Thank you for reporting, we will resolve it shortly
Q.
If $a =\hat{ i }+2 \hat{ j }+2 \hat{ k },| b |=5$ and the angle between
a and $b$ is $\frac{\pi}{6}$, then the area of the triangle formed by these two vectors as two sides is
We have, $a =\hat{ i }+2 \hat{ j }+2 \hat{ k } $
$\Rightarrow | a |=\sqrt{(1)^{2}+(2)^{2}+(2)^{2}}=3 $
and $| b |=5$
$\therefore $ Required area $=\frac{1}{2}| a \times b | $
$=\frac{1}{2}| a || b | \sin \theta $
$=\frac{1}{2} \times 3 \times 5 \times \sin \frac{\pi}{6} $
$\left[\because \theta=\frac{\pi}{6}\right] $
$=\frac{15}{4}$