ABCD is a parallelogram with $\vec{AC}=\hat i-2\hat j+\hat k$ and $\vec{BD}=-\hat i+2\hat j-5\hat k$. Area of this parallelogram is equal to:
Answer & explanation
Correct answer: option 2
Area $=\frac{1}{2}|\vec{AC}×\vec{BD}|=\frac{1}{2}\begin{vmatrix}\hat i&\hat j&\hat k\\1&-2&1\\-1&2&-5\end{vmatrix}=\frac{1}{2}|8\hat i+4\hat j|=2\sqrt{5}$sq. units