Converting a Plane Equation into Vector Form

[Mentallic]

Homework Statement

How do I convert ax_1+bx_2+cx_3+d=0 into vector form?

The Attempt at a Solution

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

 

[HallsofIvy]

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".

 

[Mentallic]

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

 

[Mentallic]

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.

 

[Raskolnikov]

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

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

<br /> &lt;a,b,c&gt; \cdot &lt;x_1,x_2,x_3&gt; = ax_1+bx_2+cx_3.<br />

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}} = &lt;a,b,c&gt; is a normal vector to the plane and \vec{\textbf{x}} = &lt;x_1,x_2,x_3&gt; 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.