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The vector equation of plane containing the pint $(1,-1,2)$ and perpendicular to planes $2 x+3 y-2 z=5$ and $x+2 y-3 z=8$ is
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Verified Answer
The correct answer is:
$\vec{r} \cdot(-5 \hat{i}+4 \hat{j}+\hat{k})=-7$
D.R's of the required plane can be obtained by
$\begin{aligned} & \frac{a}{3 \times(-3)-2 \times(-2)}=\frac{b}{1 \times(-2)-2 \times(-3)}=\frac{c}{2 \times 2-1 \times 3} \\ & \Rightarrow \frac{a}{-5}=\frac{b}{4}=\frac{c}{1}\end{aligned}$
$\Rightarrow$ the required equation is
$\begin{aligned} & -5(x-1)+4(y+1)+1(z-2)=0 \\ & \Rightarrow-5 x+4 y+z=-7 \\ & \Rightarrow \vec{r} \cdot(-5 \hat{i}+4 \hat{j}+\hat{k})=-7\end{aligned}$
$\begin{aligned} & \frac{a}{3 \times(-3)-2 \times(-2)}=\frac{b}{1 \times(-2)-2 \times(-3)}=\frac{c}{2 \times 2-1 \times 3} \\ & \Rightarrow \frac{a}{-5}=\frac{b}{4}=\frac{c}{1}\end{aligned}$
$\Rightarrow$ the required equation is
$\begin{aligned} & -5(x-1)+4(y+1)+1(z-2)=0 \\ & \Rightarrow-5 x+4 y+z=-7 \\ & \Rightarrow \vec{r} \cdot(-5 \hat{i}+4 \hat{j}+\hat{k})=-7\end{aligned}$
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