The area of a triangle whose adjacent sides are represented by the vectors $\vec{a}=3\hat{i}+\hat{j}+2\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$ is

Solution

1. The area of a triangle with adjacent sides $\vec{a}$ and $\vec{b}$ is given by the formula $\text{Area}=\frac{1}{2}|\vec{a}\times \vec{b}|$.
2. Calculate the cross product: $\vec{a}\times \vec{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 1& 2\\ 1& -1& 1\end{array}\right|=\hat{i}(1-(-2))-\hat{j}(3-2)+\hat{k}(-3-1)=3\hat{i}-\hat{j}-4\hat{k}$.
3. Find the magnitude of the resulting vector: $$\begin{array}{rl}|\vec{a}\times \vec{b}|& =\sqrt{3^2+(-1{)}^2+(-4{)}^2}\\ & =\sqrt{9+1+16}\\ & =\sqrt{26}\end{array}$$.
4. Divide the magnitude by two to find the area: $\text{Area}=\frac{\sqrt{26}}{2}$.
5. Hence the answer is (A).