$\vec{a}$ and $\vec{b}$ are two vectors such that $\vec{a}$ is not parallel to $\vec{b}$. If $\overrightarrow{\mathrm{p}}=(\mathrm{x}+2 \mathrm{y}+3) \overrightarrow{\mathrm{a}}+(5 \mathrm{x}-\mathrm{y}+2) \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{q}}=(2 \mathrm{x}+3 \mathrm{y}+5)$…

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$\vec{a}$ and $\vec{b}$ are two vectors such that $\vec{a}$ is not parallel to $\vec{b}$. If $\overrightarrow{\mathrm{p}}=(\mathrm{x}+2 \mathrm{y}+3) \overrightarrow{\mathrm{a}}+(5 \mathrm{x}-\mathrm{y}+2) \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{q}}=(2 \mathrm{x}+3 \mathrm{y}+5)$ $\vec{a}+(x-5 y-2) \vec{b}$ are two vectors such that $\vec{p}=2 \vec{q}$, then $\mathrm{x}-2 \mathrm{y}=$

Aicharya · Accepted Answer

$ \begin{aligned} & 	ext { } \bar{p}=2 \bar{q} \ & \Rightarrow(x+2 y+3) \bar{a}+(5 x-y+2) \bar{b} \ & (4 x+6 y+10) \vec{a}+(2 x-10 y-9) \vec{b} \end{aligned} $ Comparing both sides we get $ \begin{aligned} & 3 x+4 y+7=0 ...(1)\ & 	ext { and } 3 x+9 y+6=0 ...(2)\end{aligned} $ Solving (1) \& (2) we get, $ x=-\frac{13}{5}, \quad y=\frac{1}{5} $ Hence $x-2 y=-3$

$\vec{a}$ and $\vec{b}$ are two vectors such that $\vec{a}$ is not parallel to $\vec{b}$. If…

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