# Convert into vector form

1. Jul 11, 2010

### Mentallic

1. The problem statement, all variables and given/known data
How do I convert $$ax_1+bx_2+cx_3+d=0$$ into vector form?

3. The attempt at a solution
I am completely at a loss here, mainly because I don't quite understand vector geometry.

2. Jul 11, 2010

### HallsofIvy

Re: Vectors

If we are to assume that "$x_1$", "$x_2$", and "$x_3$" are components of a vector then that equation would be written $<a, b, c> \cdot <x_1, x_2, x_3>+ d= 0$ where the first term is a "dot product".

3. Jul 11, 2010

### Mentallic

Re: Vectors

I better go read up on dot products then. Thanks.

4. Jul 11, 2010

### Mentallic

Re: Vectors

Before I go on, x1, x2 and x3 are just variables in 3 dimensions such as x,y,z. Not exactly sure if that is what you were assuming.

Ok so given the formula for a dot product of two vector a and b is $$|a||b|cos\theta$$ then we have $$\sqrt{(a^2+b^2+c^2)(x_1^2+x_2^2+x_3^2)}cos\theta+d=0$$

This doesn't seem right... I don't know how to find the angle between each vector and this isn't anywhere near the kind of answer I'm looking for, it should be of a form similar to this:

$$<x_1,x_2,x_3>=<0,0,d>+\lamda<a,0,0>$$

Although I'm possibly just using the dot product all wrong.

5. Jul 11, 2010

Re: Vectors

That's true; however, there's a much simpler definition of the dot product in this case:

$$<a,b,c> \cdot <x_1,x_2,x_3> = ax_1+bx_2+cx_3.$$

As an additional remark, note that, for a plane in $$R^3$$, we have the following:

$$\vec{\textbf{n}} \cdot \vec{\textbf{x}} = 0,$$ where $$\vec{\textbf{n}} = <a,b,c>$$ is a normal vector to the plane and $$\vec{\textbf{x}} = <x_1,x_2,x_3>$$ is any point on the plane. This is intuitive when we consider the definition of the dot product that you provided. The angle between any point on the plane and a corresponding normal vector is 90 degrees. Thus, $$\cos(\theta) = \cos(90) = 0.$$

I hope this helps.