Question

Find the area of the triangle with the following vertices: $(1,0,0),(0,1,0),(1,1,1)$

• A. $\frac{\sqrt{3}}{4}$
• B. $\frac{\sqrt{5}}{2}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{\sqrt{5}}{4}$

Answer 1

Answer: (C)

$\frac{\sqrt{3}}{2}$

Let $A(1,0,0),B(0,1,0)$ and $C(1,1,1)$
Then $\overrightarrow{AB}=B-A=(0,1,0)-(1,0,0)=(-1,1,0)$
And $\overrightarrow{AC}=C-A=(1,1,1)-(1,0,0)=(0,1,1)$

Now $\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}i& j& k\\ -1& 1& 0\\ 0& 1& 1\end{array}\right|=i(1-0)-j(-1-0)+k(-1-0)=i+j-k$
Therefore are of $\Delta ABC=\frac{|\overrightarrow{AB}\times \overrightarrow{AC}|}{2}=\frac{\sqrt{1^2+{\left(1\right)}^2+{(-1)}^2}}{2}=\frac{\sqrt{3}}{2}$