Finding the Equation of a plane

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In summary, to find the equation of the plane that passes through the line of intersection of the planes x-z=1 and y+2z=3 and is perpendicular to the plane x+y-2z=1, find the normal of the third plane, <1, 1, -2>, and use it in the equation A(x - 1) + B(y - 3) + Cz = 0, with the point (1, 3, 0) on the line of intersection substituted in for x and y. This will give you the equation 2x + 2y - z = 9.
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fireb
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Homework Statement


The plane that passes through the line of intersection of the plane x-z=1 and y+2z=3 and is perpendicular to the plane x+y-2z=1

Homework Equations


The Attempt at a Solution



I found the intersection point of the 2 planes by setting z=0 yielding the point (1,3,0). Next I found the normal of the 2 planes that is intersecting, (1,0,-1)x(0,1,2)= (1,-2,1). the normal of the perpendicular plane is (1,1,-2). since the 2 normal are not parallel, i found another normal (1,-2,1) x (1,1,-2) which yields(3,3,3). Then substituting the point (1,3,0) and (3,3,3) into the plane equation. So 3(x-1)+3(y-3)+3z=d

I am unsure if this is the correct answer or this is how you do it? It would be nice if someone would confirm this with me or lead me to the right direction on how to solve this.

Thanks in advance
 
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  • #2
fireb said:

Homework Statement


The plane that passes through the line of intersection of the plane x-z=1 and y+2z=3 and is perpendicular to the plane x+y-2z=1


The Attempt at a Solution



I found the intersection point of the 2 planes by setting z=0 yielding the point (1,3,0). Next I found the normal of the 2 planes that is intersecting, (1,0,-1)x(0,1,2)= (1,-2,1). the normal of the perpendicular plane is (1,1,-2). since the 2 normal are not parallel, i found another normal (1,-2,1) x (1,1,-2) which yields(3,3,3). Then substituting the point (1,3,0) and (3,3,3) into the plane equation. So 3(x-1)+3(y-3)+3z=d

I am unsure if this is the correct answer or this is how you do it? It would be nice if someone would confirm this with me or lead me to the right direction on how to solve this.

Thanks in advance
The equation you found is incorrect. The normal to this plane is <3, 3, 3>. In your other work, you found the normal to the plane in question to be <1, 1, -2>.

If the normal to a plane if n = <A, B, C> and a point on the plane is (x0, y0, z0), then the equation of the plane is A(x - x0) + B(y - y0) + C(z - z0) = 0.

The point you found, (1, 3, 0) is on the line of intersection of the two given planes, and the normal to the third plane is <1, 1, -2>. These two things are all you need to find the equation of the plane you want.
 

1. What is the equation of a plane?

The equation of a plane is a mathematical representation that describes the relationship between three variables in a 3-dimensional space. It is expressed in the form of Ax + By + Cz + D = 0, where A, B, and C are the coefficients of the variables x, y, and z, and D is a constant term.

2. How do you find the equation of a plane given three points?

To find the equation of a plane given three points, you can use the point-normal form of the equation which states that (x - x0, y - y0, z - z0) • n = 0, where (x0, y0, z0) is one of the points on the plane and n is the normal vector perpendicular to the plane. This equation can be rearranged to the standard form by solving for n and substituting it into the equation.

3. What is the normal vector of a plane?

The normal vector of a plane is a vector that is perpendicular to the plane and has a magnitude of 1. It is represented by n = (A, B, C), where A, B, and C are the coefficients of the variables in the equation of the plane. The normal vector is important in finding the equation of a plane as it helps determine the orientation of the plane in 3-dimensional space.

4. Can you find the equation of a plane using only two points?

No, you cannot find the equation of a plane using only two points. This is because there are an infinite number of planes that can pass through two given points. To determine the equation of a plane, you need at least three non-collinear points.

5. How can you use the cross product to find the equation of a plane?

The cross product can be used to find the normal vector of a plane, which is needed to find the equation of the plane. The cross product of two non-parallel vectors in the plane will result in a vector that is perpendicular to both of them. By taking the cross product of two vectors in the plane, you can obtain the normal vector and then use it to find the equation of the plane in standard form.

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