Question

If the vertices $A,B$ and $C$ of a triangle $ABC$ are $(1,2,3),(-1,0,0),(0,1,2)$ respectively, then the $\angle ABC$ is

• A. ${60}^{\circ }$
• B. $0^{\circ }$
• C. ${\cos }^{-1}\left(\frac{5}{\sqrt{51}}\right)$
• D. ${\cos }^{-1}\left(\frac{10}{\sqrt{102}}\right)$

Answer 1

The correct option is D ${\cos }^{-1}\left(\frac{10}{\sqrt{102}}\right)$

Let $O$ be the origin.
$P.V.$ of $A=\hat{i}+2\hat{j}+3\hat{k}$
$P.V.$ of $B=-\hat{i}$
$P.V.$ of $C=\hat{j}+2\hat{k}$
Now $\overrightarrow{BA}=2\hat{i}+2\hat{j}+3\hat{k}$
$\Rightarrow \overrightarrow{BC}=\hat{i}+\hat{j}+2\hat{k}$
$\therefore \cos \angle ABC=\frac{\overrightarrow{BA}\cdot \overrightarrow{BC}}{|\overrightarrow{BA}||\overrightarrow{BC}|}$
$\Rightarrow \cos \angle ABC=\frac{10}{\sqrt{6}\sqrt{17}}=\frac{10}{\sqrt{102}}$
$\therefore \angle ABC={\cos }^{-1}\left(\frac{10}{\sqrt{102}}\right)$