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 \[\angle ABC?\]

Answer:

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 \[\overrightarrow{BC}=\{0-(-1)\,\hat{i}+(1-0)\hat{j}+(2-0)\,\hat{k}=\hat{i}+\hat{j}+2\hat{k}\] \[\therefore \cos \theta =\frac{\overrightarrow{BA}\cdot }{|\overrightarrow{BA}\parallel \overrightarrow{BC|}}=\frac{(2\hat{i}+2\hat{j}+3\hat{k})\cdot (\hat{i}+\hat{j}+2\hat{k})}{\sqrt{{{2}^{2}}+{{2}^{2}}+{{3}^{2}}}\sqrt{{{1}^{2}}+{{1}^{2}}+{{2}^{2}}}}\]\[=\frac{2+2+6}{\sqrt{4+4+9}\sqrt{1+1+4}}=\frac{10}{\sqrt{17}\sqrt{6}}=\frac{10}{\sqrt{102}}\] \[\Rightarrow \]   \[\theta ={{\cos }^{-1}}\left( \frac{12}{\sqrt{102}} \right)\]