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Algebra of Vectors — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Vectors5 Marks
Question
If $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are non-coplanar vectors, prove that the point having the following position vectors is collinear:$\vec{\text{a}},\ \vec{\text{b}},\ 3\vec{\text{a}}-2\vec{\text{b}}$
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Answer

Let the points be A, B, C
Position vector of $\text{A}=\vec{\text{a}}$
Position vector of $\text{B}=\vec{\text{b}}$
Position vector of $\text{C}=3\vec{\text{a}}-2\vec{\text{b}}$
$\overrightarrow{\text{AB}}=$Position vector of B - Postion vector of A
$=\vec{\text{b}}+\vec{\text{a}}$
$\overrightarrow{\text{BC}}=$Position vector of C - Postion vector of B
$=3\vec{\text{a}}-2\vec{\text{b}}-\vec{\text{b}}$
$=3\vec{\text{a}}-3\vec{\text{b}}$
Using $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$
Let $\overrightarrow{\text{BC}}=\lambda\Big(\overrightarrow{\text{AB}}\Big)$ [where $\lambda$ is and scalar]
$3\vec{\text{a}}-3\vec{\text{b}}=\lambda\big(\vec{\text{b}}-\vec{\text{a}}\big)$
$3\vec{\text{a}}-3\vec{\text{b}}=\lambda\vec{\text{b}}-\lambda\vec{\text{a}}$
$3\vec{\text{a}}-3\vec{\text{b}}=\lambda\vec{\text{a}}+\lambda\vec{\text{b}}$
Comparing the coefficients of LHS and RHS, we get,
$-\lambda=3$
$\lambda=3$
$\lambda=-3$
Since the value of $\lambda$ are different.
Therefore,
A, B, C are not collinear.

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