k)=\frac{\sqrt{17}}{9}$