Question

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

• A. $\frac{15}{2}$
• B. $15$
• C. $\frac{15}{4}$
• D. $\frac{15\sqrt{3}}{2}$

Answer 1

Answer: (C)

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 $=\frac{1}{2}|a\times b|$
$=\frac{1}{2}|a||b|\sin \theta$
$=\frac{1}{2}\times 3\times 5\times \sin \frac{\pi }{6}$
$\left[\because \theta =\frac{\pi }{6}\right]$
$=\frac{15}{4}$