Question

The diagonals of a parallelogram are represented by vectors $\vec{p}=5\hat{i}-4\hat{j}+3\hat{k}$ and $\vec{q}=3\hat{i}+2\hat{j}-k.$ Then the area of the parallelogram is

• A. $\sqrt{171}$ units
• B. $\sqrt{72}$ units
• C. 171 units
• D. 72 units

Answer 1

Answer: (A)

Area of parallelogram $=\frac{1}{2}\mid \vec{p}\times \vec{q}$
$=\frac{1}{2}\mid (5\hat{i}-4\hat{j}+3\hat{k})\times (3\hat{i}+2\hat{j}-\hat{k})$
$=\frac{1}{2}|-2\hat{i}+14\hat{j}+22\hat{k}|=\frac{\sqrt{684}}{2}=\sqrt{171}$ units