Question

$\overline{a}$ and $\overline{b}$ are non collinear vectors. If $\overline{c}=(x-2)\overline{a}+\overline{b}$ and $\overline{d}=(2x+1)\overline{a}-\overline{b}$ are collinear vectors, then find the value of $x$.

Answer 1

$\overline{c}=(x-2)\overline{a}+\overline{b}...\left(1\right)$

$\overline{d}=(2x+1)\overline{a}-\overline{b}...\left(2\right)$

since $\overline{a}$ & $\overline{b}$ are co liner vectors

$\overline{a}\times \overline{b}=\overline{0}...\left(3\right)$

$\therefore$ By (1) & (2) we get

$\overline{a}=\frac{(\overline{c}+\overline{d})}{(3x-1)}$

$\overline{b}=\left(\frac{2x+1}{3x-1}\right)\overline{c}-\left(\frac{x-2}{3x-1}\right)\overline{d}$

$\therefore$ putting $\overline{a}$ & $\overline{b}$ in (3)

$0=(\frac{\overline{e}}{3x-1}+\frac{\overline{d}}{3x-1})\times \left[\right(\frac{2x+1}{3x-1})\overline{c}-(\frac{x-2}{3x-1}\left)\overline{d}\right]$

$0=[\frac{2-x}{(3x-1{)}^2}-(\frac{2x+1}{(3x-1{)}^2}\left)\right](\overline{c}\times \overline{d})$

$\therefore (2-x)-(2x+1)=0$

$-3x+1=0$

$x=\frac{1}{3}$

But $x=\frac{1}{3}$ does't satisfy $\overline{a}$ & $\overline{b}$

$\therefore$ no such of x exist