Finding an equation for a plane

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In summary, the conversation discusses finding an equation for a plane that contains a given line and is parallel to the line of intersection of two other planes. The answer is provided in an attachment, but the asker has a question about a step in the solution process. The expert explains that the step in question appears to be a typo and suggests trying to fix it to see if it resolves the issue.
  • #1
pvpkillerx
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Find an equation for the plane that contains the line x = 1+t, y = 3t, z = 2t and is parallel to the line of intersection of the planes -x+2y+z = 0 and x + z + 1 = 0.

On the attachment, the answer is there. My question is, (on the first page) right after he wrote, x,y, leading, z free, he wrote down the matrix form of the equation. Where did he get the values of the matrix right after z. The values are -1, -1 and 1. How did he find them, or where did he get them? Please explain, thanks!
 

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  • #2
If you write out the equations that step represents they are:

x + z = -1
y + z = -1/2 - z
z = z

Solve for (x,y,z) written as a column vector.
 
  • #3
Wait, how exactly do you get that though? I see how you get x + z = -1, but how do you get y + z = -1/2 -z? I also understand z = z. Can you explain the second line? thanks.
 
  • #4
LCKurtz said:
If you write out the equations that step represents they are:

x + z = -1
y + z = -1/2 - z
z = z

Solve for (x,y,z) written as a column vector.

pvpkillerx said:
Wait, how exactly do you get that though? I see how you get x + z = -1, but how do you get y + z = -1/2 -z? I also understand z = z. Can you explain the second line? thanks.

Isn't it obvious to you that it is just a typo? Did you try fixing it to see if it answered your question?
 

FAQ: Finding an equation for a plane

How do you find the equation for a plane?

The equation for a plane can be found using the point-normal form, which is A(x-x0) + B(y-y0) + C(z-z0) = 0, where A, B, and C are the direction numbers of the normal vector to the plane, and (x0, y0, z0) is a point on the plane.

What is the normal vector of a plane?

The normal vector of a plane is a vector that is perpendicular to the plane. It is represented by the direction numbers (A, B, C) in the point-normal form of the plane's equation.

How many points are needed to determine a plane?

Three non-collinear points are needed to determine a unique plane. These points can be used to find the normal vector and then the equation of the plane.

Can the equation of a plane be written in different forms?

Yes, the equation of a plane can also be written in the standard form, which is Ax + By + Cz = D, where A, B, and C are the direction numbers of the normal vector, and D is a constant. It can also be written in the intercept form, which is x/a + y/b + z/c = 1, where a, b, and c are the x-, y-, and z-intercepts of the plane.

Can the equation of a plane be used to find the distance between a point and the plane?

Yes, the distance between a point and a plane can be found using the formula d = |Ax0 + By0 + Cz0 - D| / √(A2 + B2 + C2), where (x0, y0, z0) is the coordinates of the point and Ax + By + Cz = D is the equation of the plane.

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