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 angle between the vectors $\overrightarrow{BA}$ and $\overrightarrow{BC}$] .

Answer 1

The vertices of $\Delta ABC$ are given as $A(1,2,3),B(-1,0,0)$, and $C(0,1,2)$.

Also, it is given that $\angle ABC$ is the angle between the vectors $\overrightarrow{BA}$ and $\overrightarrow{BC}$.

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

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

$\therefore \overrightarrow{BA}\cdot \overrightarrow{BC}=(2\hat{i}+2\hat{j}+3\hat{k})\cdot (\hat{i}+\hat{j}+2\hat{k})=2\times 1+2\times 1+3\times 2=2+2+6=10$

$|\overrightarrow{BA}|=\sqrt{2^2+2^2+3^2}=\sqrt{4+4+9}=\sqrt{17}$

$|\overrightarrow{BC}|=\sqrt{1+1+2^2}=\sqrt{6}$

Now, it is known that:

$\overrightarrow{BA}\cdot \overrightarrow{BC}=|\overrightarrow{BA}||\overrightarrow{BC}|\cos \left(\angle ABC\right)$

$\therefore 10=\sqrt{17}\times \sqrt{6}\cos \left(\angle ABC\right)$

$\Rightarrow \cos \left(\angle ABC\right)=\frac{10}{\sqrt{17}\times \sqrt{6}}$

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