Find equations of the plane (confirmation).

[NATURE.M]

Homework Statement

Find the vector, parametric and scalar equation of the plane passing through A(3, 5, 2), B(0, 5, −1) and C(1, 5, −3).

Homework Equations

Vector, scalar, and parametric equations of plane.

The Attempt at a Solution

For my solution:

vector equation:
[x,y,z]=[3,5,2]+s[-3,0,-3]+t[-2,0,-5], s,t εℝ (from this the parametric equations are quite intuitive.

i really would like to confirm the scalar equation i have for the plane:

-9y+45=0

 

[Mark44]

Homework Statement

Find the vector, parametric and scalar equation of the plane passing through A(3, 5, 2), B(0, 5, −1) and C(1, 5, −3).

Homework Equations

Vector, scalar, and parametric equations of plane.

The Attempt at a Solution

For my solution:

vector equation:
[x,y,z]=[3,5,2]+s[-3,0,-3]+t[-2,0,-5], s,t εℝ (from this the parametric equations are quite intuitive.

i really would like to confirm the scalar equation i have for the plane:

-9y+45=0

This is the same as y = 5, and your three given points satisfy that equation. If you want to convince yourself absolutely, form vectors AB and AC, and dot them with the normal to your plane, which happens to be <0, 5, 0> (or any multiple of this vector).

Edit: Make that <0, 1, 0> for a normal. Fortuanately for me, <0, 5, 0> is a scalar multiple of <0, 1, 0>.

 

[NATURE.M]

This is the same as y = 5, and your three given points satisfy that equation. If you want to convince yourself absolutely, form vectors AB and AC, and dot them with the normal to your plane, which happens to be <0, 5, 0> (or any multiple of this vector).

Thanks a lot. And for vectors AB and AC, the cross product I obtain from them is [0,-9,0], which is a scalar multiple of [0,5,0], so that makes sense.

 

[Mark44]

From the standard equation for a plane, Ax + By + Cz = D, you can pick off the coordinates of a normal to the plane: <A, B, C>.

For your plane, the standard equation is 0x + 1y + 0z = 5, so a normal would be <0, 1, 0>. I picked off the wrong number to get <0, 5, 0>, but as luck would have it, the two are multiples of each other.

 

[NATURE.M]

Okay, I just used the normal [0,-9,0] and the point A(3, 5, 2) to solve for the scalar equation, so 0x-9y+0z+d=0, substitute point A and we get y=5. But either way,the answer is the same.