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MHT CET · Maths · Vector Algebra

\(\bar{a}\) 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 \(x=\)

  1. A \(-3\)
  2. B \(\frac{1}{3}\)
  3. C \(-\frac{1}{3}\)
  4. D 3
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{1}{3}\)

Step-by-step Solution

Detailed explanation

(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 \(-1=y \Rightarrow y=-1\)
\(2 x+1=-(x-2) \Rightarrow 3 x=1 \Rightarrow x=\frac{1}{3}\)