The position vectors of 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 area of triangle OAB is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{1}{2}\sqrt{229} \) Sq. unit
The area of triangle OAB is \( \frac{1}{2}|\vec{a}\times\vec{b}| \).
\( \displaystyle \vec{a}\times\vec{b}=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\2 & -3 & 2\\2 & 3 & 1\end{vmatrix} \)
\( \displaystyle =(-3-6)\hat{\imath}-(2-4)\hat{\jmath}+(6+6)\hat{k} \)
\( \displaystyle =-9\hat{\imath}+2\hat{\jmath}+12\hat{k} \)
\( \displaystyle |\vec{a}\times\vec{b}|=\sqrt{81+4+144}=\sqrt{229} \)
\( \displaystyle \text{Area}=\frac{1}{2}\sqrt{229}\ \text{sq. unit} \)
Why the other options are incorrect
Option (2) \( \sqrt{229} \) Sq. unit — Incorrect. This is the area of the parallelogram formed by the two vectors; the triangle has half of it.
Option (3) \( \sqrt{225} \) Sq. unit — Incorrect. \( |\vec{a}\times\vec{b}|^2 \) is 81 + 4 + 144 = 229, not 225.
Option (4) \( \displaystyle\frac{1}{2}\sqrt{225} \) Sq. unit — Incorrect. \( |\vec{a}\times\vec{b}|^2 \) is 229, not 225.