The vectors a and b are non-collinear. The va | vedclass

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(C) Since vectors $c = (x - 2)a + b$ and $d = (2x + 1)a - b$ are collinear,there exists a scalar $\lambda$ such that $c = \lambda d$.Substituting the

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$\frac{1}{3}$

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(C) Since vectors $c = (x - 2)a + b$ and $d = (2x + 1)a - b$ are collinear,there exists a scalar $\lambda$ such that $c = \lambda d$.Substituting the expressions for $c$ and $d$:$(x - 2)a + b = \lambda ((2x + 1)a - b)$Rearranging the terms:$(x - 2)a + b = \lambda(2x + 1)a - \lambda b$$((x - 2) - \lambda(2x + 1))a + (1 + \lambda)b = 0$Since $a$ and $b$ are non-collinear,they are linearly independent. Therefore,the coefficients of $a$ and $b$ must be zero:$1 + \lambda = 0 \Rightarrow \lambda = -1$$(x - 2) - \lambda(2x + 1) = 0$Substituting $\lambda = -1$ into the second equation:$(x - 2) - (-1)(2x + 1) = 0$$x - 2 + 2x + 1 = 0$$3x - 1 = 0$$x = \frac{1}{3}$

The vectors $a$ and $b$ are non-collinear. The value of $x$ for which the vectors $c = (x - 2)a + b$ and $d = (2x + 1)a - b$ are collinear,is

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