February 13th, 2017, 06:07 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 11:46 AM. 
February 16th, 2017, 05:05 PM  #2 
Member Joined: Jan 2016 From: Athens, OH Posts: 45 Thanks: 26 
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. 

Tags 
3dimensional, geometry, problem 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Geometry problem  benmuskler  Geometry  1  August 4th, 2014 08:48 AM 
geometry problem  Chrisnew1  Geometry  0  January 26th, 2014 02:19 PM 
Another Geometry Problem  Eminem_Recovery  Geometry  3  January 15th, 2011 01:28 AM 
Geometry problem  alikn  Geometry  5  September 28th, 2010 12:11 PM 
can someone help me with this geometry problem?  joshua.tse  Geometry  1  June 26th, 2009 04:10 PM 