Question

if $\hat{i}+2\hat{j}+3\hat{k}$ and $3\hat{i}-2\hat{j}+\hat{k}$ are the adjacent sides of a parallelogram, then its area will be

• A. $8\sqrt{3}$
• B. $5\sqrt{3}$
• C. $16\sqrt{3}$
• D. $6\sqrt{3}$

Answer 1

The correct option is A $8\sqrt{3}$

We have,

$\overrightarrow{a}=\hat{i}+2\hat{j}+3\hat{k}$

$\overrightarrow{b}=3\hat{i}-2\hat{j}+\hat{k}$

We know that,

Area of parallelogram $=\left|\begin{array}{ccc}i& j& k\\ 1& 2& 3\\ 3& -2& 1\end{array}\right|$

$=(-6-2)\hat{i}-\hat{j}(9-1)+\hat{k}(6+2)$

$=-8\hat{i}-8\hat{j}+8\hat{k}$

$Now,$

$|\overrightarrow{a}\times \overrightarrow{b}|=\sqrt{{(-8)}^2+{(-8)}^2+8^2}$

$=\sqrt{64+64+64}$

$=\sqrt{3\times 64}$

$=8\sqrt{3}sq.unit$

Hence, this is the answer.