Question

The position vectors of the points A, B, C are \[2 \hat{i} + \hat{j} - \hat{k} , 3 \hat{i} - 2 \hat{j} + \hat{k}\text{ and }\hat{i} + 4 \hat{j} - 3 \hat{k}\] respectively.
These points

Options

•  form an isosceles triangle

• form a right triangle

• are collinear

• form a scalene triangle

Answer 1

form an isosceles triangle
Given: Position vectors of A, B, C and \[\hat{i} + 4 \hat{j} - 3 \hat{k}\].
Then,

\[\overrightarrow{AB} = \left( 3 \hat{i} - 2 \hat{j} + \hat{k} \right) - \left( 2 \hat{i} + \hat{j} - \hat{k} \right) = \hat{i} - 3 \hat{j} + 2 \hat{k} \]

\[ \overrightarrow{BC} = \left( \hat{i} + 4 \hat{j} - 3 \hat{k} \right) - \left( 3 \hat{i} - 2 \hat{j} + \hat{k} \right) = - 2 \hat{i} + 6 \hat{j} - 4 \hat{k} \]

\[ \overrightarrow{CA} = \left( 2 \hat{i} + \hat{j} - \hat{k} \right) - \left( \hat{i} + 4 \hat{j} - 3 \hat{k} \right) = \hat{i} - 3 \hat{j} + 2 \hat{k}\]
\[\text{ Now, }\overrightarrow{\left| AB \right|} = \sqrt{1^2 + \left( - 3 \right)^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14}\]
\[ \left| \overrightarrow{CA} \right| = \sqrt{1^2 + \left( - 3 \right)^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14}\]
\[\left| \overrightarrow{BC} \right| = \sqrt{\left( - 2 \right)^2 + 6^2 + \left( - 4^2 \right)} = \sqrt{4 + 36 + 16} = \sqrt{56}\]
\[ \therefore \left| AB \right| = \left| \overrightarrow{CA} \right|\]
Hence, the triangle is isosceles as two of its sides are equal.