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Algebra of Vectors — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Vectors4 Marks
Question
Prove that the points having position vectors $\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}},\ 3\hat{\text{i}}+4\hat{\text{j}}+7\hat{\text{k}}$ and $-3\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}$ are collinear.
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Answer

Let A, B, C be the points with position vectors $\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}},\ 3\hat{\text{i}}+4\hat{\text{j}}+7\hat{\text{k}}$ and $-3\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}$
Then,
$\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A
$=3\hat{\text{i}}+4\hat{\text{j}}+7\hat{\text{k}}-\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
$=2\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}$
$\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B
$=-3\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}-3\hat{\text{i}}-4\hat{\text{j}}-7\hat{\text{k}}$
$=-6\hat{\text{i}}-6\hat{\text{j}}-12\hat{\text{k}}$
$=-3\big(2\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}\big)$
$\therefore\ \overrightarrow{\text{BC}}=-3\overrightarrow{\text{AB}}$
So, $\overrightarrow{\text{AB}}$ 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, A, B, C are collinear.

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