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Question
Show that the vectors \[\vec{a} = 3 \hat{i} - 2 \hat{j} + \hat{k} , \vec{b} = \hat{i} - 3 \hat{j} + 5 \hat{k} , \vec{c} = 2 \hat{i} + \hat{j} - 4 \hat{k}\] form a right-angled triangle.
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Solution
\[\text{ Let ABC be the given triangle and }\]
\[ \vec{AC} = \vec{b} = \hat{i} - 3 \hat{j} + 5 \hat{k} \]
\[ \vec{CB} = \vec{a} = 3 \hat{i} - 2 \hat{j} + \hat{k} \]
\[ \vec{AB} = \vec{c} = 2 \hat{i} + \hat{j} - 4 \hat{k} \]
\[ \vec{a} . \vec{b} = 3 + 6 + 5 = 14\]
\[ \vec{b} . \vec{c} = 2 - 3 - 20 = - 21\]
\[ \vec{c} . \vec{a} = 6 - 2 - 4 = 0\]
\[\text{ So }, \vec{AB} \text{ is perpendicular to } \vec{CB} . \]
\[\text{ Thus }, ∆ABC\text{ is a right-angled triangle. }\]
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