Find an equation of the plane that passes through the....

[space-time]

Homework Statement

Find an equation of the plane that passes through the line of intersection of the two planes 3x + 7y + 2z= 0 and -9x - 8y - 6z = 9, and is perpendicular to the plane -2x - 6y + 3z = -10.

Homework Equations

The Attempt at a Solution

Here is what I did:

I first took the two normal vectors (n₁ and n₂) of the first two planes mentioned.

n₁ = (3, 7, 2)
n₂ = (-9, -8, -6)

I then took the cross product of these two vectors:

n₁ × n₂ = (-26, 0, 39) We can call this vector V.

I then found the normal vector of the plane that is mentioned in the end of the problem (which I will denote as n₃)

n₃ = (-2, -6, 3)

Next I took the cross product between V and n₃.

V × n₃ = (234, 0, 156)

so then my normal vector for the plane that will be my solution is:

n_final = (234, 0, 156)

I then found a point on the line of intersection in the plane by first setting z=0 and then solving a system of equations for x and y. Here is the system:

3x + 7y = 0
-9x - 8y = 9

solving for x and y yields:

x = -189/117 , y= 9/13 (and of course z = 0)

Now that I have my point and my direction vector, I can find the equation of the plane. I got:

(234, 0, 156) ⋅ ( x + 189/117 , y - 9/13 , z) = 0

which expands to:

(234x +378) + (0 - 0) + 156z = 0

which simplifies to

234x + 378 + 156z =0

or simply

234x + 156z = -378

That was the final answer that I plugged into the software for the equation of the desired plane. There were two answers that the problem wanted: They wanted the normal vector of the desired plane and the equation of the desired plane.

When I plugged in my normal vector of (234, 0, 156) and my equation 234x + 156z = -378, the software said that I got the normal vector right, but got the equation wrong.

Why?! I don't know what I did wrong here. Furthermore, I worked another example of this type of problem that simply had different numbers and equations with a friend earlier, and I got it right using the exact same process I used here! I've also looked up other examples of this online and every other example used the exact same process!

Can someone please help me with this (because I am really ticked off and don't know what is wrong)?

 

[andrewkirk]

I don't understand why you are taking normal vectors to the first two planes, so I don't understand your approach.

Here's what I'd do:

1. Find the equation for the line L of intersection of the first two planes.

2. Identify two points P and Q on that line, eg by setting each of the two parameters to zero in turn.

3. Find the normal vector for the 3rd plane.

4. Add that normal vector to P to get a 3rd point R that is not on the line L. That plane must be in the target plane, since L is and the target plane is parallel to the normal vector.

5. Find the equation of the plane containing P, Q and R.