Find the point of intersection of three planes

cluivee

1. The problem statement, all variables and given/known data
The plane P1 contains the points A,B,C, which have position vectors a=(0,0,0), b=(1,1,8) and c=(0,1,5) respectively. Plane P2 passes through A and is orthogonal to the line BC, whilst plane P3 passes through B an is orthogonal to the line AC. Find the coordinates of r, the point of intersection of the three planes



3. The attempt at a solution
I have found the equation of plane P1 using the vector product of vectors AB and AC, but I can't find the equation of the line perpendicular to BC which passes through A to find plane P2.
Can anyone help?
Thanks

tiny-tim

hi cluivee! welcome to pf!

does the point of intersection lie in the same plane as ABC ?

cluivee

It must do surely, otherwise it wouldn't intersect all 3 planes?

tiny-tim

i don't following your reasoning , but yes, it must be in the plane of ABC

so doesn't that make it fairly easy to find?

HallsofIvy

1. Find the equation of plane p1.

2. Find the equations of plane p2 and p3.

3. Find the equation of the line of intersection of any two of those three planes.

4. Find the point at which that line intersects the third plane.

blackjack4401

Okay, I can't guarantee if I have understood the question perfectly, but here's what I came up with:

Plane 1
normal vector = CA x BA = (0,1,5) x (1,1,8) = (3,5,-1)

So the equation can be written as 3x+5y-z+k=0.

Since it passes point B, 3(1)+5(1)-(8)+k=0, so k=0. (3+5-8+k=0)
The equation I got is therefore 3x+5y-z=0.

Plane 2
This plane is perpendicular to the line passing points B and C as you mentioned. Therefore, the line joining B and C is a normal to this plane.

BC = C - B = (0,1,5) - (1,1,8) = (-1,0,-3) = normal vector of B

The equation can be written as -x-3z+k=0, or x+3z+k=0 if you'd like.
Since it also passes point A, (0)+3(0)+k=0, so k=0.
The equation I got is therefore x+3z=0.

Plane 3
This plane is perpendicular to the line passing A and C as you mentioned. Therefore, the vector AC is a normal to this plane.

AC = C - A = (0,1,5) - (0,0,0) = (0,1,5) = normal vector of C

The equation can be written as y+5z+k=0.
Since it also passes point B, (1)+5(8)+k=0, so k= -41 (1+40+k=0)
The equation I got is therefore y+5z-41=0

So let's solve for the variables now, given point r = (x,y,z).

(A) 3x+5y-z=0
(B) x+3z=0
(C) y+5z-41=0

I will eliminate 'z' as it appears nonzero in all three equations.

3(A)+(B)
(9x+15y-3z) + (x+3z) = 10x+15y=0 (1)

and

5(A)+(C)
(15x+25y-5z) + (y+5z-41) = 15x+26y-41=0 (2)

Solve for 'x' and 'y' given these two equations.
(1) 10x+15y=0
(2) 15x+26y-41=0

Eliminate x.

3(1) - 2(2)
(30x+45y) - (30x+52y-82) = 0
-7y+82=0

7y=82
y = 82/7

Substitute to either equation.
(1) 10x+15(82/7)=0
10x+1230/7=0
10x=-1230/7
x = -123/7

Substitute x and y into plane (A).

3(-123/7) + 5(82/7)- z = 0
-369/7 + 410/7 - z = 0
41/7 - z = 0
z = 41/7

Point 'r' is therefore r(-123/7,82/7,41/7).