If the foot of the perpendicular drawn from the origin to a plane is $M(-1,-2,2)$, then the vector equation…

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
  1. $\bar{r} \cdot(-\hat{i}-2 \hat{j}+2 \widehat{k})=9$
  2. $\bar{r} \cdot(\widehat{i}+2 \widehat{j}+2 \widehat{k})=9$
  3. $\bar{r} \cdot(-\hat{i}-2 \hat{j}-2 \widehat{k})=9$
  4. $\bar{r} \cdot(\hat{i}+2 \hat{j}-2 \widehat{k})=9$

Solution

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 $\vec{r} \cdot\left(-\frac{1}{3} \hat{i}-\frac{2}{3} \hat{j}+\frac{2}{3} \hat{k}\right)=3$ $\Rightarrow \vec{r} \cdot(-\hat{i}-2 \hat{j}+2 \widehat{k})=9$

Asked in: MHT CET 2022 (10 Aug Shift 1)

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