The area of the parallelogram whose adjacent sides are given by the vectors \(2\hat{\imath}+\hat{\jmath}\) and \(\hat{\imath}-\hat{k}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\sqrt{6}\) square units
The area of a parallelogram with adjacent sides \(\vec{a}\) and \(\vec{b}\) is \(|\vec{a}\times\vec{b}|\). Here \(\vec{a}=2\hat{\imath}+\hat{\jmath}\) and \(\vec{b}=\hat{\imath}-\hat{k}\).
\( \displaystyle \vec{a}\times\vec{b} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 1 & 0 \\ 1 & 0 & -1 \end{vmatrix} \)
\( = \hat{\imath}(-1-0)-\hat{\jmath}(-2-0)+\hat{k}(0-1)\)
\( = -\hat{\imath}+2\hat{\jmath}-\hat{k}\)
Its magnitude is
\(|\vec{a}\times\vec{b}| = \sqrt{1+4+1} = \sqrt{6}\)