Question

The area of a parallelogram with diagonals $\mathbf{a} = 3\mathbf{i} + \mathbf{j} - 2\mathbf{k}$ and $\mathbf{b} = \mathbf{i} - 3\mathbf{j} + 4\mathbf{k}$ is

• A. $10\sqrt{3}$
• B. $5\sqrt{3}$
• C. $5\sqrt{3}$
• D. $\dfrac{5}{\sqrt{3}}$

Hint:

Area of parallelogram with diagonals $\mathbf{d_1}$ and $\mathbf{d_2}$: Area $= \dfrac{1}{2}|\mathbf{d_1}\times\mathbf{d_2}|$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Area of parallelogram when diagonals are given $= \dfrac{1}{2}|\mathbf{d_1} \times \mathbf{d_2}|$. 
Step 2: Detailed Explanation: 
$\mathbf{a}\times\mathbf{b} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\3&1&-2\\1&-3&4\end{vmatrix} = \mathbf{i}(4-6)-\mathbf{j}(12+2)+\mathbf{k}(-9-1) = -2\mathbf{i}-14\mathbf{j}-10\mathbf{k}$. 
$|\mathbf{a}\times\mathbf{b}| = \sqrt{4+196+100} = \sqrt{300} = 10\sqrt{3}$. 
Area $= \dfrac{1}{2}\times10\sqrt{3} = 5\sqrt{3}$. 
Step 3: Final Answer: 
Area $= 5\sqrt{3}$.