Question

If $a =\hat{ i }+2 \hat{ j }+2 \hat{ k },| b |=5$ and the angle between $a$ and $b$ is $\pi / 6$, then the area of the triangle formed by these two vectors as two sides is

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

Answer 1

Answer: (A)

$\frac{15}{4}$

Answer 2

The correct answer is A:$\frac{15}{4}$
Given that;
$\vec{a}=\hat{i}+2\hat{j}+2\hat{k}$
$|b|=5$
$\therefore |a|=\sqrt{1^2+2^2+2^2}$
$=3$
Area of the triangle
$=\frac{1}{2}| a \times b |$
$=\frac{1}{2} \,\| a |\,| b |\sin \theta \,\hat{ n }|$
$=\frac{1}{2}\left[3 \times 5 \times \sin \frac{\pi}{6}\right]$
$\left(\therefore| a |=\sqrt{1+2^{2}+2^{2}}=3\right)$
$=\frac{1}{2}\left[15 \times \frac{1}{2}\right]=\frac{15}{4}$