Plane through point and intersection of 2 other planes

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Homework Statement


Find the plane that passes through the point (-1,2,1) and contains the line of intersection of the planes:
x+y+z=2
2x-y+3z=1


The Attempt at a Solution



First, I know that I need to find the line of intersection of the 2 planes. To do this, I used the cross product of the normal vectors to get <4,-1,-3>. This is the direction vector, but now I don't know exactly what to do. I know I need to find a point in both planes, but I'm not really sure how. I found another post which will help after I find the intersection line.
 

Answers and Replies

  • #2
rock.freak667
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x+y+z=2
2x-y+3z=1


eliminate one variable (say z)

then put either x=0 or y=0
 
  • #3
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Homework Statement


Find the plane that passes through the point (-1,2,1) and contains the line of intersection of the planes:
x+y+z=2
2x-y+3z=1


The Attempt at a Solution



First, I know that I need to find the line of intersection of the 2 planes. To do this, I used the cross product of the normal vectors to get <4,-1,-3>. This is the direction vector, but now I don't know exactly what to do. I know I need to find a point in both planes, but I'm not really sure how. I found another post which will help after I find the intersection line.
One form of the equation of a line in space is called the symmetric form (I think that's what this form is called. A line in the direction of <a, b, c> and passing through the point (x1, y1, z1) can be described by these equations:
[tex]\frac{x - x_1}{a} ~=~ \frac{y - y_1}{b} ~=~\frac{z - z_1}{c} [/tex]
 
  • #4
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Good. You found the direction vector of the line. Now, because the plane contains the line, the dot product of the directional vector and the normal vector would be zero, because the angle between them is 900.

If the equation of plane is: Ax+By+Cz+D=0 then <A,B,C> is the vector normal to the plane.

<A,B,C>.<4,1,-3>=0

4A+B-3C=0

Also substitute for (-1,2,1) in the equation of the plane so that:

-A+2B+C+D=0
 
  • #5
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njama,
I follow most of your response. But what does 4A+B-3C=0 represent?
 
  • #6
35,028
6,774
Good. You found the direction vector of the line. Now, because the plane contains the line, the dot product of the directional vector and the normal vector would be zero, because the angle between them is 900.

If the equation of plane is: Ax+By+Cz+D=0 then <A,B,C> is the vector normal to the plane.

<A,B,C>.<4,1,-3>=0

4A+B-3C=0

Also substitute for (-1,2,1) in the equation of the plane so that:

-A+2B+C+D=0
Njama, please pay attention. The advice you give here is not helpful, since the OP is aware of this concept and has already done it to find the normals to the two planes he has.
 
  • #7
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musicmar said:
njama,
I follow most of your response. But what does 4A+B-3C=0 represent?

@musicmar the equation represents the dot product of the normal vector of the plane and the directional vector of the line.

Njama, please pay attention. The advice you give here is not helpful, since the OP is aware of this concept and has already done it to find the normals to the two planes he has.
@Mark44 I don't know what are you referring to. He only found the directional vector of the line. :confused:
 
  • #8
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@musicmar the equation represents the dot product of the normal vector of the plane and the directional vector of the line.


@Mark44 I don't know what are you referring to. He only found the directional vector of the line. :confused:
He found the direction vector for the line from the normals to the two planes, so he didn't need to "refind" the normals to the planes. In other words, he was way ahead of you. The OP crossed the normals to the two planesthem to find the direction of the line. He was given a point on the line, so all he needed to do was use the line's direction vector and the given point, either in the suggestion I gave in an earlier post or in some other way.
 
  • #9
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He found the direction vector for the line from the normals to the two planes, so he didn't need to "refind" the normals to the planes. In other words, he was way ahead of you. The OP crossed the normals to the two planesthem to find the direction of the line. He was given a point on the line, so all he needed to do was use the line's direction vector and the given point, either in the suggestion I gave in an earlier post or in some other way.
@Mark44 there was much easier way if I was asked.

Mark44 he need to find the equation of PLANE not LINE.

Just find two points on the line (even without finding the directional vector), and using the third point, you can easily make plane using the 3 points by solving the determinant:

[tex] \begin{vmatrix}x-x_1 & y-y_1 & z-z_1\\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-x_1 & z_3-z_1 \end{vmatrix}=0 [/tex]

I just didn't want to start from the beginning.
 
Last edited:
  • #10
35,028
6,774
Mea culpa. The OP clearly said he wanted the plane, not the line. I misinterpreted the problem statement.
Sorry,
Mark
 

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