Question

The area (in sq. units) of the parallelogram whose diagonals are along the vectors $8\hat{i}-6\hat{j}$ and $3\hat{i}+4\hat{j}-12\hat{k}$, is.

• A. $65$
• B. $20$
• C. $26$
• D. $52$

Answer 1

The correct option is A $65$

$\overrightarrow{d_1}\times \overrightarrow{d_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 8& -6& 0\\ 3& 4& -12\end{array}\right|$

$=72\hat{i}-(-96)\hat{j}+50\hat{k}$

$|\overrightarrow{d_1}\times \overrightarrow{d_2}|=5184+9216+2500=\sqrt{16900}=130$

$Area=\frac{1}{2}|\overrightarrow{d_1}\times \overrightarrow{d_2}|=\frac{1}{2}\times 130=65$.