Question

If $A(3,2,6),B(1,4,5)$ and $C(3,5,5)$ are the vertices of $\Delta ABC$, then find $m\angle ABC$

Answer 1

we have,

$\begin{array}{c}D{r}^{\prime }sofAB=(2,-2,1)\\ D{r}^{\prime }sofBC=(2,1,0)\\ \cos \angle ABC=\frac{12-2-1}{\sqrt{4+4+1}\sqrt{4+1}}=0\\ \angle ABC={90}^0\end{array}$

so,$m\angle ABC={90}^0$