- Messages
- 438
- Reaction score
- 3,645
- Points
- 253
Ok nevermind that. If you want a plane that is perpendicular to both, then the line of intersection of the two planes must act as a perpedicular to the third plane correct?
basically want to find the equation of that line and then we can proceed from there. We have point A on the line with y coordinate of 2 so lets find its x and y components
x+2+3z=8....x+3z=6
2x-2(2)+z=3....2x+z=7
rearrage the first equation to make x the subject
x=6-3z
substitute x into the second equation
2(6-3z)+z=7
12-5z=7
z=1
x=3
so A: (3,2,1)
Thats one part down. Now we need to find ANY other point on that line so we can figure out its direction. Let us take y=3
x+3+3z=8....x+3z=5
2x-2(3)+z=3....2x+z=9
rearrage the first equation to make x the subject
x=5-3z
substitute x into the second equataion
2(5-3z)+z=9
10-5z=9
z=0.2
x=4.4
so the other point is (4.4,3,0.2)
Let us find the direction of the line then:
[4.4i+3j+0.2k]-[3i+2j+k]=[1.4i+j-0.8k] This is the normal to the plane in question
setting up the cartesian equation of the plane: 1.4x+y-0.8z=[1.4i+j-0.8k].[3i,2j,k]
1.4x+y-0.8z=5.4
1.75x+1.25y-z=6.75
7x+5y-4z=27
Thanks boyz