# Thread: Vectors in 3 space

1. ## Vectors in 3 space

I have a quick conceptual question regarding vectors in 3-space. Say we are given vector A, and vector B, and we are told the dot product of said vectors is 0. We could deduce that they are at 90 degrees relative to each other. If the question takes this one step further and asks for a vector C that is also perpendicular to both A and B, we would merely take the cross product of AxB=C, and C would be the answer. If we were to multiply C by -1, would we also get another vector that is also at 90 degrees to those vectors? I would presume there would be two vectors at 90 degrees to A and B, one up and one down.

2. ## Re: Vectors in 3 space

Originally Posted by quantoembryo
I have a quick conceptual question regarding vectors in 3-space. Say we are given vector A, and vector B, and we are told the dot product of said vectors is 0. We could deduce that they are at 90 degrees relative to each other. If the question takes this one step further and asks for a vector C that is also perpendicular to both A and B, we would merely take the cross product of AxB=C, and C would be the answer.

This would only be true if the vectors AxB and C have the same magnitude and direction. That is,

$\displaystyle |A\times B|=|C|=|A||B|\sin\left(\frac{\pi}{2}\right)=|A||B|$ and $\displaystyle A,B,C$ must be a Right hand system of vectors.

If we were to multiply C by -1, would we also get another vector that is also at 90 degrees to those vectors? I would presume there would be two vectors at 90 degrees to A and B, one up and one down.
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3. ## Re: Vectors in 3 space

I thought that the cross product of two vectors automatically gave the result being perpendicular to the plane those vectors are in, and hence those vectors?

4. ## Re: Vectors in 3 space

Originally Posted by quantoembryo
I thought that the cross product of two vectors automatically gave the result being perpendicular to the plane those vectors are in, and hence those vectors?
There are infinitely many vectors perpendicular to the two vectors and the cross product gives one of them. I suggest you carefully refer the definition of the Cross Product.

5. ## Re: Vectors in 3 space

Okay, I see what you're saying. Lets assume we know the magnitude of the resulting product is 2. AxB=C, where |AxB|=2. Now would there be two vectors? Magnitude is always positive, so you could essentially multiply results by -1 to find the other vector in the opposite direction?

6. ## Re: Vectors in 3 space

Never mind, I think I get it now. With those given vector quantities, only one vector with a specific magnitude will result from them.

7. ## Re: Vectors in 3 space

Originally Posted by quantoembryo
Okay, I see what you're saying. Lets assume we know the magnitude of the resulting product is 2. AxB=C, where |AxB|=2. Now would there be two vectors? Magnitude is always positive, so you could essentially multiply results by -1 to find the other vector in the opposite direction?
No. The cross product of two vectors does not give two answers. Let $\displaystyle \underline{A}\mbox{ and }\underline{B}$ are two vectors with an angle $\displaystyle 0\leq\theta\leq\pi$ between them. Then the cross product is defined by,

$\displaystyle \underline{A}\times\underline{B}=\underline{\hat{n }}|\underline{A}||\underline{B}|\sin(\theta)$ where $\displaystyle \underline{\hat{n}}$ is unit vector whose direction is given by the Right hand rule.

The magnitude of $\displaystyle \underline{A}\times\underline{B}$ is; $\displaystyle |A\times B|=|\underline{A}||\underline{B}|\sin(\theta)$.

The direction of $\displaystyle \underline{A}\times\underline{B}$ is that of $\displaystyle \underline{\hat{n}}$.

Hope this clears all doubts.

8. ## Re: Vectors in 3 space

Sudharaka, I may be understanding quantumembryo's question differently than you because I would have simply said "yes, you are right". quantumembryo said " If the question takes this one step further and asks for a vector C that is also perpendicular to both A and B, we would merely take the cross product of AxB=C, and C would be the answer." I added the bold face. It is NOT necessary that A and B be at right angles to each other, $\displaystyle A\time B$ will be perpendicular to both. Also, the lengths of A and B are irrelevant because quantumembryo did not ask for any particular perpendicular, just a perpendicular.

quantumembryo said "Now would there be two vectors? Magnitude is always positive, so you could essentially multiply results by -1 to find the other vector in the opposite direction?" and sudharaka answered "No. The cross product of two vectors does not give two answers."

Well, $\displaystyle A\times B$ and $\displaystyle B\times A$ are two different vectors, one the negative of the other. I think that was what quantumembryo was referring to.

9. ## Re: Vectors in 3 space

Originally Posted by HallsofIvy
Sudharaka, I may be understanding quantumembryo's question differently than you because I would have simply said "yes, you are right". quantumembryo said " If the question takes this one step further and asks for a vector C that is also perpendicular to both A and B, we would merely take the cross product of AxB=C, and C would be the answer." I added the bold face. It is NOT necessary that A and B be at right angles to each other, $\displaystyle A\time B$ will be perpendicular to both. Also, the lengths of A and B are irrelevant because quantumembryo did not ask for any particular perpendicular, just a perpendicular.
Yes I think I had misunderstood the question as you have said. I thought that vector C is another particular vector perpendicular to both A and B. When I read it now I see that quantumembryo asks for a vector C perpendicular to both A and B.

quantumembryo said "Now would there be two vectors? Magnitude is always positive, so you could essentially multiply results by -1 to find the other vector in the opposite direction?" and sudharaka answered "No. The cross product of two vectors does not give two answers."

Well, $\displaystyle A\times B$ and $\displaystyle B\times A$ are two different vectors, one the negative of the other. I think that was what quantumembryo was referring to.
Yes again. I think I did not read the problem properly in this case. I thought he was referring to the cross product when he said, "Now would there be two vectors?"

Thanks for the clarification. I should learn to read more carefully.