Question

The vertices of a triangle are $A(1,4,2),B(-2,1,2),C(2,3,-4)$. Find $\angle A,\angle B,$ and $\angle C$.

Answer 1

$\to$ $\overline{AB}=-3\hat{i}-3\hat{j}+0\hat{k}\overline{AC}=\hat{i}-\hat{j}-6\hat{k}$

$\overline{BC}=4\hat{i}+2\hat{j}-6\hat{k}$

$\Rightarrow \cos A=\left|\frac{\overline{AB}.\overline{AC}}{|\overline{AB}||\overline{AC}|}\right|=\left|\frac{-3+3}{\sqrt{9+9}.\sqrt{1+1+12}}\right|=0\Rightarrow A={90}^o$

$\Rightarrow \cos B=\left|\frac{\overline{AB}.\overline{BC}}{|\overline{AB}||\overline{BC}|}\right|=\left|\frac{-12-6}{\sqrt{18}\times \sqrt{16+4+36}}\right|=\frac{-18}{\sqrt{18\times 56}}$

$\cos B=\sqrt{\frac{18}{56}}=\sqrt{\frac{9}{28}}\to B={\cos }^{-1}\sqrt{\frac{9}{28}}$ and $C={90}^o-{\cos }^{-1}\sqrt{\frac{9}{28}}$