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February 13th, 2017, 07:07 AM   #1
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Geometry 3D problem

Please help me answer the following question:

A variable plane passes through a fixed point (1, 2, 3). Show that the locus of the foot of the perpendicular drawn from origin to this plane is the sphere given by the equation:
x² + y² + z² - x - 2y - 3z = 0

Last edited by skipjack; February 17th, 2017 at 12:46 PM.
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February 16th, 2017, 06:05 PM   #2
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I think this question is poorly stated. If the origin (0,0,0) is on one of the planes, there is no perpendicular from the origin to this plane. Here's an answer to a rephrased question:

A variable plane not containing the origin passes through a fixed point $(1, 2, 3)$. Show that the locus of the foot of the perpendicular drawn from the origin to this plane is the sphere S, except for the origin; here S has the equation:
$$x^2 + y^2 + z^2 - x - 2y - 3z = 0$$
Proof.
Any plane has equation of the form $ax+by+cz=d$ where $(a,b,c)$ is a normal to the plane (hence $\neq (0,0,0)$). Assume the normal is normalized; i.e. $a^2+b^2+c^2=1$. Since (1,2,3) is on the plane, $d=a+2b+3c$. Now the line through the origin and perpendicular to the plane has parametric representation $(0,0,0)+t(a,b,c)=t(a,b,c)$. This line then meets the plane at the foot of the perpendicular when $t(a,b,c)\bullet (a.b.c)=d$; i.e. $t=d$. Thus the point of intersection is $(da,db,dc)$. Let $(x_1,y_1,z_1)=(da,db,dc)$. Then $$x_1^2+y_1^2+z_1^2=d^2=d(a+2b+3c)=da+2db+3dc=x_1+ 2y_1+3z_1$$. So $(x_1,y_1,z_1)$ is on S.
Conversely, suppose $(x_1,y_1,z_1)\neq(0,0,0)$ is on S. Then $$0<x_1^2+y_1^2+z_1^2=x_1+2y_1+3z_1$$. Let $$d=\sqrt{x_1+2y_1+3z_1}, a=x_1/d, b=y_1/d \text{ and }c=z_1/d$$. Then $$a^2+b^2+c^2={x_1^2+y_1^2+z_1^2\over d^2}=1$$. Consider the plane $ax+by+cz=d$; then $$a+2b+3c=x_1/d+2y_1/d+3z_1/d=d^2/d=d$$ and (1,2,3) is on this plane. By the paragraph above, the foot of the perpendicular to this plane is the point $(da,db,dc)=(x_1,y_1,z_1)$. Hence any point of S, unequal to (0,0,0), is on the locus.
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