Question

If three points A, B and C have position vectors \[\hat{i} + x \hat{j} + 3 \hat{k} , 3 \hat{i} + 4 \hat{j} + 7 \hat{k}\text{ and }y \hat{i} - 2 \hat{j} - 5 \hat{k}\] respectively are collinear, then (x, y) =

Options

•  (2, −3)

• (−2, 3)

•  (−2, −3)

• (2, 3)

Answer 1

(2, −3)
Given position vectors of A, B and C are \[\hat{i} + x \hat{j} + 3 \hat{k} , 3 \hat{i} + 4 \hat{j} + 7 \hat{k}\] and \[y \hat{i} - 2 \hat{j} - 5 \hat{k} .\]
 Then,
\[\overrightarrow{AB} = 3 \hat{i} + 4 \hat{j} + 7 \hat{k} - \hat{i} - x \hat{j} - 3 \hat{k} = 2 \hat{i} + \left( 4 - x \right) \hat{j} + 4 \hat{k} \]
\[ \overrightarrow{BC} = y \hat{i} - 2 \hat{j} - 5 \hat{k} - 3 \hat{i} - 4 \hat{j} - 7 \hat{k} = \left( y - 3 \right) \hat{i} - 6 \hat{j} - 12 \hat{k}\]
Since, the given vectors are collinear.

\[\therefore \overrightarrow{AB} = \lambda \overrightarrow{BC} \]

\[ \Rightarrow 2 \hat{i} + \left( 4 - x \right) \hat{j} + 4 \hat{k} = \lambda \left( y - 3 \right) \hat{i} - 6\lambda \hat{j} - 12\lambda \hat{k} \]

\[ \Rightarrow 2 = \lambda \left( y - 3 \right) , \left( 4 - x \right) = - 6\lambda, 4 = - 12\lambda \]

\[ \Rightarrow 2 = \lambda \left( y - 3 \right) , \left( 4 - x \right) = - 6\lambda, \lambda = - \frac{1}{3}\]

\[ \Rightarrow 2 = - \frac{1}{3}\left( y - 3 \right) , \left( 4 - x \right) = - 6 \times \left( - \frac{1}{3} \right)\]

\[ \Rightarrow - 6 = y - 3 , 4 - x = 2\]

\[ \Rightarrow y = - 3 , x = 2\]