• # question_answer If the vertices A,B,C of a $\Delta ABC$ have position vectors (1, 2, 3),  $(-\,1,\,\,0,\,\,0)$ and (0, 1, 2) respectively, what is the magnitude of $\angle ABC?$

Given, A (1, 2, 3), $B\,(-1,\,\,0,\,\,0)$ and C (0, 1, 2). $\angle ABC$=angle between $\overrightarrow{BA}$ and$\overrightarrow{BC}$ We use formula, $\overrightarrow{a}\cdot \vec{b}=|\overrightarrow{a}|\cdot |\vec{b}|cos\theta$where$\theta$is the angle between $\vec{a}$and $\vec{b}$ Now, $\overrightarrow{BA}=\{1-(-\,1)\,\hat{i}+(2-0)\hat{j}+(3-0)\,\hat{k}=2\hat{i}+2\hat{j}+3\hat{k}$and $\overrightarrow{BC}=\{0-(-1)\,\hat{i}+(1-0)\hat{j}+(2-0)\,\hat{k}=\hat{i}+\hat{j}+2\hat{k}$ $\therefore \cos \theta =\frac{\overrightarrow{BA}\cdot }{|\overrightarrow{BA}\parallel \overrightarrow{BC|}}=\frac{(2\hat{i}+2\hat{j}+3\hat{k})\cdot (\hat{i}+\hat{j}+2\hat{k})}{\sqrt{{{2}^{2}}+{{2}^{2}}+{{3}^{2}}}\sqrt{{{1}^{2}}+{{1}^{2}}+{{2}^{2}}}}$$=\frac{2+2+6}{\sqrt{4+4+9}\sqrt{1+1+4}}=\frac{10}{\sqrt{17}\sqrt{6}}=\frac{10}{\sqrt{102}}$ $\Rightarrow$   $\theta ={{\cos }^{-1}}\left( \frac{12}{\sqrt{102}} \right)$