Question

The area of a triangle whose vertices are $A(1,-1,2)$, $B(2,1,-1)$, $C(3,-1,2)$

• A. $\sqrt{5}$
• B. $\sqrt{7}$
• C. $\sqrt{12}$
• D. $\sqrt{13}$

Answer 1

The correct option is D $\sqrt{13}$

Let $A(1,-1,1),B(2,1,-1)$ and $C(3,-1,2)$

$\overrightarrow{AB}=B-A=(2,1,-1)-(1,-1,2)=(1,2,-3)\overrightarrow{AC}=C-A=(3,-1,2)=(1,-1,2)=(2,0,0)$

Now, $\overrightarrow{AB}\times \overrightarrow{AC}=i(0-0)-j(0-(-6))+4(0-4)=0i-6j-4k$

Now area of triangle $=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|$

$=\frac{\sqrt{{\left(0\right)}^2+{(-6)}^2+{(-4)}^2}}{2}=\frac{\sqrt{36+6}}{2}=\sqrt{13}$