Question

Using vector method, prove that the following points are collinear:
A (1, 2, 7), B (2, 6, 3) and C (3, 10, −1)

Answer 1

Given the points \[A\left( 1, 2, 7 \right), B\left( 2, 6, 3 \right)\] and \[C\left( 3, 10, - 1 \right)\]. Then,
\[\overrightarrow{AB} =\]  Position vector of B - Position vector of A
\[= 2 \hat{i} + 6 \hat{j} + 3 \hat{k} - \hat{i} - 2 \hat{j} - 7 \hat{k} \]
\[ = \hat{i} + 4 \hat{j} - 4 \hat{k}\]
\[\overrightarrow{BC} =\] Position vector of C -  Position vector of B.
\[= 3 \hat{i} + 10 \hat{j} - \hat{k} - 2 \hat{i} - 6 \hat{j} - 3 \hat{k} \]
\[ = \hat{i} + 4 \hat{j} - 4 \hat{k}\]
\[\therefore \overrightarrow{AB} = \overrightarrow{BC}\]
\[\text { So }, \overrightarrow{AB} , \overrightarrow{BC}\]  are parallel vectors. But B  is a point common to them.
Hence, the given points A, B  and C are collinear.