Question

The area (in sq. units) of the parallelogram whose diagonals are along the vectors $8 \hat{i} - 6 \hat{j}$ and $3 \hat{i} + 4 \hat{j} - 12 \hat{k}$, is :

• A. 26
• B. 65
• C. 20
• D. 52

Answer 1

Answer: (B)

$d_{1}\times d_{2} = \begin{vmatrix}i&j&k\\ 8&-6&0\\ 3&4&-12\end{vmatrix}$
$= 72\hat{i}-\left(-96\right)\hat{j}+50\hat{k}$
$= 5178 + 7056 + 2500$
$\left|d_{1}\times d_{2}\right| = \sqrt{16900} = 130$
$A = \frac{1}{2}\left|d_{1}\times d_{2}\right| = \frac{1}{2}\times130$
$= 65$