Question

The area of the parallelogram whose diagonals are $\hat{i}-3\hat{j}+2\hat{k},-\hat{i}+2\hat{j}$ is

• A. $4\sqrt{29}sq$.units
• B. $\frac{1}{2}\sqrt{21}sq$.units
• C. $10\sqrt{3}sq$.units
• D. $\frac{1}{2}\sqrt{270}sq$.units

Answer 1

The correct option is B $\frac{1}{2}\sqrt{21}sq$.units

Vector area =$\frac{1}{2}(a\times b)=\frac{1}{2}\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -3& 2\\ 1& 2& 0\end{array}\right|=\frac{1}{2}\left[\right(0-4)-\hat{j}(0+2)+\hat{k}(2-3\left)\right]=\frac{1}{2}(-4\hat{i}-2\hat{j}-\hat{k})$
Area =$\frac{1}{2}\sqrt{16+4+1}=\frac{1}{2}\sqrt{21}sq.$ units