Question

Find the area of the triangle with the following vertices: $(0,0,0),(1,2,3),(2,-1,4)$

• A. $\frac{\sqrt{150}}{2}$
• B. $\frac{\sqrt{130}}{2}$
• C. $\frac{\sqrt{130}}{4}$
• D. $\frac{\sqrt{110}}{2}$

Answer 1

The correct option is A $\frac{\sqrt{150}}{2}$

Let $A(0,0,0),B(1,2,3)$ and $C(2,-1,4)$
Then $\overrightarrow{AB}=B-A=(1,2,3)-(0,0,0)=(1,2,3)$
And $\overrightarrow{AC}=C-A=(2,-1,4)-(0,0,0)=(2,-1,4)$
Now $\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}i& j& k\\ 1& 2& 3\\ 2& -1& 4\end{array}\right|=i(8+3)-j(4-6)+k(-4-1)=11i+2j-5k$
Therefore area of triangle $\Delta ABC=\frac{|\overrightarrow{AB}\times \overrightarrow{AC}|}{2}=\frac{\sqrt{{\left(11\right)}^2+{\left(2\right)}^2+{\left(5\right)}^2}}{2}=\frac{\sqrt{150}}{2}$