Question

The area of the triangle formed by $3\hat{i}+\hat{j}-\hat{k}$ and $\hat{i}+\hat{j}+\hat{k}$ is

• A. $3sq.units$
• B. $2\sqrt{3}sq.units$
• C. $2\sqrt{14}sq.units$
• D. $\sqrt{6}sq.units$

Answer 1

The correct option is D $\sqrt{6}sq.units$

$\overrightarrow{A}=3\hat{i}+\hat{j}-\hat{k},\overrightarrow{B}=\hat{i}+\hat{j}+\hat{k}$
Area of the triangle $=\frac{1}{2}|\overrightarrow{A}\times \overrightarrow{B}|$

Where$|\overrightarrow{A}\times \overrightarrow{B}|$ is
$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 1& -1\\ 1& 1& 1\end{array}\right|$

Finding
$\frac{1}{2}|\overrightarrow{A}\times \overrightarrow{B}|$ by solving matrix,

$\frac{1}{2}|\overrightarrow{A}\times \overrightarrow{B}|=\frac{1}{2}|2\hat{i}-4\hat{j}+2\hat{k}|=\frac{1}{2}\sqrt{4+16+4}$
$=\sqrt{6}sq.unit$