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.

Answer 1

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.