Question

If $\mathbf{a} = \mathbf{i} + \mathbf{j} + \mathbf{k},\mathbf{b} = \mathbf{i} + 3\mathbf{j} + 5\mathbf{k}$ and $\mathbf{c} = 7\mathbf{i} + 9\mathbf{j} + 11\mathbf{k}$, then the area of the parallelogram having diagonals $\mathbf{a} + \mathbf{b}$ and $\mathbf{b} + \mathbf{c}$ is

• A. $4\sqrt{6}$
• B. $\frac{1}{2}\sqrt{21}$
• C. $\frac{\sqrt{6}}{2}$
• D. $\sqrt{6}$

Answer 1

Answer: (A)

$4\sqrt{6}$

Answer 2

Area of the parallelogram with diagonals $\mathbf{a} + \mathbf{b}$ and $\mathbf{b} + \mathbf{c}$ = $\frac{1}{2}|(\mathbf{a} + \mathbf{b}) \times (\mathbf{b} + \mathbf{c})|$

=$\frac{1}{2}|\{(\mathbf{i} + \mathbf{j} + \mathbf{k}) + (\mathbf{i} + 3\mathbf{j} + 5\mathbf{k})\} \times \{(\mathbf{i} + 3\mathbf{j} + 5\mathbf{k}) + (7\mathbf{i} + 9\mathbf{j} + 11\mathbf{k})\}|$

=$\frac{1}{2}|\{(2\mathbf{i} + 4\mathbf{j} + 6\mathbf{k}) \times (8\mathbf{i} + 12\mathbf{j} + 16\mathbf{k})\}|$

= $4|(\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) \times (2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k})|$

= $4|\left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 2 & 3 & 4 \end{matrix} \right|| = 4| - \mathbf{i} + 2\mathbf{j} - \mathbf{k}|$ = $4\sqrt{6}$