Question

The area of the triangle formed by the adjacent sides with $\vec{A}=-3\hat{i}+2\hat{j}-4\hat{k}$ and $\vec{B}=-\hat{i}+2\hat{j}+\hat{k}$ is

• A. $\frac{\sqrt{165}}{2}$ units
• B. $\frac{\sqrt{137}}{2}$ units
• C. $\sqrt{165}$ units
• D. $\sqrt{137}$ units

Answer 1

Answer: (A)

$\vec{A}\times \vec{B}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -3& 2& -4\\ -1& 2& 1\end{array}\right|$
$=\hat{i}\left[2+8\right]+\hat{j}\left[4+3\right]+\hat{k}\left[-6+2\right]$
$=10\hat{i}+7\hat{j}-4\hat{k}$
$\therefore \left|\vec{A}\times \vec{B}\right|=\sqrt{10^2+7+4^2}$
$=\sqrt{100+49+16}=\sqrt{165}$
Area of triangle $=\frac{1}{2}\left|\vec{A}\times \vec{B}\right|$
$=\frac{\sqrt{165}}{2}$ units