Question

If the diagonals of a parallelogram are $3\hat{i}+\hat{j}-2\hat{k}$ and $\hat{i}-3\hat{j}+4\hat{k}$, then its area is

• A. $5\sqrt{3}$
• B. $4\sqrt{3}$
• C. $3\sqrt{3}$
• D. $2\sqrt{3}$

Answer 1

Answer: (A)

$5\sqrt{3}$

Area of parellelogram when diagonals are given is $\frac{1}{2}{d}_1\times d_2$

we have $d_1=3\overline{i}+\overline{j}-2\overline{k}$

and $d_2=\overline{i}-3\overline{j}+4\overline{k}$

So $d_1\times d_2=3(4-6)\overline{i}-1(12+2)\overline{j}-2(-9-1)\overline{k}$

$=-6\overline{i}-14\overline{j}+20\overline{k}$

So $\frac{1}{2}|d_1\times d_2|$

$=5\sqrt{3}$