Vector Projection?

Question

I'm trying to find the projection of the vector (1,1,1) onto the affine plane 2x+3y-z=12, but I am having some problems.

I know that the standard form of the vector should be written like:
1(2x-0)+1(3y-0)-1(z-0), but that's all I can figure out.

Thanks for your help.

Answer 1

Project the vector v = <1, 1, 1> onto the affine plane
2x + 3y - z = 12.
_________

Find the magnitude of the projection of v onto the plane.

The normal vector n, of the plane is n = <2, 3, -1>.

|| v || = √[1² + 1² + 1²] = √(1 + 1 + 1) = √3
|| n || = √[2² + 3² + (-1)²] = √(4 + 9 + 1) = √14

Take the dot product of v and n.

v • n = <1, 1, 1> • <2, 3, -1> = 2 + 3 - 1 = 4

Another definition of the dot product is:

v • n = || v || || n || cosθ
where θ is the angle between the two vectors

cosθ = (v • n) / (|| v || || n ||) = 4 / [(√3)(√14)] = 4/√42

sinθ = √[1 - cos²θ] = √[1 - 16/42] = √(26/42)

The magnitude of the projection of v onto the plane is:

Magnitude = || v || sinθ = (√3)(√(26/42) = √(26/14)
Magnitude = √(13/7)

Now find the direction of the vector u, of the projection.

n X v = <1, 1, 1> X <2, 3, -1> = <-4, 3, 1>

u = n X (n X v) = <2, 3, -1> X <-4, 3, 1> = <6, 2, 18>

Any positive multiple of u is also a directional vector. Divide by 2.

u = <3, 1, 9>

|| u || = √[3² + 1² + 9²] = √(9 + 1 + 81) = √90

We need to adjust the magnitude to √(13/7). Multiply by
√(13/7) / √90 = √(13/630).

u = [√(13/630)]<3, 1, 9>

Answer 2

Your answer depends on how the projection is carried out.

(Think of it as the sun shining on an object. Depending on what time of the day, the sun will shine from different directions so the shadow on the ground will correspondingly fall in different directions. Your plane is like the ground on which the shadow of the vector/object falls.)

Is the projection maybe orthogonal to the plane? Then it would make sense.