Finding solution for three sets of planes

[sushichan]

Homework Statement

(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

Homework Equations

Scalar triple product: n₁⋅(n₂ × n₃)

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)

 

[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.

 

[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.

 

[sushichan]

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.

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