# Perpendicular Vector

1. Oct 9, 2004

### Spectre32

If you had two vectors, and you wanted to find a vector perpendicular to those useing th dot product, what would be needed to be done. I alreaded Doted A *dot* B and have a vector. I'm just stuck on the last part

2. Oct 9, 2004

### faust9

let the unknow vector have variables u=xi+yj+zk.

The dot product of perpendicular vectors is zero so you dot vectors a and b with U and set the results equal to zero.

You'll have two equations with three unknows which is expected because the component of U perpendicular the a and b can be any length you want it you be.
lets say a=(1,2,3) and b=(3,5,7)

$$x1+y2+z3=0$$
$$x3+y5+z7=0$$

I can choose to solve for x and y I get:

z=-1/2y
z=-x

Now I can let z be anything I want--I'll choose -2 to elliminate the -1/2.

x=2
y=1
z=-2

thus u=(2,1,-2) is perp to a and b.

I did the above in my head so it might not be right... The process is though. Check the above to see if I did it correctly.

Good luck.

3. Oct 9, 2004

### Spectre32

yeah it looks ok.. I see whats going on now... Soo was me figureing out the dot product useless? Or can i use that as my x1 + y2 +z3 = ?"

4. Oct 10, 2004

### eddo

Just as a sidenote, another way to approach this problem if you don't "have to" use dot products, is to use the cross product. This works because the cross product of two vectors is perpendicular to both. The vector you get as an answer can than me multiplied by any scalar to make the answer look neater, although this isn't necessary.

5. Oct 10, 2004

### Spectre32

Yeah I took notice to that, but i had doing these when I wasn't funny aware that a x b if noy equal to 0 means this.

6. Oct 10, 2004

### faust9

This is only true for 3D space. You can't use the cross product in spaces greater that 3D or less than 3D.