Question

Using vector method, prove that the following points are collinear:
A (6, −7, −1), B (2, −3, 1) and C (4, −5, 0)

Answer 1

Given  the points \[A\left( 6, - 7, - 1 \right), B\left( 2, - 3, 1 \right)\] and \[C\left( 4, - 5, 0 \right)\]. Then,
\[\overrightarrow{AB} =\]Position vector of B - Position vector of A
\[= 2 \hat{i} - 3 \hat{j} + \hat{k} - 6 \hat{i} + 7 \hat{j} + \hat{k} \]
\[ = - 4 \hat{i} + 4 \hat{j} + 2 \hat{k} \]
\[ = - 2\left( 2 \hat{i} - 2 \hat{j} - \hat{k} \right)\]
\[\overrightarrow{BC} =\] Position vector of C - Position vector of B
\[= 4 \hat{i} - 5 \hat{j} - 2 \hat{i} + 3 \hat{j} - \hat{k} \]
\[ = 2 \hat{i} - 2 \hat{j} - \hat{k}\]
\[\therefore \overrightarrow{AB} = - 2 \overrightarrow{BC}\]
\[So, \overrightarrow{AB} , \overrightarrow{BC}\]  are parallel vectors. But B is a point common to them.
Hence, the given points A, B  and C  are collinear.