# vector analysis zero dot product proof

• Aug 31st 2012, 08:29 AM
OneMileCrash
vector analysis zero dot product proof
My professor gave the following theorem today:

Let A and B be vectors. Then A * B =0 if and only if A and B are perpendicular.

Now, in the proof for this my professor gave the following reasoning in the case that A or B are the zero vector, in the "iff direction" of if 0 => perpendicular:

|A| = 0 or |B| = 0 => A=0 or B=0 => Give A are direction perp. To B or give B a direction perp to A => A is perp. To B.

I understand that we can attribute any direction to the zero vector, but does anyone else feel it is a bit sketchy to "choose" A to be perp. to B in order to prove that A is perp to B?

In other words.. the z ero vector's direction is anything. I don't have to "choose" it to be perpendicular to anything for the dot product to be zero. Therefore, as the theorem reads,"if and only if A and B are perpendicular" is a bit... not "wrong" but at the same time, could be more accurate.
• Aug 31st 2012, 10:19 AM
Plato
Re: vector analysis zero dot product proof
Quote:

Originally Posted by OneMileCrash
My professor gave the following theorem today:
Let A and B be vectors. Then A * B =0 if and only if A and B are perpendicular.
Now, in the proof for this my professor gave the following reasoning in the case that A or B are the zero vector, in the "iff direction" of if 0 => perpendicular: |A| = 0 or |B| = 0 => A=0 or B=0 => Give A are direction perp. To B or give B a direction perp to A => A is perp. To B. I understand that we can attribute any direction to the zero vector, but does anyone else feel it is a bit sketchy to "choose" A to be perp. to B in order to prove that A is perp to B? In other words.. the z ero vector's direction is anything. I don't have to "choose" it to be perpendicular to anything for the dot product to be zero. Therefore, as the theorem reads,"if and only if A and B are perpendicular" is a bit... not "wrong" but at the same time, could be more accurate.

Without knowing what text/set of notes your professor follows, it is very hard to comment.
But most Vector Calculus textbooks give that theorem as: Two non-zero vectors are perpendicular if and only if their scalar product(dot product) is zero. At the same time it is usual to define an angle between two non-zero vectors. Moreover, the scalar products is defined as $\displaystyle A\cdot B=\|A\|\|B\|\cos(\theta)$, where $\displaystyle \theta$ is the angle between vectors $\displaystyle A~\&~B$. Thus the question you have does not even come up.
• Aug 31st 2012, 02:41 PM
OneMileCrash
Re: vector analysis zero dot product proof
My question is about the proof given, not the dot product.

I just feel like if I wrote, in proving A is perpendicular to B, "Choose A to be perpendicular to B. Therefore A is perpendicular to B" the professor who taught me how to write proofs would chokeslam me.
• Aug 31st 2012, 02:54 PM
Plato
Re: vector analysis zero dot product proof
Quote:

Originally Posted by OneMileCrash
My question is about the proof given, not the dot product.
I just feel like if I wrote, in proving A is perpendicular to B, "Choose A to be perpendicular to B. Therefore A is perpendicular to B" the professor who taught me how to write proofs would chokeslam me.

Did you read my reply carefully? I think you missed the point.
There is no reason to even consider a zero vector in that theorem. I don't understand why your professor would do that.
• Aug 31st 2012, 02:56 PM
OneMileCrash
Re: vector analysis zero dot product proof
Ah, I see. I actually, after making my point suggested putting "nonzero" in the theorem.