# If $\overrightarrow {AB}=3\hat i-2\hat j+2\hat k$ and $\overrightarrow {BC}=-\hat i-2\hat k$ are adjacent sides of a parallelogram, then the angle between the diagonals is ?

$\begin{array}{1 1} \large\frac{\pi}{4} \\ \large\frac{\pi}{3} \\ \large\frac{2\pi}{3} \\ \large\frac{\pi}{6}\end{array}$

The Diagonals are $\overrightarrow {AC}=\overrightarrow {AB}+\overrightarrow {BC}$ and $\overrightarrow {AB}-\overrightarrow {BC}$
$i.e., 2\hat i-2\hat j\:\: and\:\:4\hat i-2\hat j+4\hat k$
$\therefore$ The angle between the diagonals $\theta$ is given by
$cos\theta=\large\frac{(2\hat i-2\hat j).(4\hat i-2\hat j+4\hat k)}{\sqrt 8.\sqrt {36}}=\frac{12}{12.\sqrt 2}=\frac{1}{\sqrt 2}$
$\therefore\:\theta=\large\frac{\pi}{4}$