# The dot product

1. Nov 4, 2004

### bezgin

Let
n = (a,b,c)
v = (x,y,z)

Whatever the dot product of these vectors equal to, lets call d, the vector n is perpendicular to v. Again, I cannot stay calm and ask WHY?

If we call n = (a,b); v= (x,y) ==> ax+by= d and the slope of this line is -a/b whereas the slope of the vector n is b/a. Yes, they are perpendicular since -a/b * b/a = -1 (tanx * tan(90+x) = -1)
I can visualize and experiment it in 2-D but in 3-D I can't.

Also, how do we define a line that passes through the origin and, for instance (1,1,1). I have trouble transforming my logic from x-y plane to space.

2. Nov 4, 2004

### Muzza

Where did you get this faulty statement from? n and v are perpendicular if and only if their dot product is 0 (i.e. d = 0).

3. Nov 4, 2004

### HallsofIvy

Staff Emeritus
The line that passes throught (0,0,0) and (1,1,1) is given by:
x= t, y= t, z= t.

You know that the equations will be linear and taking t= 0 gives (0,0,0) while taking t= 1 gives (1,1,1).

In general: any straight line, in 3 dimensions, can be written x= at+ b, y= ct+ d, z= et+ f for some choice of a, b, c, d, e, f. If you are given two points, (x00,z0) and (x1,y1,z1), choose either one to be t= 0, the other t= 1 and plug the values into the equations. That gives 6 equations for the 6 letters.

If you understand that one, try this: what's the equation of the line passing through (0,0,0,0) and (1, 1, 1, 1) in 4 dimensions?

4. Nov 4, 2004

### bezgin

What I meant was, the vector (a,b,c) is always perpendicular to the plane ax+by+cz = d for any value of d. WHY?

5. Nov 4, 2004

### matt grime

What defines a plane? It is a point in the plane, p and the normal, n. Why? A point, x, is in the plane if and only if the displacement vector from p to x is zero, ie

n.(p-x)=0, or n.x=n.p

we set n.p = d.

6. Nov 4, 2004

### JonF

no.

First the dot product gives a scaler, not a vector
second, the two vectors are only perpendicular if their dot is 0

let u = n
let k = v - (<v,u>/||u||^2)*u

Where <> is the dot product, and ||u|| is u’s norm

u and k are perpendicular

Last edited: Nov 4, 2004