Q.The vectors from origin to the points A and B are $\vec{a}=2\hat{i}-3\hat{j}+2\hat{k}$ and $\vec{b}=2\hat{i}+3\hat{j}+\hat{k}$, respectively, then the area of triangle OAB is
(A) $\sqrt{340}$
(B) $\sqrt{25}$
(C) $\sqrt{229}$
(D) $\dfrac{1}{2}\sqrt{229}$

Area of $\triangle OAB=\dfrac{1}{2}|\vec{a}\times\vec{b}|$, where $\vec{a}=\overrightarrow{OA}$ and $\vec{b}=\overrightarrow{OB}$.

$\vec{a}\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&-3&2\\2&3&1\end{vmatrix}=-9\hat{i}+2\hat{j}+12\hat{k},$ …

Method

For a triangle whose two sides are given as vectors from the same vertex, the area is half of the parallelogram those vectors span:

Here $\vec{a}=\overrightarrow{OA}=2\hat{i}-3\hat{j}+2\hat{k}$ and $\vec{b}=\overrightarrow{OB}=2\hat{i}+3\hat{j}+\hat{k}$.

Cross product

• $\hat{i}$: $(-3)(1)-(2)(3)=-3-6=-9$
• $\hat{j}$: $-\big[(2)(1)-(2)(2)\big]=-(2-4)=2$
• $\hat{k}$: $(2)(3)-(-3)(2)=6+6=12$

So $\vec{a}\times\vec{b}=-9\hat{i}+2\hat{j}+12\hat{k}$.

Magnitude and area …

Method: Area of a triangle from two side-vectors via the cross product

Use this whenever two sides of a triangle are given as vectors from a common vertex (here $\overrightarrow{OA}$ and $\overrightarrow{OB}$).

Steps

Step 1: Identify the two side-vectors sharing a vertex

The cross product measures the area of the parallelogram spanned by two vectors placed tail-to-tail, so both must start at the same vertex of the triangle.

Step 2: Compute the cross product as a determinant

Watch the middle ($\hat{j}$) term — it carries a minus sign in the cofactor expansion.

Step 3: Take the magnitude …

Common Mistakes

Mistake 1: Forgetting the factor of $\tfrac{1}{2}$

Why it's wrong: $|\vec{a}\times\vec{b}|=\sqrt{229}$ is the parallelogram area (option C here), not the triangle's. Correct approach: the triangle is half the parallelogram, so the area is $\tfrac{1}{2}\sqrt{229}$.

Mistake 2: Sign slip on the $\hat{j}$ component of the cross product

Why it's wrong: the cofactor for $\hat{j}$ is subtracted, so $-\big[(2)(1)-(2)(2)\big]=+2$, not $-2$. Dropping that minus corrupts the magnitude. Correct approach: expand as $\hat{i}(\cdots)-\hat{j}(\cdots)+\hat{k}(\cdots)$, keeping the middle sign negative. …