The foot of the perpendicular drawn from origin to a plane is $\mathrm{M}(2,1,-2)$, then vector equation of…

The foot of the perpendicular drawn from origin to a plane is $\mathrm{M}(2,1,-2)$, then vector equation of the plane is
  1. $\bar{r} \cdot(2 \hat{i}+\hat{j}-2 \hat{k})=9$
  2. $\overline{\mathrm{r}} \cdot(-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}})=7$
  3. $\quad \overline{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}})=9$
  4. $\bar{r} \cdot(2 \hat{i}-\hat{j}-\hat{k})=7$

Solution

The plane passes through $(2,1,-2)$ This point satisfies the equation of plane in option (A). Also, it has d.r.s. 2, 1, -2 $\therefore \quad$ Option (A) is the correct answer.

Asked in: MHT CET 2024 (15 May Shift 2)

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