Is A always equal to zero if it is perpendicular to every vector X?

[Jbreezy]

Homework Statement

Let A be a vector perpendicular to every vector X. Show that A = O Edit: it is O not 0. (OH not zero) ha

Homework Equations

So, we know if A and X are perpendicular then A(dot)X = 0
I see no reason why A would have to be equal to 0.
Could X (not equal) 0? Could it be both?

The Attempt at a Solution

Above is my attempt. I get a hint please?

 

[1MileCrash]

Think about the trigonometric definition of the dot product.

 

[LCKurtz]

Homework Statement

Let A be a vector perpendicular to every vector X. Show that A = O Edit: it is O not 0. (OH not zero) ha

Homework Equations

So, we know if A and X are perpendicular then A(dot)X = 0
I see no reason why A would have to be equal to 0.
Could X (not equal) 0? Could it be both?

The Attempt at a Solution

Above is my attempt. I get a hint please?

You might start by telling us what vector space your vectors are from. Is it ${\mathbb R}^n$? What happens if you dot A with the standard basis vectors?

 

[Dick]

If a vector A is perpendicular to EVERY vector, it must be perpendicular to itself. So?

 

[Simon Bridge]

... assuming A is in the same space as X.
X could be a set of vectors lying in a plane.

But I agree - they key to this is to realize that that A must be perpendicular to EVERY vector.

 

[Jbreezy]

You might start by telling us what vector space your vectors are from. Is it ${\mathbb R}^n$? What happens if you dot A with the standard basis vectors?

I wrote the questions exactly as is. Doesn't say anything about the vector space. If I dot A with the standard basis vectors then...I'm not sure because all that I'm told is that A is perpendicular to X I do not know anything else.

If a vector A is perpendicular to EVERY vector, it must be perpendicular to itself. So?

How can a vector be perpendicular to itself?

 

[1MileCrash]

I wrote the questions exactly as is. Doesn't say anything about the vector space. If I dot A with the standard vectors then...I'm not sure because all that I'm told is that A is perpendicular to X I do not know anything else.

You are told that A is perpendicular to every vector. Not just some vector.

How can a vector be perpendicular to itself?

There is only one way, and it proves your result.

|A||X|cos(theta) = A (dot) B

Is the trigonometric definition.

If A is perpendicular to all vectors, it is perpendicular to itself. All vectors means all vectors, A is a vector, therefore A is perpendicular to A. Therefore

|A||X|cos(theta) = 0

Even whenever X = A.

What is the angle between X and A if X=A?

How can this equation always be 0 (perpendicular to any vector) if I allow one magnitude and theta to vary freely?

 

[Jbreezy]

Only if A = 0. But wouldn't theta not vary freely if you are told it is perpindicular then the angle must always be 90 degree.
So why do you say theta varies freely. It would seem to me that X would vary and theta would always be 90.

 

[Jbreezy]

How about this...can you just say ..The only way that A can be perpindicular to it's self is if it is the 0 vector?

 

[1MileCrash]

But wouldn't theta not vary freely if you are told it is perpindicular then the angle must always be 90 degree.

No.

Two vectors are perpendicular if and only if their dot product vanishes. That is the definition. If I tell you that two vectors are perpendicular, then you can tell me that their dot product vanishes, and that is all. The dot product of O (dot) X = 0 for any X, it doesn't matter what I call the angle between O and X, nor can I say what it is (because it is arbitrary).

That two perpendicular nonzero vectors have an angle of 90* is a direct corollary to that definition. It has no effect on this proof, because we aren't explicitly talking about nonzero vectors.

If you do not let theta vary freely then you are not considering every vector. The theorem you are proving states that X is perpendicular to every vector, not just ones that make an angle of 90* with it.

How about this...can you just say ..The only way that A can be perpindicular to it's self is if it is the 0 vector?

You have to show why that assertion is true.

 

[Jbreezy]

What is the angle between X and A if X=A?

0?

So

If you do not let theta vary freely then you are not considering every vector. The theorem you are proving states that X is perpendicular to every vector, not just ones that make an angle of 90* with it.



How about this...can you just say ..The only way that A can be perpindicular to it's self is if it is the 0 vector?

You have to show why that assertion is true.

So just from the trig defination like you say the only way that it can be 0 is if A is 0 because X and theta vary.
I don't know how to show that the only way to be perpindiuclar to it's self is if it is 0.
Your proof is mostly words like stating conditons not really like showing it. (you are though if you get what I mean)
So I suppose that saying A can only be perpindicular to itself if it is the zero vector would be sort of the same. I can't think of it yet.
Thanks

 

[Dick]

0?

So

So just from the trig defination like you say the only way that it can be 0 is if A is 0 because X and theta vary.
I don't know how to show that the only way to be perpindiuclar to it's self is if it is 0.
Your proof is mostly words like stating conditons not really like showing it. (you are though if you get what I mean)
So I suppose that saying A can only be perpindicular to itself if it is the zero vector would be sort of the same. I can't think of it yet.
Thanks

What's your definition of 'dot product'?

 

[1MileCrash]

I don't know how to show that the only way to be perpindiuclar to it's self is if it is 0.

If A is perpendicular to B, then

|A||B|cos(theta) = 0

Consider the case where A = B.If I say anything else, I've done it for you completely.

 

[Jbreezy]

X dot A = |X||A|cos(theta)

 

[1MileCrash]

X dot A = |X||A|cos(theta)

Yes, so now write that for X (dot) X and show, algebraically, that |X| is 0 if the dot product is 0.

 

[Jbreezy]

Is just
A dot A = |A||A|cos(theta)
A dot A = |A||A|cos(0)
cos(0) = 1 so A must be 0
I don't like this
?

 

[1MileCrash]

You need to set the equation equal to 0.

|A||A|cos(0)=0
|A|^2 =0
|A|=0

What do you mean you don't like it? It concisely proves the theorem by literally, simplifying and dividing once. Itdoesn't get much cleaner than that.

 

[1MileCrash]

"Show that if A is perpendicular to all vectors, A is the 0 vector"

Proof:

If A is perpendicular to all vectors, then it must be perpendicular to itself, thus

A(dot)A= 0
|A||A|cos(0) = 0
|A|^2 = 0
|A|=0

Since the magnitude of A is 0, A is the zero vector.

That's all there is to it.

 

[Jbreezy]

What is wrong iwth how I said it?

 

[1MileCrash]

What is wrong iwth how I said it?

You didn't set the equation equal to 0, all you said was that A(dot)A = |A||A| and then asserted that A was 0. Reading that, no one would know what you were talking about. You can't solve for A in the equation A(dot)A = |A||A| because it is tautological.

 

[Jbreezy]

Yes they would because I said cos(0) =1 so if it is 1 then A must be 0 for it to be 0
Thanks,

 

[1MileCrash]

Yes they would because I said cos(0) =1 so if it is 1 then A must be 0 for it to be 0
Thanks,

No, they wouldn't, because you never mentioned that the dot product is equal to 0 in the first place.

Is just
A dot A = |A||A|cos(theta)
A dot A = |A||A|cos(0)
cos(0) = 1 so A must be 0
I don't like this
?

No where did you say that A dot A is 0. You need to say that the equation is equal to 0 in order to solve for A.

What you wrote doesn't say anything about A. You said it must be O, but based on what you wrote, it doesn't have to be O. A dot A = |A||A|cos(0) is true for any vector A, even one that is not O, and that is what you wrote.

A dot A = |A||A|cos(0) = 0

Is only true for A = O, so that is what you must use. Without setting it equal to 0, you aren't saying anything.

 

[Jbreezy]

Opps.
Thanks