Homework Help: Finding solution for three sets of planes

1. Apr 22, 2015

sushichan

1. The problem statement, all variables and given/known data
(I did not copy the problem statement, but basically solve the system of equations if there is solution and give a geometrical interpretation)

P1: 2x - y + 6z = 7
P2: 3x + 4y + 3z = -8
P3: x - 2y - 4z = 9

2. Relevant equations

Scalar triple product: n1(n2 × n3)

3. The attempt at a solution
(Step 1): Checked that they are not parallel

(Step 2): Checked that they are not coplanar

(Step 3): Find the unique solution
R2 - 3R3
⇒ 0x + 10y + 15z = -35
⇒ 6y + 9z = -21
R1 - 2R3
⇒ 0x + 3y + 14z = -11
⇒ 6y + 28z = -22

⇒ -21 - 9z = -22 - 28z
⇒ z = -1/19
⇒ y = -65/19
⇒ x = 37/19

(Edited: i double checked values for y & x)

Although the answer is that they intersect at (3, -5, 1)

Last edited: Apr 22, 2015
2. Apr 22, 2015

HallsofIvy

It looks to me like the part you left out (the actual problem statement) was the crucial part.

If the equations really are
P1: 2x - y + 6z = 7
P2: 3x + 4y + 3z = -8
P3: x - 2y - 4z = 9

Then x= 3, y= -5, z= 1 clearly is not the solution!
2(3)- (-5)+ 6(1)= 6+ 5+ 6= 17 NOT 7.

Since (3, -5, 1) do satisfy the other two equations, I suspect you have the first equation wrong.

3. Apr 22, 2015

haruspex

Did you try substituting the given answer in the three plane equations? I think you will discover a typo.

Edit: strange... On two threads, I see a post by Halls posted half an hour before mine that wasn't visible to me until half an hour after mine.

Last edited: Apr 22, 2015
4. Apr 22, 2015

sushichan

Thank you!!! I re-did the question where the equation for my first plane is 2x - y + 6z = 17 and I got the answer :D