Question

If the vertices $A,B,C$ of a triangle $ABC$ are$(1,2,3),(-1,0,0),(0,1,2)$, respectively,then find $\angle{ABC}$.[$\angle{ABC}$ is the triangle between the vectors$\overrightarrow{BA}$and $\overrightarrow{BC}$].

Answer 1

The vertices $△ABC$are given as $A(1,2,3),B(-1,0,0),C(0,1,2).$
Also,it is given that $\angle ABC$ is the angle between the vectors $\overrightarrow{BA}\space and\space \overrightarrow{ BC}.$
\overrightarrow{BA}$$={1-(-1)}\hat{i}+(2-0)\hat{j}+(3-0)\hat{k}=2\hat{i+}2\hat{j}+3\hat{k}
$\overrightarrow{BC}={0-(-1)}\hat{i}+(1-0)\hat{j}+(2-0)\hat{k}=\hat{i}+\hat{j}+2\hat{k}$
$∴\overrightarrow{BA}.\overrightarrow{BC}→=(2\hat{i}+2\hat{j}+3\hat{k}).(\hat{i}+\hat{j}+2\hat{k})=2×1+2×1+3×2=10$
$|\overrightarrow{BA}|=\sqrt{22+22+32}=\sqrt{4+4+9}=\sqrt{17}$
$|\overrightarrow{BC}|=\sqrt{1+1+22}=\sqrt{6}$
Now,It is known that:
$\overrightarrow{BA}.\overrightarrow{BC}=|\overrightarrow{BA}||\overrightarrow{BC}|cos(∠ABC)$
$∴10=\sqrt{17}×\sqrt{6}cos(∠ABC)$
$⇒cos(\angle{ABC})=\frac{10}{\sqrt{17}\times\sqrt{6}}$
$⇒∠ABC=cos^{-1}(\frac{10}{\sqrt{102}})$