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# Solution for Show that the Points Whose Position Vectors Are → a = 4 ^ I − 3 ^ J + ^ K , → B = 2 ^ I − 4 ^ J + 5 ^ K , → C = ^ I − ^ J Form a Right Triangle. - CBSE (Science) Class 12 - Mathematics

ConceptBasic Concepts of Vector Algebra

#### Question

Show that the points whose position vectors are $\vec{a} = 4 \hat{i} - 3 \hat{j} + \hat{k} , \vec{b} = 2 \hat{i} - 4 \hat{j} + 5 \hat{k} , \vec{c} = \hat{i} - \hat{j}$ form a right triangle.

#### Solution

$Given that$
$\vec{a} = \vec{OA} = 4 \hat{i} - 3 \hat{j} + \hat{k} ; \vec{b} = \vec{OB} = 2 \hat{i} - 4 \hat{j} + 5 \hat{k} ; \vec{c} = \vec{OC} = \hat{i} - \hat{j} + 0 \hat{k}$
$\vec{AB} = \vec{OB} - \vec{OA} = - 2 \hat{i} - \hat{j} + 4 \hat{k}$
$\vec{BC} = \vec{OC} - \vec{OB} = - \hat{i} + 3\hat{j} - 5 \hat{k}$
$\vec{CA} = \vec{OA} - \vec{OC} = 3\hat{i} - 2 \hat{j} + \hat{k}]$
$\vec{AB} . \vec{BC} = 2 - 3 - 20 = - 21 \neq 0$
$\vec{BC} . \vec{CA} = - 3 - 6 - 5 = - 14 \neq 0$
$\vec{AB} . \vec{CA} = - 6 + 2 + 4 = 0$
$\text{So}, \vec{AB} \text{ is perpendicular to } \vec{CA} .$
$\text{So}, ∆ABC\hspace{0.167em}\text{ is a right-angled triangle. }$

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Solution Show that the Points Whose Position Vectors Are → a = 4 ^ I − 3 ^ J + ^ K , → B = 2 ^ I − 4 ^ J + 5 ^ K , → C = ^ I − ^ J Form a Right Triangle. Concept: Basic Concepts of Vector Algebra.
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