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 February 13th, 2017, 07:07 AM #1 Newbie   Joined: Feb 2017 From: manipur Posts: 1 Thanks: 0 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: xÂ² + yÂ² + zÂ² - x - 2y - 3z = 0 Last edited by skipjack; February 17th, 2017 at 12:46 PM.
 February 16th, 2017, 06:05 PM #2 Member   Joined: Jan 2016 From: Athens, OH Posts: 40 Thanks: 24 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

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