Question-31 The area of a triangle having vertices as \hat{i}-2 \hat{j}+3 \hat{k},-2 \hat{i}+3 \hat{j}-\hat{k} \& 4 \hat{i}-7 \hat{j}+7 \hat{k} is
A) 36 sq. units
B) 0 sq. units
C) 39 sq. units
D) 11 sq. units
Answer: B) 0 sq. units
Explanation
Let \vec{A}=\hat{i}-2 \hat{j}+3 \hat{k}, \quad \vec{B}=-2 \hat{i}+3 \hat{j}-\hat{k}
and \vec{C}=4 \hat{i}-7 \hat{j}+7 \hat{k}
\overrightarrow{A B}=-3 \hat{\dot{i}}+5 \hat{\dot{j}}-4 \hat{k}and \overrightarrow{A C}=3 \hat{\dot{i}}-5 \hat{\dot{j}}+4 \hat{k}
Area of \triangle A B C=\frac{1}{2}\|\overrightarrow{A B} \times \overrightarrow{A C}\|
=\frac{1}{2}\left|\begin{array}{ccc}\hat{i} & \hat{j} & \widehat{\mathbf{k}} \\ -3 & 5 & -4 \\ 3 & -5 & 4\end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}\hat{i} & \hat{j} & \widehat{\mathbf{k}} \\ -3 & 5 & -4 \\ 0 & 0 & 0\end{array}\right|[operating R_{3} \rightarrow R_{2}+R_{3} ]
=\frac{1}{2}[0]=0