Question

If vertices of triangle ABC are A(1,2,3), B(-1,0,0) and C (0,1,2) then find $\angle$ ABC .If $\theta$ be any angle between unit vector

Answer 1

Given,

The vertices of $△ABC$ are $A(1,2,3),B(-1,0,0)$ and $C(0,1,2)$ then we have to find $\angle ABC$.

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

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

$\cos \theta =\frac{\overrightarrow{BA}\cdot \overrightarrow{BC}}{\mid \overrightarrow{BA}\mid \mid \overrightarrow{BC}\mid }=\frac{2+2+6}{\sqrt{4+4+1}\sqrt{4+4+9}}=\frac{10}{\sqrt{6}\sqrt{7}}=\frac{10}{\sqrt{102}}$

$\therefore \theta ={\cos }^{-1}\frac{10}{\sqrt{102}}$