Question

Vertices of a triangle are given by $\hat{i}+3\hat{j}+2\hat{k},2\hat{i}-\hat{j}+\hat{k}$ and $-\hat{i}+2\hat{j}+3\hat{k}$, then area of triangle is (in units)

• A. $\frac{\sqrt{107}}{2}$
• B. $\sqrt{\frac{107}{6}}$
• C. $\sqrt{\frac{107}{2}}$
• D. $\frac{\sqrt{207}}{2}$

Answer 1

The correct option is A $\frac{\sqrt{107}}{2}$

Vertices of a triangle are given by $\hat{i}+3\hat{j}+2\hat{k},2\hat{i}-\hat{j}+\hat{k},-\hat{i}+2\hat{j}+3\hat{k}$

Let, the vertices of the triangle be A, B and C . and O be the fixed point.

Then, $\overrightarrow{0A}=\hat{i}+3\hat{j}+2\hat{k}=(1,3,2)\overrightarrow{OB}=2\hat{i}-\hat{j}+\hat{k}=(2,-1,1)\overrightarrow{OC}=-\hat{i}+2\hat{j}+3\hat{k}=(-1,2,3)$

Now, $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(1,-4,-1)\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=(-2,-1,1)$

Hence, $\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -4& -1\\ -2& -1& 1\end{array}\right|=\hat{i}(-4-1)-\hat{j}(1-2)+\hat{k}(-1-8)=-5\hat{i}+\hat{j}-9\hat{k}$

$\therefore |\overrightarrow{AB}\times \overrightarrow{AC}|=\sqrt{(-5{)}^2+(1{)}^2+(-9{)}^2}=\sqrt{107}units$

And, the area if the triangle is given by, $\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|=\frac{1}{2}\sqrt{107}=\frac{\sqrt{107}}{2}unit{s}^2$

$\therefore$ option A is correct.