Orthogonal Projection Matrices for Points on Subspaces in R^3

[forty]

Find the matrices of the transformations T which orthogonally project a point (x,y,z) on to the following subspaces of R^3.

(a) The z-axis
(b) the straight line x=y=2z
(c) the plane x+y+z=0

(a) is easy just the matrix [0 0 0;0 0 0;0 0 1]

as for (b) and (c) i have no idea how to work them out. I think (b) might have something to do with projection of a vector on to another... ((u.v)/(|u|^2))u

So maybe v = (x,y,z) and u = a(1,1,2) (a is any real number)

But I'm really stuck on what to do

Any help like usual greatly appreciated :)

 

[HallsofIvy]

Find the matrices of the transformations T which orthogonally project a point (x,y,z) on to the following subspaces of R^3.

(a) The z-axis
(b) the straight line x=y=2z
(c) the plane x+y+z=0

(a) is easy just the matrix [0 0 0;0 0 0;0 0 1]

as for (b) and (c) i have no idea how to work them out. I think (b) might have something to do with projection of a vector on to another... ((u.v)/(|u|^2))u

So maybe v = (x,y,z) and u = a(1,1,2) (a is any real number)

But I'm really stuck on what to do

Any help like usual greatly appreciated :)

The simplest way to find the matrix of a linear transformation is to apply the transformation to each basis vector in turn. That gives the columns of the matrix.
For example, projecting $\vec{i}= (1, 0, 0)$ onto the z-axis gives (0, 0, 0). The first column of that matrix is [0 0 0].

Yes, it has everything to do with the projection of one vector onto another. One vector on the line x= y= 2z is (2, 2, 1) (take z= 1). The projection of (1, 0, 0) onto that is, using the formula you give, (2)/(3)(2,2,1)= (4/3, 4/3, 2/3). The first column of the matrix is [4/3, 4/3, 2/3]. The projection of (0, 1, 0) onto (2, 2, 1) is the same and the projection of (0, 0, 1) onto it is (1)/(3)(2, 2, 1)= (2/3, 2/3, 1/3).

To project onto a plane, project onto its normal line and then subtract. The normal vector of the plane x+ y+ z= 0 is (1, 1, 1). The projection of (1, 0, 0) onto that line is $(1)/\sqrt{3}(1, 1, 1)= (\sqrt{3},\sqrt{3},\sqrt{3})$ so the projection onto the plane is $(1, 0, 0)- (\sqrt{3},\sqrt{3},\sqrt{3})= (1- \sqrt{3}, -\sqrt{3}, - \sqrt{3})$. That is the first column of the matrix. Do the same with (0, 1, 0) and (0, 0, 1).

 

[forty]

Amazing! You always make things so simple, Thanks a lot :)

by the way you seem to have a 'thing' for linear algebra...