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

  1. Sep 7, 2015 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations

    3. 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:

    nfinal = (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)?
  2. jcsd
  3. Sep 7, 2015 #2


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    Science Advisor
    Homework Helper
    Gold Member

    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.
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