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Algebra of Vectors — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Vectors4 Marks
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 $\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}$
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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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