The vectors $\vec a=3\hat i-2\hat j+2\hat k$ and $\vec b =-\hat i-2\hat k$ are the adjacent sides of a parallelogram. Then, the acute angle between its diagonals is

$π/4$

The diagonals of the parallelogram are given by

$\vec α =\vec a +\vec b$ and $\vec β= ±(\vec a -\vec b)$

i.e. $\vec α =2\hat i-2\hat j$ and $\vec β=±(4\hat i-2\hat j+4\hat k)$

Let θ be the angle between the diagonals. Then,

$\cos θ=\frac{\vec α.\vec β}{|\vec α||\vec β|}$

$⇒\cos θ=\frac{1}{\sqrt{2}}$ or, $\cos θ=-\frac{1}{\sqrt{2}}⇒θ=π/4$ oг, $θ=3π/4$