Let's look at eq.3 again:
[tex]\vec{F}\cdot\vec{B}=0[/tex]
Now, we agree that this means that F is orthogonal to B.
Now, B is a fixed vector.
Do you agree then, that we can say that every vector which is orthogonal to B is a solution to (eq.3)?
(That is, "can play the role of F"?)
Let us now look a bit closer at the "solution set" of (eq.3)
(that is, the set of those vectors which are orthogonal to vector B).
Do you agree that:
a) One solution of (eq.3) is the choice of [tex]\vec{F}=\vec{0}[/tex]?
(That is, putting the zero vector into F's place in (eq.3) gives us 0 at the left-hand-side of (eq.3), and hence, that (eq.3) is fulfilled?
In other words, that the zero vector is a SOLUTION of (eq.3))
b) A plane is a geometric surface characterized by, that at all points on the plane, the vector normal is the same?
c) Hence, from b), that the solution set of (eq.3) is, in fact the plane containing the origin and with vector B as its vector normal?
Ok, so rewriting (eq.2) a bit, we have:
[tex]\vec{W}=\vec{F}+\vec{B}[/tex]
(You've seen this one before..)
We have already established, that all acceptable choices of [tex]\vec{F}[/tex] together constitutes a plane, with [tex]\vec{B}[/tex] as the normal vector.
Now then, if you combine this insight with the above equation, what sort of geometric surface must all acceptable choices for [tex]\vec{W}[/tex] taken together constitute?
It sure is!
Let us argue like this:
1) Take any acceptable choice of [tex]\vec{F}[/tex]
(Such a point lies in the plane containing the origin, with [tex]\vec{B}[/tex] as normal vector.
2) To find the corresponding [tex]\vec{W}[/tex] we go straight up along [tex]\vec{B}[/tex]
(Adding that is, [tex]\vec{B}[/tex] to our [tex]\vec{F}[/tex])
3). Now, the DISTANCE that [tex]\vec{W}[/tex] is removed from the plane in which [tex]\vec{F}[/tex] lies, is how far along the vector normal to the plane [tex]\vec{W}[/tex] is.
4) But this distance is simply the magnitude of [tex]\vec{B}[/tex], since [tex]\vec{B}[/tex] IS normal to the plane in which [tex]\vec{F}[/tex] lies.
5) Clearly, therefore, EVERY ACCEPTABLE CHOICE OF [tex]\vec{W}[/tex] LIES AN EQUAL DISTANCE FROM OUR PLANE (that is given by the length of [tex]\vec{B}[/tex]
6) But this means, that the set of acceptable [tex]\vec{W}[/tex] is A PLANE PARALLELL TO THE PLANE OF ACCEPTABLE [tex]\vec{F}[/tex]!!
1.Each acceptable choice of [tex]\vec{W}[/tex] corresponds to an acceptable choice of [tex]\vec{F}[/tex]
2. The distance of a point to a plane is found by measuring the length of the perpendicular (normal) joining that point to the plane.
3. In our case, the perpendicular to the plane in which our [tex]\vec{F}[/tex] lies, is given by [tex]\vec{B}[/tex]
yes, but this is due in another twenty minutes if you do not mind could you walk me through the steps. Another thing this is a physics assingment, but I placed it in the math due to it being math oriented. I took intro to linear algebra and I realize that in linear algebra it has to be more in dept, but for physics not as much. But I really appreciate your help, nothing like understanding a problem fully through.
But now we're really done!
Because:
1) We've shown that the set of acceptable [tex]\vec{W}[/tex] is a plane.
2) In addition, since [tex]\vec{F}=\vec{0}[/tex] was a solution of (eq.3), we know that one acceptable [tex]\vec{W}=\vec{0}+\vec{B}=\vec{B}[/tex]
But the physical point corresponding to [tex]\vec{B}[/tex] is D..
This was what you had to show:
That your original equation represents a plane with normal vector given by [tex]\vec{B}[/tex] containing D.