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If the foot of the perpendicular drawn from the origin to a plane is $M(-1,-2,2)$, then the vector equation of the plane is
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The correct answer is:
$\bar{r} \cdot(-\hat{i}-2 \hat{j}+2 \widehat{k})=9$
D.r's of normal to the plane $<-1,-2,2>$
D.c's of Normal to the plane $<-\frac{1}{3},-\frac{2}{3}, \frac{2}{3}>$
length of perpendicular $=\sqrt{1^2+2^2+(-2)^2}=3$
Hence, Equation of plane $\vec{r} \cdot\left(-\frac{1}{3} \hat{i}-\frac{2}{3} \hat{j}+\frac{2}{3} \hat{k}\right)=3$
$\Rightarrow \vec{r} \cdot(-\hat{i}-2 \hat{j}+2 \widehat{k})=9$
D.c's of Normal to the plane $<-\frac{1}{3},-\frac{2}{3}, \frac{2}{3}>$
length of perpendicular $=\sqrt{1^2+2^2+(-2)^2}=3$
Hence, Equation of plane $\vec{r} \cdot\left(-\frac{1}{3} \hat{i}-\frac{2}{3} \hat{j}+\frac{2}{3} \hat{k}\right)=3$
$\Rightarrow \vec{r} \cdot(-\hat{i}-2 \hat{j}+2 \widehat{k})=9$
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