The vectors from origin to the points A and B are \( \vec{a}=2\hat{\imath}-3\hat{\jmath}+2\hat{k} \) and \( \vec{b}=2\hat{\imath}+3\hat{\jmath}+\hat{k} \) respectively , then the area (in Sq. unit) of triangle OAB is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{1}{2}\sqrt{229} \)
OA and OB are two sides of the triangle, so the area of triangle OAB is \( \frac{1}{2}|\vec{a}\times\vec{b}| \).
\( \displaystyle \vec{a}\times\vec{b}=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\2 & -3 & 2\\2 & 3 & 1\end{vmatrix} \)
\( \displaystyle =\hat{\imath}(-3-6)-\hat{\jmath}(2-4)+\hat{k}(6+6) \)
\( \displaystyle =-9\hat{\imath}+2\hat{\jmath}+12\hat{k} \)
\( \displaystyle |\vec{a}\times\vec{b}|=\sqrt{81+4+144} \)
\( \displaystyle =\sqrt{229} \)
\( \displaystyle \text{Area}=\frac{1}{2}\sqrt{229} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{1}{2}\sqrt{225} \) — Incorrect. \( |\vec{a}\times\vec{b}|^2 \) is 81 + 4 + 144 = 229, not 225.
Option (2) \( \displaystyle\frac{25}{2} \) — Incorrect. \( \frac{25}{2}=\frac{1}{2}\sqrt{625} \), not \( \frac{1}{2}\sqrt{229} \).
Option (3) \( \sqrt{229} \) — Incorrect. This is the area of the parallelogram with sides OA and OB; the triangle is half of it.