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

Answer

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

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