The area of the parallelogram determined by the vectors \(\hat{\imath}+2\hat{\jmath}+3\hat{k}\) and \(3\hat{\imath}-2\hat{\jmath}+\hat{k}\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(8\sqrt{3}\)
The area of the parallelogram with adjacent sides \(\vec{a}\) and \(\vec{b}\) is \(|\vec{a} \times \vec{b}|\). Here \(\vec{a} = \hat{\imath}+2\hat{\jmath}+3\hat{k}\) and \(\vec{b} = 3\hat{\imath}-2\hat{\jmath}+\hat{k}\).
The cross product:
\( \vec{a} \times \vec{b} = \begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\ 1 & 2 & 3\\ 3 & -2 & 1\end{vmatrix} \)
\( = \hat{\imath}\,(2 + 6) - \hat{\jmath}\,(1 - 9) + \hat{k}\,(-2 - 6) \)
\( = 8\hat{\imath} + 8\hat{\jmath} - 8\hat{k} \)
Its magnitude:
\( |\vec{a} \times \vec{b}| = \sqrt{64 + 64 + 64} \)
\( = \sqrt{192} = 8\sqrt{3} \)