Question 293 of 379EasyMCQ
The area (in sq. units) of the parallelogram whose diagonals are along the vectors 8i^−6j^ and 3i^+4j^−12k^, is: Solutions · 1
Official solution
Let d1=8i^−6j^, d2=3i^+4j^−12k^.
\[\vec d_1\times\vec d_2 = \begin{vmatrix}\hat i&\hat j&\hat k\\8&-6&0\\3&4&-12\end{vmatrix} = 72\hat i+96\hat j+50\hat k,\qquad |\vec d_1\times\vec d_2|=\sqrt{16900}=130\]
\[\text{Area} = \frac12|\vec d_1\times\vec d_2| = 65\] Answer the question to unlock