# Project vector onto a plane

1. Jul 7, 2013

### Niles

1. The problem statement, all variables and given/known data
Hi

Say I have a vector v=(vx, vy, vz) that makes an angle α with the x-axis. The amount of v projected onto the x-axis I find from the dot product

v.x = √(vx2 + vy2 + vz2)cos(α)

I am also interested in finding the amount of v in the plane normal to the x-axis, i.e. I want to project v onto the y-z plane. Is this simply given by v.(y + z)?

2. Jul 7, 2013

### HallsofIvy

Staff Emeritus
The components of a vector, v, at angle $\theta$ to a line, parallel and perpendicular to the line, form the legs of a right triangle having the vector as hypotenuse. Yes, the component parallel to the line is $|v|cos(\theta)$. And just as obviously, the component perpendicular to the line is $|v|sin(\theta))$.

3. Jul 7, 2013

### Niles

Thanks for your reply. OK, I took an example to test this out. My (unit) vector is $v = (0, v_y, v_z)/\sqrt{2}$ and the vector I want to project v onto is
$$u = (u_x, u_y, 0)$$
The projection v onto u is then (disregarding the √2...)
$$u_y = \sqrt{u_x^2 + u_y^2}\cos \theta$$
The perpendicular part is then given by $\sin\theta$, which must be given by (since sin2 + cos2 = 1...)
$$u_x = \sqrt{u_x^2 + u_y^2}\sin \theta$$

There is something I can't figure out - how can it be that there is no reference to any z-component in these expressions? I mean the above (perpendicular) sin-part only accounts for how much of vy is perpendicular to u, it does not include the fact that vz is also perpendicular. Is this something I have to manually add somehow? I hope you understand my question.

Last edited: Jul 7, 2013
4. Jul 8, 2013

### HallsofIvy

Staff Emeritus
Well, I avoided mentioning components at all for exactly that reason. if $\vec{u}= u_x\vec{ i}+ u_y\vec{j}+ u_z\vec{k}$ then its projection onto any line with which it makes angle $\theta$ is $|u|cos(\theta)= \sqrt{u_x^2+ u_y^2+ u_z^2} cos(\theta)$ and the length of the perpendicular is $|u|sin(\theta)= \sqrt{u_x^2+ u_y^2+ u_z^2} sin(\theta)$

Last edited: Jul 8, 2013
5. Jul 9, 2013

### Niles

OK, so just a generalization from 2D to 3D. Thanks for your help and time