1. ## Cross Product Problem

Basically, I've drawn the diagram and considered the fact that the geometric definition of the cross product is: U x V = |u||v|sin(theta).

Given that the angle between them is pi/6, then sin(theta) is equal to 1/2. So, I see where they get the 1/2 in the answer. However, these are my questions:

1. I'm unsure how to relate the 1/2 to the vector (e3) perpendicular to the plane spanned by e1 and e2. How do they relate?

2. Does normal basis mean each of the vectors in the basis is a unit vector and that therefore there magnitude's are all 1? This would simplify the problem but i'm not sure if it's correct.

Any ideas would be appreciated and I will definitely click the little thank you button for you

2. ## Re: Cross Product Problem

Originally Posted by divinelogos
Basically, I've drawn the diagram and considered the fact that the geometric definition of the cross product is: U x V = |u||v|sin(theta).

Given that the angle between them is pi/6, then sin(theta) is equal to 1/2. So, I see where they get the 1/2 in the answer. However, these are my questions:

1. I'm unsure how to relate the 1/2 to the vector (e3) perpendicular to the plane spanned by e1 and e2. How do they relate?

2. Does normal basis mean each of the vectors in the basis is a unit vector and that therefore there magnitude's are all 1? This would simplify the problem but i'm not sure if it's correct.

Correct.

Any ideas would be appreciated and I will definitely click the little thank you button for you
Hi divinelogos,

If $\displaystyle \underline{a}\mbox{ and }\underline{b}$ are two vectors the cross product is defined by,

$\displaystyle \underline{a}\times\underline{b}=\underline{\hat{n }}|\underline{a}||\underline{b}|\sin\theta~;~\mbox { where }\theta\mbox{ is the angle between }\underline{a}\mbox{ and }\underline{b}~;~\underline{\hat{n}}\mbox{ is the unit vector perpendicular to }\underline{a}\mbox{ and }\underline{b}$

The direction of $\displaystyle \underline{\hat{n}}$ is given by the Right hand rule.

Therefore, $\displaystyle e_{1}\times e_{2}=\underline{\hat{n}}|e_{1}||e_{2}|\sin \frac{\pi}{6}=\frac{1}{2}\underline{\hat{n}}$

Since $\displaystyle e_{3}$ is perpendicular to $\displaystyle e_{1}\mbox{ and }e_{2}$;

$\displaystyle \underline{\hat{n}}=e_{3}$

$\displaystyle \therefore e_{1}\times e_{2}=\frac{1}{2}e_{3}$

It should be noted that, we have to assume that $\displaystyle \underline{\hat{n}}\neq -e_{3}$ to get the given answer.

3. ## Re: Cross Product Problem

Originally Posted by Sudharaka
Hi divinelogos,

If $\displaystyle \underline{a}\mbox{ and }\underline{b}$ are two vectors the cross product is defined by,

$\displaystyle \underline{a}\times\underline{b}=\underline{\hat{n }}|\underline{a}||\underline{b}|\sin\theta~;~\mbox { where }\theta\mbox{ is the angle between }\underline{a}\mbox{ and }\underline{b}~;~\underline{\hat{n}}\mbox{ is the unit vector perpendicular to }\underline{a}\mbox{ and }\underline{b}$

The direction of $\displaystyle \underline{\hat{n}}$ is given by the Right hand rule.

Therefore, $\displaystyle e_{1}\times e_{2}=\underline{\hat{n}}|e_{1}||e_{2}|\sin \frac{\pi}{6}=\frac{1}{2}\underline{\hat{n}}$

Since $\displaystyle e_{3}$ is perpendicular to $\displaystyle e_{1}\mbox{ and }e_{2}$;

$\displaystyle \underline{\hat{n}}=e_{3}$

$\displaystyle \therefore e_{1}\times e_{2}=\frac{1}{2}e_{3}$

It should be noted that, we have to assume that $\displaystyle \underline{\hat{n}}\neq -e_{3}$ to get the given answer.

Why can we assume n=e3 if n is a unit vector? Does "normalized basis" in the problem statement mean e1,e2, and e3 are unit vectors as well? If this is the case, your solution looks correct.

4. ## Re: Cross Product Problem

Originally Posted by divinelogos
Basically, I've drawn the diagram and considered the fact that the geometric definition of the cross product is: U x V = |u||v|sin(theta).
This is not true. The length of the vector $\displaystyle U\times V$ is $\displaystyle |U||V| sin(\theta)$

Given that the angle between them is pi/6, then sin(theta) is equal to 1/2. So, I see where they get the 1/2 in the answer. However, these are my questions:

1. I'm unsure how to relate the 1/2 to the vector (e3) perpendicular to the plane spanned by e1 and e2. How do they relate?

2. Does normal basis mean each of the vectors in the basis is a unit vector and that therefore there magnitude's are all 1? This would simplify the problem but i'm not sure if it's correct.
Yes, to "normalize" a vector means to divide by its length to get a vector in the same direction with length 1. A "normalized" basis is a basis of vectors, each of length 1. (An "orthonormal" basis would be a basis consisting of vectors,each of length 1, each at right angles to the others- but you are told that the distance between e1 and e2 so this is NOT an orthonormal vector.)

The length of the vector $\displaystyle e_1\times e_2$ would be $\displaystyle |e_1||e_2|sin(\theta)= (1)(1)(1/2)= 1/2$. Since the cross product of two vectors is perpendicular to both, in the "right hand rule" direction, and we are given that $\displaystyle e_3$ is perpendiculat to both $\displaystyle e_1$ and $\displaystyle e_2$, $\displaystyle e_1\times e_2$ is either $\displaystyle \frac{1}{2}e_3$ or $\displaystyle -\frac{1}{2}e_3$ but we don't have enough information to decide which. That would depend upon in which of two possible directions, perpendicular to the plane determined by the two vectors, $\displaystyle e_3$. They are apparently assuming that the three vectors form a "right hand frame"- that is, that if you curl the fingers of your right hand from $\displaystyle e_1$ to $\displaystyle e_2$, your thumb would be pointing in the direction of $\displaystyle e_3$ but I don't see that explicitely stated.

Any ideas would be appreciated and I will definitely click the little thank you button for you

5. ## Re: Cross Product Problem

Originally Posted by HallsofIvy
This is not true. The length of the vector $\displaystyle U\times V$ is $\displaystyle |U||V| sin(\theta)$
I just missed the normal vector on the end of the definition... my bad.

6. ## Re: Cross Product Problem

I understand now how to get e1xe2, but what about e2xe3 and e1xe3?

e1xe2 generates a vector perpendicular to the plane spanned by e1 and e2, but i'm not sure how they got 2e1-rad(3)e2?