Question

The area of the parallelogram having a diagonal $3\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k}$ and a side $\overrightarrow{i}-3\overrightarrow{j}-4\overrightarrow{k}$ is

• A. $10\sqrt{3}$
• B. $6\sqrt{30}$
• C. $\frac{3}{2}\sqrt{30}$
• D. $3\sqrt{30}$

Answer 1

The correct option is D $3\sqrt{30}$

The required area of the parallelogram will be
$=2\times {\text{Area}}_{\text{traingle}}$
$=2\times \frac{1}{2}|({\overrightarrow{d}}_{\text{diagonal}}\times {\overrightarrow{a}}_{\text{side}})|$
$=|{\overrightarrow{d}}_{\text{diagonal}}\times {\overrightarrow{a}}_{\text{side}}|$
$=|(3i+j-k)\times (i-3j-4k)|$
$=|-9k+12j-k-4i-j-3i|$
$=|-7i+11j-10k|$
$=\sqrt{49+100+121}$
$=\sqrt{270}$
$=3\sqrt{30}$.