Question

Side and diagonal of a parallellogram are given by vectors $\overrightarrow{a}=3\hat{i}+2\hat{j}-\hat{k}$ and ${\overrightarrow{d}}_1=2\hat{i}+4\hat{j}+2\hat{k}$.

• A. Another side of parallelogram may be $-\hat{i}+2\hat{j}+3\hat{k}$
• B. Another diagonal may be $4\hat{i}-4\hat{j}-4\hat{k}$
• C. Area of parallelogram is $8\sqrt{3}$ unit
• D. Area of parallelogram is $8\hat{i}+64\hat{j}$

Answer 1

Answer: (A), (C)

Area of parallelogram is $8\sqrt{3}$ unit

Another side is $\overrightarrow{b}$.
Depending on whether $\overrightarrow{d_1}$ is the larger or the smaller diagonal, there are two possibilities.
$\overrightarrow{a}+\overrightarrow{b}={\overrightarrow{d}}_1$ or $\overrightarrow{a}-\overrightarrow{b}={\overrightarrow{d}}_1$
$\overrightarrow{b}={\overrightarrow{d}}_1-\overrightarrow{a}$ or $\overrightarrow{b}=-({\overrightarrow{d}}_1-\overrightarrow{a})$
$\overrightarrow{b}=-\hat{i}+2\hat{j}+3\hat{k}$ or $\overrightarrow{b}=\hat{i}-2\hat{j}-3\hat{k}$
$\overrightarrow{a}-\overrightarrow{b}={\overrightarrow{d}}_2\Rightarrow {\overrightarrow{d}}_2=4\hat{i}-4\hat{k}$
$Area=\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 2& -1\\ -1& 2& 3\end{array}\right|=8\hat{i}-8\hat{j}+8\hat{k}$
$\Rightarrow |Area|=\sqrt{8^2+8^2+8^2}=8\sqrt{3}$