Projection of a point

1. Nov 15, 2008

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 :)

2. Nov 16, 2008

HallsofIvy

Staff Emeritus
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).

3. Nov 16, 2008

forty

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

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