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Algebra of Vectors — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Vectors4 Marks
Question
If $\vec{\text{a}},\vec{\text{b}}$ are two non-collinear vectors, prove that the points with position vectors $\vec{\text{a}}+\vec{\text{b}},\ \vec{\text{a}}-\vec{\text{b}}$ and $\vec{\text{a}}+\lambda\vec{\text{b}}$ are collinear for all real values of $\lambda$.
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Answer

Let A, B, C be the points then,

Position vector of $\text{A}=\vec{\text{a}}+\vec{\text{b}}$

Position vector of $\text{B}=\vec{\text{a}}-\vec{\text{b}}$

Position vector of $\text{C}=\vec{\text{a}}+\lambda\vec{\text{b}}$

$\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A

$=\big(\vec{\text{a}}-\vec{\text{b}}\big)-\big(\vec{\text{a}}+\vec{\text{b}}\big)$

$=\vec{\text{a}}-\vec{\text{b}}-\vec{\text{a}}-\vec{\text{b}}$

$\overrightarrow{\text{AB}}=-2\vec{\text{b}}$

$\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B

$=\big(\vec{\text{a}}+\lambda\vec{\text{b}}\big)-\big(\vec{\text{a}}-\vec{\text{b}}\big)$

$=\vec{\text{a}}+\lambda\vec{\text{b}}-\vec{\text{a}}+\vec{\text{b}}$

$\overrightarrow{\text{BC}}=(\lambda+1)\vec{\text{b}}$

Using $\overrightarrow{\text{AB}}$ and $\overrightarrow{\text{BC}}$, we get

$\overrightarrow{\text{AB}}=\Big[\frac{(\lambda+1)}{2}\Big]\Big(\overrightarrow{\text{BC}}\Big)$

Let $\Big(\frac{\lambda+1}{2}\Big)=\mu$

Since $\lambda$ is a real number. So,

$\mu$ is also a real number.

So, $\overrightarrow{\text{AB}}$ is parallel to $\overrightarrow{\text{BC}}$, but $\vec{\text{B}}$ is a common vector. Hence,

A, B, C are collinear.

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