If \(\vec{a} = 2\hat{\jmath} - \hat{k}, \vec{b} = 2\hat{\imath} - 3\hat{\jmath} + \hat{k}\) and \(\vec{c} = -\hat{\imath} + \hat{k}\) are three vectors, then the area (in sq. units) of the parallelogram whose diagonals are \((\vec{b} + \vec{c})\) and \((\vec{a} + \vec{c})\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{1}{2}\sqrt{21}\)
The diagonals are:
\(\vec{d_1} = \vec{b} + \vec{c} = \hat{\imath} - 3\hat{\jmath} + 2\hat{k}\)
\(\vec{d_2} = \vec{a} + \vec{c} = -\hat{\imath} + 2\hat{\jmath}\)
Their cross product:
\( \displaystyle \vec{d_1} \times \vec{d_2} = \begin{vmatrix}\hat{\imath}&\hat{\jmath}&\hat{k}\\1&-3&2\\-1&2&0\end{vmatrix} \)
\(= \hat{\imath}(0 - 4) - \hat{\jmath}(0 + 2) + \hat{k}(2 - 3)\)
\(= -4\hat{\imath} - 2\hat{\jmath} - \hat{k}\)
\(|\vec{d_1} \times \vec{d_2}| = \sqrt{16 + 4 + 1} = \sqrt{21}\)
The area of a parallelogram with diagonals \(\vec{d_1}\) and \(\vec{d_2}\) is \(\frac{1}{2}|\vec{d_1} \times \vec{d_2}|\):
\( \displaystyle \text{Area} = \frac{1}{2}\sqrt{21} \text{ sq. units} \)
Tip: Area of a parallelogram = \(|\vec{p} \times \vec{q}|\) when \(\vec{p}, \vec{q}\) are adjacent sides, but \(\frac{1}{2}|\vec{d_1} \times \vec{d_2}|\) when \(\vec{d_1}, \vec{d_2}\) are its diagonals. Forgetting the \(\frac{1}{2}\) gives \(\sqrt{21}\), Option (1).