I am struggling with the following problem and would appreciate any help I can get.
Given the points $A=(1,2,3)$, $B=(-2,1,0)$ and $C=(2,0,1)$, I need to find:
(a) the area of the triangle formed by these points
(b) the angle at point $A$
Here is my attempt at solving the problem:
(a) To find the area of the triangle, I first need to find the length of its sides. Using the distance formula, I find that $AB = \sqrt{18}$, $AC = \sqrt{14}$, and $BC = \sqrt{26}$. Next, I use Heron's formula to find the area of the triangle:
$$ \begin{aligned} s & =\frac{1}{2}(A B+A C+B C) \\ & =\frac{1}{2}(\sqrt{18}+\sqrt{14}+\sqrt{26}) \\ & =\frac{1}{2} \sqrt{2(9+7+13+2 \sqrt{126}+2 \sqrt{117}+2 \sqrt{39})} \\ & \approx 6.222 \end{aligned} $$ Therefore, the area of the triangle is approximately 6.222 square units.
(b) To find the angle at point $A$, I need to use the dot product. Let $\vec{AB}$ and $\vec{AC}$ be the vectors from $A$ to $B$ and $C$, respectively. Then, the angle at point $A$ is given by:
$$ \theta=\cos ^{-1}\left(\frac{\overrightarrow{A B} \cdot \overrightarrow{A C}}{|\overrightarrow{A B}||\overrightarrow{A C}|}\right) $$
Computing the dot product and magnitudes, I get:
$$ \overrightarrow{A B} \cdot \overrightarrow{A C}=-5, \quad|\overrightarrow{A B}|=\sqrt{18}, \quad|\overrightarrow{A C}|=\sqrt{14} $$
Substituting these values into the formula, I get: $$ \theta=\cos ^{-1}\left(\frac{-5}{\sqrt{18} \sqrt{14}}\right) \approx 1.744 $$