Question

The vectors $\overrightarrow{a}\mathrm{and}\overrightarrow{b}$ are non-collinear. If vectors $(x-2)\overrightarrow{a}+\overrightarrow{b}\mathrm{and}(2x+1)\overrightarrow{a}-\overrightarrow{b}$ are collinear, then x = _________________.

Answer 1

Given:
The vectors $\overrightarrow{a}\mathrm{and}\overrightarrow{b}$ are non-collinear.
Vectors $(x-2)\overrightarrow{a}+\overrightarrow{b}\mathrm{and}(2x+1)\overrightarrow{a}-\overrightarrow{b}$ are collinear.


Let $\overrightarrow{p}=(x-2)\overrightarrow{a}+\overrightarrow{b}\mathrm{and}\overrightarrow{q}=(2x+1)\overrightarrow{a}-\overrightarrow{b}$ are collinear
Then, $\overrightarrow{p}=\lambda \overrightarrow{q},\mathrm{where}\lambda \mathrm{is}\mathrm{some}\mathrm{scalar}.$

$$\begin{gathered} \left(x-2\right)\overrightarrow{a}+\overrightarrow{b}=\lambda \left[\left(2x+1\right)\overrightarrow{a}-\overrightarrow{b}\right] \\ \Rightarrow \left(x-2\right)\overrightarrow{a}+\overrightarrow{b}=\lambda \left(2x+1\right)\overrightarrow{a}+\left(-\lambda \right)\overrightarrow{b} \\ \Rightarrow x-2=\lambda \left(2x+1\right)\mathrm{and}1=\left(-\lambda \right) \\ \Rightarrow x-2=\lambda \left(2x+1\right)\mathrm{and}-1=\lambda \\ \Rightarrow x-2=-1\left(2x+1\right) \\ \Rightarrow x-2=-2x-1 \\ \Rightarrow x+2x=-1+2 \\ \Rightarrow 3x=1 \\ \Rightarrow x=\frac{1}{3} \end{gathered}$$


Hence, x = $\underline{\frac{1}{3}}$.