bar{a} and bar{b} are non-collinear vectors. | vedclass

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(B) Since $\bar{a}$ and $\bar{b}$ are non-collinear vectors,two vectors $\bar{p} = m_1\bar{a} + n_1\bar{b}$ and $\bar{q} = m_2\bar{a} + n_2\bar{b}$ ar

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$\frac{1}{3}$

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(B) Since $\bar{a}$ and $\bar{b}$ are non-collinear vectors,two vectors $\bar{p} = m_1\bar{a} + n_1\bar{b}$ and $\bar{q} = m_2\bar{a} + n_2\bar{b}$ are collinear if and only if their components are proportional,i.e.,$\frac{m_1}{m_2} = \frac{n_1}{n_2} = k$. Given $\bar{p} = (2x + 1)\bar{a} - 1\bar{b}$ and $\bar{q} = (x - 2)\bar{a} + 1\bar{b}$. For collinearity,we have: $\frac{2x + 1}{x - 2} = \frac{-1}{1}$ $2x + 1 = -(x - 2)$ $2x + 1 = -x + 2$ $3x = 1$ $x = \frac{1}{3}$

$\bar{a}$ and $\bar{b}$ are non-collinear vectors. If $\bar{p} = (2x + 1)\bar{a} - \bar{b}$ and $\bar{q} = (x - 2)\bar{a} + \bar{b}$ are collinear vectors,then $x =$

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