The area (in sq. units) of a triangle formed by vertices O, A and B where \(\vec{OA} =\hat{\imath}+2\hat{\jmath}+3\hat{k}\) and \(\vec{OB} =-3\hat{\imath}-2\hat{\jmath}+\hat{k}\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(3\sqrt{5}\)
The area of a triangle with two sides along \(\vec{OA}\) and \(\vec{OB}\) is \(\frac{1}{2}|\vec{OA}\times\vec{OB}|\), half the magnitude of the cross product.
\( \displaystyle \vec{OA}\times\vec{OB} = \begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\ 1 & 2 & 3\\ -3 & -2 & 1\end{vmatrix} \)
\(= (2 + 6)\hat{\imath} - (1 + 9)\hat{\jmath} + (-2 + 6)\hat{k}\)
\(= 8\hat{\imath} - 10\hat{\jmath} + 4\hat{k}\)
\( \displaystyle |\vec{OA}\times\vec{OB}| = \sqrt{64 + 100 + 16} = \sqrt{180} = 6\sqrt{5} \)
\( \displaystyle \text{Area} = \frac{1}{2}\times 6\sqrt{5} = 3\sqrt{5} \) sq. units
Why the other options are incorrect
Option (2) \(4\sqrt{5}\) — Incorrect. The area is \(\frac{1}{2}\times 6\sqrt{5} = 3\sqrt{5}\), not \(4\sqrt{5}\).
Option (3) \(6\sqrt{5}\) — Incorrect. This is \(|\vec{OA}\times\vec{OB}|\) itself, the area of the parallelogram on OA and OB. The triangle is half of it.
Option (4) \(2\sqrt{5}\) — Incorrect. The area is \(3\sqrt{5}\), not \(2\sqrt{5}\).