Question

Calculate the area of the triangle determined by the two vectors $\overrightarrow{A}=3\hat{i}+4\hat{j}$ and $\overrightarrow{B}=-3\hat{i}+7\hat{j}$

• A. 33 sq. unit
• B. 21 sq. unit
• C. 0 sq. unit
• D. $\frac{33}{2}$ sq. unit

Answer 1

The correct option is D

$\frac{33}{2}$ sq. unit

We know that the half of magnitude of the cross product of two vectors gives the area of the triangle.

$\overrightarrow{A}\times \overrightarrow{B}$ = $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 4& 0\\ -3& 7& 0\end{array}\right|$ = $\hat{i}(0-0)-\hat{j}(0-0)+\hat{k}(21+12)$ = $33\hat{k}$

Taking magnitude $|\overrightarrow{A}\times \overrightarrow{B}|$ = $\sqrt{{33}^2}$ = 33.

So area of triangle = $\frac{1}{2}|\overrightarrow{A}\times \overrightarrow{B}|$ = $\frac{33}{2}sq.unit$