Question
Using vector method, prove that the point is collinear:
A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1)
A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1)
✓
Answer
Given the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1). Then, $\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A$=2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}-\hat{\text{i}}-2\hat{\text{j}}-7\hat{\text{k}}$
$=\hat{\text{i}}+4\hat{\text{j}}-4\hat{\text{k}}$
$\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B
$=3\hat{\text{i}}+10\hat{\text{j}}-\hat{\text{k}}-2\hat{\text{i}}-6\hat{\text{j}}-3\hat{\text{k}}$
$=\hat{\text{i}}+4\hat{\text{j}}-4\hat{\text{k}}$
$\therefore\ \overrightarrow{\text{AB}}=\overrightarrow{\text{BC}}$
So, $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$ are parallel vectors. But B is a point common to them. Hence, the given points A, B, and C are collinear.
$=\hat{\text{i}}+4\hat{\text{j}}-4\hat{\text{k}}$
$\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B
$=3\hat{\text{i}}+10\hat{\text{j}}-\hat{\text{k}}-2\hat{\text{i}}-6\hat{\text{j}}-3\hat{\text{k}}$
$=\hat{\text{i}}+4\hat{\text{j}}-4\hat{\text{k}}$
$\therefore\ \overrightarrow{\text{AB}}=\overrightarrow{\text{BC}}$
So, $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$ are parallel vectors. But B is a point common to them. Hence, the given points A, B, and C are collinear.
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