Question

Show that the points whose position vectors are as given below are collinear:
\[2 \hat{i} + \hat{j} - \hat{k} , 3 \hat{i} - 2 \hat{j} + \hat{k} \text{ and }\hat{i} + 4 \hat{j} - 3 \hat{k}\]

Answer 1

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