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If $(2,3,-1)$ is the foot of the perpendicular from $(4,2,1)$ to a plane, then the equation of the plane is
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Verified Answer
The correct answer is:
$2 x-y+2 z+1=0$
Since, the line joining the two points is
perpendicular to the plane, it's DR's will given the normal to the plane.
DR's of the normal $=(4-2,2-3,1+1)=(2,-1,2)$
Hence, $a=2, b=-1$ and $c=2$
Since, the plane passes through $(2,3,-1)$.
$\begin{aligned} & d=2(2)+3(-1)-1(2) \\ & d=-1\end{aligned}$
Hence, the required equation of plane is
$2 x-y+2 z=-1$
perpendicular to the plane, it's DR's will given the normal to the plane.
DR's of the normal $=(4-2,2-3,1+1)=(2,-1,2)$
Hence, $a=2, b=-1$ and $c=2$
Since, the plane passes through $(2,3,-1)$.
$\begin{aligned} & d=2(2)+3(-1)-1(2) \\ & d=-1\end{aligned}$
Hence, the required equation of plane is
$2 x-y+2 z=-1$
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