The distance of a point $(1,2,-1)$ from the plane $x-2 y+4 z+10=0$ is

The distance of a point $(1,2,-1)$ from the plane $x-2 y+4 z+10=0$ is
  1. $\frac{3}{\sqrt{7}}$ units
  2. $\frac{\sqrt{3}}{7}$ units
  3. $\sqrt{\frac{7}{3}}$ units
  4. $\sqrt{\frac{3}{7}}$ units

Solution

$\begin{aligned} \text { Required distance } &=\frac{|1(1)+-2(2)+4(-1)+10|}{\sqrt{1^{2}+(-2)^{2}+(4)^{2}}} \\ &=\frac{3}{\sqrt{21}}=\frac{\sqrt{3}}{\sqrt{7}} \text { units } \end{aligned}$

Asked in: MHT CET 2020 (19 Oct Shift 1)

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