Question

The area of the triangle with vertices $A(3,4,-1)$, $B(2,2,1)$ and $C(3,4,-3)$ is:

• A. $4\sqrt{5}$
• B. $3$
• C. $6$
• D. None of the above

Answer 1

Answer: (D)

None of the above

Given points are

$A(3,4,-1),B(2,2,1),C(3,4,-3)$

$\overrightarrow{A}=3\hat{i}+4\hat{j}-\hat{k}$

$\overrightarrow{B}=2\hat{i}+2\hat{j}+\hat{k}$

$\overrightarrow{C}=3\hat{i}+4\hat{j}-3\hat{k}$

$\overrightarrow{AB}=\overrightarrow{B}-\overrightarrow{A}=2\hat{i}+2\hat{j}+\hat{k}-(3\hat{i}+4\hat{j}-\hat{k})$

$\overrightarrow{AB}=-\hat{i}-2\hat{j}+2\hat{k}$

$\overrightarrow{BC}=\overrightarrow{C}-\overrightarrow{B}=3\hat{i}+4\hat{j}-3\hat{k}-(2\hat{i}+2\hat{j}+\hat{k})$

$\overrightarrow{AB}=\hat{i}+2\hat{j}-4\hat{k}$

$Areaoftriangle=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{BC}|$----------(1)

$\overrightarrow{AB}\times \overrightarrow{BC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1& -2& 2\\ 1& 2& -4\end{array}\right|$

$\overrightarrow{AB}\times \overrightarrow{BC}=\hat{i}(8-4)-\hat{j}(4-2)+\hat{k}(-2+2)$

$\overrightarrow{AB}\times \overrightarrow{BC}=4\hat{i}-2\hat{j}$

$|\overrightarrow{AB}\times \overrightarrow{BC}|=\sqrt{(4{)}^2+(-2{)}^2}$

$|\overrightarrow{AB}\times \overrightarrow{BC}|=\sqrt{16+4}$

$|\overrightarrow{AB}\times \overrightarrow{BC}|=\sqrt{20}$

From eq (1)

Area of triangle $=\frac{1}{2}\sqrt{20}$

Area of triangle $=\sqrt{5}$