Surface of 3D Triangle

Question

The coordinates $A(-1,0,2), B(2,-1,3)$ and $C(4,0,1)$ are the corners in the triangle $ABC$.

• a) Find the length of the sides in the triangle.
• b) Find the area of the triangle.

Now I'm able to solve the first one, and I also did manage to find the area using the Law of Cosines, but this leaves me with a decimal answer $(A \approx 4.74)$, when my book says I should be able to get the exact answer $3\sqrt{10}/2$.

Any ideas on how I achieve this?

Answer 1

Pick two of the sides of the triangle, for example:

$$\vec{AB} = \vec{OB}-\vec{OA} = (3,-1,1), \quad \vec{AC} = \vec{OC}-\vec{OA} = (5,0,-1).$$

Then compute the area as follows:

$$ \color{green}{A} = \color{blue}{\frac{1}{2} | \vec{AB} \wedge \vec{AC} | }= \frac{1}{2} \left| \left| \begin{array}{ccc} \vec{\imath} & \vec{\jmath} & \vec{k} \\ 3 & - 1 & 1 \\ 5 & 0 & -1 \\ \end{array}\right| \right| = \frac{1}{2}|(1,8,5)|=\color{green}{\frac{\sqrt{90}}{2}} \ \blacksquare$$

The blue formula comes from here.

Hope this helps.

Cheers!