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

$Given that \vec{O A} = 0 \hat{i} + \hat{j} + \hat{k}; \vec{O B} = 3 \hat{i} + \hat{j} + 5 \hat{k}; \vec{O C} = 0 \hat{i} + 3 \hat{j} + 3 \hat{k} \vec{B C} = \vec{O C} - \vec{O B} = - 3 \hat{i} + 2 \hat{j} - 2 \hat{k} \vec{C A} = \vec{O A} - \vec{O C} = 0 \hat{i} - 2 \hat{j} - 2 \hat{k} Now, \vec{B C} . \vec{C A} = 0 - 4 + 4 = 0 So, \vec{B C} is perpendicular to \vec{C A} . So, ∆ ABC is right-angled at C .$

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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.

Given thatOA→=0i^+j^+k^; OB→=3i^+j^+5k^; OC→=0i^+3j^+3k^BC→=OC→-OB→=-3i^+2j^-2k^CA→=OA→-OC→=0i^-2j^-2k^Now,BC→. CA→=0-4+4=0So, BC→ is perpendicular to CA→.So, ∆ABC is right-angled at C.