Question

Prove that the points having position vectors \[\hat{i} + 2 \hat{j} + 3 \hat{k} , 3 \hat{i} + 4 \hat{j} + 7 \hat{k} , - 3 \hat{i} - 2 \hat{i} - 5 \hat{k}\] are collinear.

Answer 1

Let A, B, C  be the points with position vectors
\[\hat{i} + 2 \hat{j} + 3 \hat{k} , 3 \hat{i} + 4 \hat{j} + 7 \hat{k} , - 3 \hat{i} - 2 \hat{j} - 5 \hat{k} .\]
 Then,
\[\overrightarrow{AB} =\]  Position vector of B - Position vector of A 
\[= 3 \hat{i} + 4 \hat{j} + 7 \hat{k} - \hat{i} - 2 \hat{j} - 3 \hat{k} \]
\[ = 2 \hat{i} + 2 \hat{j} + 4 \hat{k}\]
\[\overrightarrow{BC} =\] Position vector of C -  Position vector of B
\[= - 3 \hat{i} - 2 \hat{j} - 5 \hat{k} - 3 \hat{i} - 4 \hat{j} - 7 \hat{k} \]
\[ = - 6 \hat{i} - 6 \hat{j} - 12 \hat{k} \]
\[ = - 3\left( 2 \hat{i} + 2 \hat{j} + 4 \hat{k} \right)\]
∴ \[\overrightarrow{BC} = - 3 \overrightarrow{AB}\]
\[\text{ So, }\overrightarrow{AB} \text{ and }\overrightarrow{BC}\]  are parallel vectors.
But B  is a point common to them.
So,
\[\overrightarrow{AB}\] and \[\overrightarrow{BC}\]  are collinear.
Hence, 
A, B, C are collinear.