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Eqn of Plane passing through a point and perpendicular to another plane's trace

by Stumbleinn
Tags: passing, perpendicular, plane, point, trace
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Stumbleinn
#1
Aug25-08, 06:46 PM
P: 1
1. The problem statement, all variables and given/known data
Here's the problem:Find the equation of the plane passing through the point (-3,1,4) and perpendicular to the trace of the plane x-3y+7z-3=0 in the xy plane.


2. Relevant equations

to me this should be as easy as finding the two coordiates of the plane's traces on the x and y axis. ie (3,0,0) and (0,-1, 0). Since these two points and the point (-3,1,4) determine the plane.

3. The attempt at a solution
since the general form of the plane is Ax+By+Cz+D=0. I should get 3 equations that look like this:
1) 3A+D=0
2) -B+D=0
3) -3A+B+4C+D=0
1) implies that -D/A=3 and thus D=-3 and A=1
substituting into eqn2 yields B=-3
and sunstituing into eqn3 gives C=9/4
so the eqn should be x-3y+9/4Z-3=0 or 4x-12y+9z-3=0
However the answer in the book is 3x+y+8=0. I find that the book is rarely wrong and usually it's me that is confused.

Please advise on where I went wrong
Thanks so much
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HallsofIvy
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Aug25-08, 07:37 PM
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Quote Quote by Stumbleinn View Post
1. The problem statement, all variables and given/known data
Here's the problem:Find the equation of the plane passing through the point (-3,1,4) and perpendicular to the trace of the plane x-3y+7z-3=0 in the xy plane.


2. Relevant equations

to me this should be as easy as finding the two coordiates of the plane's traces on the x and y axis. ie (3,0,0) and (0,-1, 0). Since these two points and the point (-3,1,4) determine the plane.
These 3 points give the plane containing the trace in the xy plane, not perpendicular to it. The trace of the plane in the xy-plane is given, of course, by x- 3y- 3= 0 and z= 0. If you use y itself as parameter, x= 3y+ 2, y= y, z= 0 are parametric equation of that line. A vector pointing in the direction of that line is <3, 1, 0>. You want the equation of a plane containing (-3, 1, 4) and having normal vector <3, 1, 0>.


3. The attempt at a solution
since the general form of the plane is Ax+By+Cz+D=0. I should get 3 equations that look like this:
1) 3A+D=0
2) -B+D=0
3) -3A+B+4C+D=0
1) implies that -D/A=3 and thus D=-3 and A=1
substituting into eqn2 yields B=-3
and sunstituing into eqn3 gives C=9/4
so the eqn should be x-3y+9/4Z-3=0 or 4x-12y+9z-3=0

However the answer in the book is 3x+y+8=0. I find that the book is rarely wrong and usually it's me that is confused.

Please advise on where I went wrong
Thanks so much


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