# Projection onto a Plane

#### ktcyper03

I am trying to project (1, 0, 0) onto the plane 6x - 3y - 2z = 0.

I guessed two additional vectors in order to complete the basis, (1, 2, 0) and (1, 0, 3), then I performed Gram-Schmidt to orthonormalize my basis. If my basis is X=(x1, x2, x3), I am confused because I have written in my notes to perform Gram-Schmidt on just x2 and x3, and not x1.

Anyways, I continued the problem as I have it in my notes. I calculated:
z1= [1/(square root 5)](1, 2, 0)
z2 = [1/(square root 245/25)](.8, -.4, 3)

Now I must find the inner product of (1, 0, 0) and z^i for i = 1,2... and this is the step I do not understand.

Next I have the Projection = Sum(inner product (1, 0, 0) , z^i))z^i.

Can someone walk me through the end of this problem? Thanks.

#### HallsofIvy

MHF Helper
Better would be to choose two vectors in the plane. 6x- 3y- 2z= 0 is the same as z= 3x- (3/2)y so that <x, y, z>= <x, y, 3x- (3/2)y>= x<1, 0, 3> + y<0, 1, -3/2> or, if you don't like fractions, x<1, 0, 3>- (y/2)<0, 2, -3> so that u= <1, 0, 3> and v= <0, 2, -3> form a basis for the subspace consisting of the plane.

I wouldn't bother with Gram-Schmidt, just project <1, 0, 0> onto each of those two vectors and add. (in fact, since <1, 0, 0>.<0, 2, -3>= 0, just project onto u!)

#### zzzoak

The straight line perpendicular to the plane is
r=(1,0,0)+(6,-3,-2)t.
Inserting this to the plane we get
t=-6/49
and projection point of vector (1,0,0) on the plane is
(1/49)(13,18,12).

#### HallsofIvy

MHF Helper
Please do not "hijack" someone else's thread for a completely different question. Also, your question has nothing to do with "Linear and Abstract Algebra". (Admittedly, neither did the original post in this thread!) Finally, there simply is not sufficient information given. There are no coordinate values in your picture.