Question

Let $A (2, 3, 5)$ and $C(-3, 4, -2)$ be opposite vertices of a parallelogram $ABCD$. If the diagonal $\overrightarrow{BD} = \hat{i} + 2 \hat{j} + 3 \hat{k}$, then the area of the parallelogram is equal to

• A. $\frac{1}{2} \sqrt{410}$
• B. $\frac{1}{2} \sqrt{474}$
• C. $\frac{1}{2} \sqrt{586}$
• D. $\frac{1}{2} \sqrt{306}$

Answer 1

Answer: (B)

$\frac{1}{2} \sqrt{474}$

Answer 2

The area is given by:

Area = $\frac{1}{2} |\overrightarrow{AC} \times \overrightarrow{BD}|$

Calculate $\overrightarrow{AC} = (-5i + j - 7k)$ and $\overrightarrow{BD} = i + 2j + 3k$ and find the cross product.

Then,

Area = $\frac{1}{2} \sqrt{474}$