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Question
If A, B and C have position vectors (0, 1, 1), (3, 1, 5) and (0, 3, 3) respectively, show that ∆ ABC is right-angled at C.
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Solution
\[\text{ Given that }\]
\[ \vec{OA} = 0 \hat{i} + \hat{j} + \hat{k} ; \vec{OB} = 3 \hat{i} + \hat{j} + 5 \hat{k} ; \vec{OC} = 0 \hat{i} + 3 \hat{j} + 3 \hat{k} \]
\[ \vec{BC} = \vec{OC} - \vec{OB} = - 3 \hat{i} + 2 \hat{j} - 2 \hat{k} \]
\[ \vec{CA} = \vec{OA} - \vec{OC} = 0 \hat{i} - 2 \hat{j} - 2 \hat{k} \]
\[\text{ Now },\]
\[ \vec{BC} . \vec{CA} = 0 - 4 + 4 = 0\]
\[\text{ So }, \vec{BC} \text{ is perpendicular to } \vec{CA} .\]
\[\text{ So }, ∆ABC\hspace{0.167em}\text{ is right-angled at C. }\]
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