Question

The area of the parallelogram whose adjacent sides are $\hat{i}+ \hat {k}$ and $2\hat {i}+\hat {j}+\hat {k}$ is

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

Answer 1

Answer: (B)

$\sqrt {3}$

Answer 2

The correct answer is B:$\sqrt{3}$
Let adjacent sides of a parallelogram are
$a =\hat{ i }+\hat{ k }$ and $b =2 \hat{ i }+\hat{ j }+\hat{ k }$
$\therefore$ Area of parallelogram $=\| a \times b \|$
$\begin{Vmatrix}\hat{i}&\hat{j}&\hat{k}\\\ 1&0&1\\\ 2&1&0\end{Vmatrix}$
$=|\hat{ i }(0-1)-\hat{ j }(1-2)+\hat{ k }(1-0)|$
$\therefore |\vec{a}\times\vec{b}|=|-\hat{ i }+\hat{ j }+\hat{ k }|$
$=\sqrt{(-1)^{2}+(1)^{2}+(1)^{2}}= \sqrt{3}\space sq.unit$