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Need solution for RD Sharma maths class 12 chapter 22 Algebra of Vectors exercise Fill in the blanks question 15

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Answer:\frac{1}{3}

Hint:-To solve this we know that two lines are collinear

Given:-The vector \vec{a} and \vec{b} are non collinear if (x-2) \vec{a}+\vec{b} \text { and }(2 x+1) \vec{a}-\vec{b} are collinear then x……

Solution:- Let a_{1} x+b_{1} y+c_{1}=0, a_{2} x+b_{2} y+c_{2}=0

For collinear

\begin{aligned} &\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} \\\\ &\vec{c}=(x-2) \vec{a}+\vec{b} \\\\ &\vec{b}=(2 x+1) \vec{a}-\vec{b} \end{aligned}

Hence

\begin{aligned} \\ &\frac{x-2}{2 x+1}=\frac{1}{-1}\\\\ &-x+2=2 x+1\\\\ &2-1=2 x+x \end{aligned}

\begin{aligned} &1=3 x \\\\ &x=\frac{1}{3} \end{aligned}

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