Aspire Faculty ID #17162 · Topic: AMU MCA 2020 · Just now
AMU MCA 2020

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

Solution

Distance formula:

$d=\frac{|ax_1+by_1+cz_1-d|}{\sqrt{a^2+b^2+c^2}}$

Plane: $x+2y+z-4=0$

$d=\frac{|1(1)+2(0)+1(-2)-4|}{\sqrt{1^2+2^2+1^2}}$

$=\frac{|1-2-4|}{\sqrt6}=\frac{5}{\sqrt6} =\frac{5\sqrt6}{6}$

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