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Show that the Points a ( 2 ^ I − ^ J + ^ K ) , B ( ^ I − 3 ^ J − 5 ^ K ) , C ( 3 ^ I − 4 ^ J − 4 ^ K ) Are the Vertices of a Right Angled Triangle.

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Question

Show that the points \[A \left( 2 \hat{i} - \hat{j} + \hat{k} \right), B \left( \hat{i} - 3 \hat{j} - 5 \hat{k} \right), C \left( 3 \hat{i} - 4 \hat{j} - 4 \hat{k} \right)\] are the vertices of a right angled triangle.

Sum
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Solution

Given the points \[A\left( 2 \hat{i} - \hat{j} + \hat{k} \right), B\left( \hat{i} - 3 \hat{j} - 5 \hat{k} \right)\] and \[C\left( 3 \hat{i} - 4 \hat{j} - 4 \hat{k} \right) .\] Then, \[\vec{AB} =\] Position vector of B - Position vector of A 
\[= \hat{i} - 3 \hat{j} - 5 \hat{k} - \left( 2 \hat{i} - \hat{j} + \hat{k} \right)\]
\[ = \hat{i} - 3 \hat{j} - 5 \hat{k} - 2 \hat{i} + \hat{j} - \hat{k} \]
\[ = - \hat{i} - 2 \hat{j} - 6 \hat{k}\]

\[\overrightarrow{BC} =\] Position vector of  C - Position vector of B
\[= 3 \hat{i} - 4 \hat{j} - 4 \hat{k} - \left( \hat{i} - 3 \hat{j} - 5 \hat{k} \right)\]
\[ = 3 \hat{i} - 4 \hat{j} - 4 \hat{k} - \hat{i} + 3 \hat{j} + 5 \hat{k} \]
\[ = 2 \hat{i} - \hat{j} + \hat{k}\]
\[\overrightarrow{CA} =\] Position vector of A- Position vector of C
\[= 2 \hat{i} - \hat{j} + \hat{k} - \left( 3 \hat{i} - 4 \hat{j} - 4 \hat{k} \right)\]
\[ = 2 \hat{i} - \hat{j} + \hat{k} - 3 \hat{i} + 4 \hat{j} + 4 \hat{k} \]
\[ = - \hat{i} + 3 \hat{j} + 5 \hat{k}\]
Clearly, 
\[\overrightarrow{AB} + \vec{BC} + \vec{CA} = \vec{0}\]
\[\text{ Now, }\overrightarrow{\left| AB \right|} = \sqrt{\left( - 1 \right)^2 + \left( - 2 \right)^2 + \left( - 6 \right)^2} = \sqrt{1 + 4 + 36} = \sqrt{41}\]
\[ \overrightarrow{\left| BC \right|} = \sqrt{\left( 2 \right)^2 + \left( - 1 \right)^2 + \left( 1 \right)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}\]
\[ \overrightarrow{\left| CA \right|} = \sqrt{\left( - 1 \right)^2 + \left( 3 \right)^2 + \left( 5 \right)^2} = \sqrt{1 + 9 + 25} = \sqrt{35}\]
\[\text{ Clearly, }\overrightarrow{\left| AB \right|}^2 = \overrightarrow{\left| BC \right|}^2 + \overrightarrow{\left| CA \right|}^2 \]
\[ \Rightarrow A B^2 = B C^2 + C A^2 \]
So, A, B, C forms a right angled triangle.
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Chapter 22: Algebra of Vectors - Exercise 23.6 [Page 49]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 22 Algebra of Vectors
Exercise 23.6 | Q 13 | Page 49

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