Finding an Equation of a Plane

[major_maths]

1. Find an equation for the plane which goes through the point (2, 2, -1) and which is parallel to the plane 2x-3y+7z = 100.

2. x = x₀+ta
y = y₀+ta
z = z₀+ta

3. First I found the parametric equations of a line parallel to the plane by using the vector
<2,-3,7> from the equation of the parallel plane and the point given:

x = 2+2t
y = 2-3t
z = -1+7t

And that's where I get lost. I think that I'm forgetting some equation, but I'm not sure.

 

[Mark44]

1. Find an equation for the plane which goes through the point (2, 2, -1) and which is parallel to the plane 2x-3y+7z = 100.

2. x = x₀+ta
y = y₀+ta
z = z₀+ta

3. First I found the parametric equations of a line parallel to the plane by using the vector
<2,-3,7> from the equation of the parallel plane and the point given:

x = 2+2t
y = 2-3t
z = -1+7t

And that's where I get lost. I think that I'm forgetting some equation, but I'm not sure.

Since the plane you're looking for is parallel to the plane 2x-3y+7z = 100, both planes have the same normal, which is <2, -3, 7>.

If you know a point P₀(x₀, y₀, z₀) on a plane and its normal N = <a, b, c>, you can find the equation of the plane by using the fact that the dot product of any vector in the plane with the normal to the plane has to be zero.

If P(x, y, z) is any point in the plane, a vector in the plane is P₀P = <x - x₀, y - y₀, z - z₀).

 

[major_maths]

Okay, so just to be clear, to get the vector in the plane I would take the dot product of <2, -3, 7> and a vector of variables, say <a ,b, c> and set it equal to 0. I would get a final equation of 2a-3b+7c = 0. And this would be the equation of a plane that goes through (2, 2, -1) and is parallel to the plane 2x-3y+7z = 100, correct?

 

[Mark44]

Okay, so just to be clear, to get the vector in the plane I would take the dot product of <2, -3, 7> and a vector of variables, say <a ,b, c> and set it equal to 0. I would get a final equation of 2a-3b+7c = 0. And this would be the equation of a plane that goes through (2, 2, -1) and is parallel to the plane 2x-3y+7z = 100, correct?

No.
Your vector <a, b, c> isn't in the plane. It extends from the origin to some point with coordinates (a, b, c).

Let P(x, y, z) be a point in your plane. The point P₀(2, 2, -1) is in your plane. Form the vector from P₀ to P, which is a vector in your plane. Take the dot product of the normal and the vector P₀P, and set it to zero. That will give you the equation of the plane.