Question
Show that the points A, B, C with position vectors $\vec{\text{a}}-2\vec{\text{b}}+3\vec{\text{c}},\ 2\vec{\text{a}}+3\vec{\text{b}}-4\vec{\text{c}}$ and $-7\vec{\text{b}}+10\vec{\text{c}}$ are collinear.

Answer

We have, A, B, C with position vectors $\vec{\text{a}}-2\vec{\text{b}}+3\vec{\text{c}},\ 2\vec{\text{a}}+3\vec{\text{b}}-4\vec{\text{c}}$ and $-7\vec{\text{b}}+10\vec{\text{c}}$ Then,
$\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A
$=2\vec{\text{a}}+3\vec{\text{b}}-4\vec{\text{c}}-\vec{\text{a}}+2\vec{\text{b}}-3\vec{\text{c}}$
$=\vec{\text{a}}+5\vec{\text{b}}-7\vec{\text{c}}$
$\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B
$=-7\vec{\text{b}}+10\vec{\text{c}}-2\vec{\text{a}}-3\vec{\text{b}}+4\vec{\text{c}}$
$=-2\vec{\text{a}}-10\vec{\text{b}}+14\vec{\text{c}}$
$=-2\big(\vec{\text{a}}+5\vec{\text{b}}-7\vec{\text{c}}\big)$
$\therefore\ \overrightarrow{\text{BC}}=-2\overrightarrow{\text{AB}}$
Hence, $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$ are parallel vectors.
But B is a point common to them.
So, $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$ are collinear.
Hence, points A, B and C are collinear.

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