Question

The area of the parallelogram whose sides are represented by the vectors $\widehat{j} + 3\widehat{k}$ and $\widehat{i} + 2\widehat{j} - \widehat{k}$ is

• A. $\sqrt{61}$sq.unit
• B. $\sqrt{59}$sq.unit
• C. $\sqrt{49}$sq.unit
• D. $\sqrt{52}$sq.unit

Answer 1

Answer: (B)

$\sqrt{59}$sq.unit

Answer 2

$\overrightarrow{A} = \widehat{j} + 3\widehat{k}$,$\overrightarrow{B} = \widehat{i} + 2\widehat{j} - \widehat{k}$

\widehat{i} & \widehat{j} & \widehat{k} \\ 0 & 1 & 3 \\ 1 & 2 & - 1 \end{matrix} ⥂ \right| = - 7\widehat{i} + 3\widehat{j} - \widehat{k}$$ Hence area = $|\overrightarrow{C}| = \sqrt{49 + 9 + 1} = \sqrt{59}squnit$