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Question
If the vertices A, B and C of ∆ABC have position vectors (1, 2, 3), (−1, 0, 0) and (0, 1, 2), respectively, what is the magnitude of ∠ABC?
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Solution
\[\text{ Given that }\]
\[ \vec{OA} = \hat{i} + 2 \hat{j} + 3 \hat{k} ; \vec{OB} = - 1 \hat{i} + 0 \hat{j} + 0 \hat{k} ; \vec{OC} = 0 \hat{i} + 1 \hat{j} + 2 \hat{k} \]
\[ \vec{AB} = \vec{OB} - \vec{OA} = - 2 \hat{i} - 2 \hat{j} - 3 \hat{k} \Rightarrow \left| \vec{AB} \right| = \sqrt{4 + 4 + 9} = \sqrt{17}\]
\[ \vec{BC} = \vec{OC} - \vec{OB} = \hat{i} + \hat{j} + 2 \hat{k} \Rightarrow \left| \vec{BC} \right| = \sqrt{1 + 1 + 4} = \sqrt{6}\]
\[ \vec{CA} = \vec{OA} - \vec{OC} = \hat{i} + \hat{j} + \hat{k} \Rightarrow \left| \vec{CA} \right| = \sqrt{1 + 1 + 1} = \sqrt{3}\]
\[\cos ∠ ABC = \frac{\left| \vec{AB} . \vec{BC} \right|}{\left| \vec{AB} \right|\left| \vec{BC} \right|} = \frac{\left| - 2 - 2 - 6 \right|}{\left( \sqrt{17} \right)\left( \sqrt{6} \right)} = \frac{10}{\sqrt{102}}\]
\[ \Rightarrow ∠ ABC = \cos^{- 1} \left( \frac{10}{\sqrt{102}} \right)\]
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