Question
Using vector method, prove that the point is collinear:
A(-3, -2, -5), B(1, 2, 3) and C(3, 4, 7)
A(-3, -2, -5), B(1, 2, 3) and C(3, 4, 7)
✓
Answer
Given the points A(-3, -2, -5), B(1, 2, 3) and C(3, 4, 7). Then, $\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A$=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}+3\hat{\text{i}}+2\hat{\text{j}}+5\hat{\text{k}}$
$=4\hat{\text{i}}+4\hat{\text{j}}+8\hat{\text{k}}$
$=2\big(2\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}\big)$$\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B
$=3\hat{\text{i}}+4\hat{\text{j}}+7\hat{\text{k}}-\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
$=2\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}$
$\therefore\ \overrightarrow{\text{AB}}=2\overrightarrow{\text{BC}}$
So, $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$ are parallel vectors. But B is a point common to them. Hence, the given points A, B, and C are collinear.
$=4\hat{\text{i}}+4\hat{\text{j}}+8\hat{\text{k}}$
$=2\big(2\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}\big)$$\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B
$=3\hat{\text{i}}+4\hat{\text{j}}+7\hat{\text{k}}-\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
$=2\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}$
$\therefore\ \overrightarrow{\text{AB}}=2\overrightarrow{\text{BC}}$
So, $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$ are parallel vectors. But B is a point common to them. Hence, the given points A, B, and C are collinear.
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