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$\bar{a}$ and $\bar{b}$ are non-collinear vectors. If $\bar{p}=(2 x+1) \bar{a}-\bar{b}$ and $\bar{q}=(x-2) \bar{a}+\bar{b}$
are collinear vectors, then $x=$
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are collinear vectors, then $x=$
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Verified Answer
The correct answer is:
$\frac{1}{3}$
(A)
The vector $\bar{q}$ is non-zero since the coefficient in $\bar{b}$ is different from zero and so the vectors $\bar{q}$ and $\bar{p}$ are collinear if for some number $y$ we have
$\overline{\mathrm{p}}=\mathrm{c} \overline{\mathrm{q}}$ that is
$(2 x+1) \bar{a}-\bar{b}=y[(x-2) \bar{a}+\bar{b}]$
$\therefore(2 x+1)=y(x-2)$ and $-1=y \Rightarrow y=-1$
$2 x+1=-(x-2) \Rightarrow 3 x=1 \Rightarrow x=\frac{1}{3}$
The vector $\bar{q}$ is non-zero since the coefficient in $\bar{b}$ is different from zero and so the vectors $\bar{q}$ and $\bar{p}$ are collinear if for some number $y$ we have
$\overline{\mathrm{p}}=\mathrm{c} \overline{\mathrm{q}}$ that is
$(2 x+1) \bar{a}-\bar{b}=y[(x-2) \bar{a}+\bar{b}]$
$\therefore(2 x+1)=y(x-2)$ and $-1=y \Rightarrow y=-1$
$2 x+1=-(x-2) \Rightarrow 3 x=1 \Rightarrow x=\frac{1}{3}$
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