Question

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

Answer 1

$\text{Given}:A=(1,2,3),B=(-1,0,0)\text{and}C=(0,1,2)$
$\angle ABC=\text{Angle between}\overrightarrow{BA}\text{and}\overrightarrow{BC}=\theta$

$\overrightarrow{BA}.\overrightarrow{BC}=\left|\overrightarrow{BA}\right|\left|\overrightarrow{BC}\right|\cos \theta$

$\left(\theta \text{is the angle between}\overrightarrow{BA}\text{and}\overrightarrow{BC}\right)$

$\overrightarrow{BA}=(1-(-1\left)\right)\hat{i}+(2-0)\hat{j}+(3-0)\hat{k}$

$\Rightarrow \overrightarrow{BA}=2\hat{i}+2\hat{j}+3\hat{k}$

$\overrightarrow{BC}=(0-(-1\left)\right)\hat{i}+(1-0)\hat{j}+(2-0)\hat{k}$

$\Rightarrow \overrightarrow{BC}=\hat{i}+\hat{j}+2\hat{k}$

$\text{Magnitude of}\overrightarrow{BA}=\sqrt{2^2+2^2+3^2}$

$=\sqrt{4+4+9}=\sqrt{17}$

$\text{Magnitude of}\overrightarrow{BC}=\sqrt{1^2+1^2+2^2}$

$=\sqrt{1+1+4}=\sqrt{6}$

$Now,$

$\overrightarrow{BA}.\overrightarrow{BC}=(2\hat{i}+2\hat{j}+3\hat{k}).(\hat{i}+\hat{j}+2\hat{k})$

$\overrightarrow{BA}.\overrightarrow{BC}=2.1+2.1+3.2$

$\overrightarrow{BA}.\overrightarrow{BC}=10$

$\text{The angle between two vectors is}$


$\cos \theta =\frac{\overrightarrow{BA}.\overrightarrow{BC}}{\left|\overrightarrow{BA}\right|\left|\overrightarrow{BC}\right|}$

$\cos \theta =\frac{10}{\sqrt{17.}\sqrt{6}}=\frac{10}{\sqrt{102}}$

$\theta =\cos {}^{-}1\left(\frac{10}{\sqrt{102}}\right)$