Dot and cross product properties?

In summary: And that scalar is 5, so we can write it as\text{Proj}_\textbf{u}(\textbf{v} \times \textbf{w}) = 5 \left( \frac{\textbf{u}}{\textbf{u}\cdot \textbf{u}} \right) \cdot (\textbf{v} \times \textbf{w}).In summary, the conversation discusses the formula for finding the projection of one vector onto another vector in terms of the dot product. The formula is given as \frac{(\textbf{v} \times \textbf{w}) \cdot \textbf{u
  • #1
Justhanging
18
0
Knowing:

u * (v x w) = 5

where u, v and w are vectors.

What is

[Proj u (v x w)] * u

Ive been staring at the dot and cross product properties in my book for a while and I don't see how to do it.
 
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  • #2
I'm guessing what Proju means, but if it is the projection of the cross product onto u, then answer should be 5.
 
  • #3
Do you have - or can you think of - a formula which expresses the projection of one vector onto another in terms of the dot product?
 
  • #4
mathman said:
I'm guessing what Proju means, but if it is the projection of the cross product onto u, then answer should be 5.

Can you elaborate?
 
  • #5
Rasalhague said:
Do you have - or can you think of - a formula which expresses the projection of one vector onto another in terms of the dot product?

All i got is

proj u (v x w) = ((v x w) * u)/(mag(u)2) * u

But I can't do anything with the magnitude of u.
 
  • #6
You're nearly there. The projection is the vector

[tex]\frac{(\textbf{v} \times \textbf{w}) \cdot \textbf{u}}{\left \| \textbf{u} \right \|^2} \; \textbf{u},[/tex]

rather than the scalar

[tex]\frac{(\textbf{v} \times \textbf{w}) \cdot \textbf{u}}{\left \| \textbf{u} \right \|^2} \cdot \textbf{u}.[/tex]

This is why it makes sense to take the dot product of the projection with u. Now, have you got any formula at all for the magnitude or the squared magnitude? You want one that expresses it in terms of the dot product, but if you have another formula, you might be able to work out how to get from that to one involving the dot product.
 
  • #7
Rasalhague said:
You're nearly there. The projection is the vector

[tex]\frac{(\textbf{v} \times \textbf{w}) \cdot \textbf{u}}{\left \| \textbf{u} \right \|^2} \; \textbf{u},[/tex]

rather than the scalar

[tex]\frac{(\textbf{v} \times \textbf{w}) \cdot \textbf{u}}{\left \| \textbf{u} \right \|^2} \cdot \textbf{u}.[/tex]

This is why it makes sense to take the dot product of the projection with u. Now, have you got any formula at all for the magnitude or the squared magnitude? You want one that expresses it in terms of the dot product, but if you have another formula, you might be able to work out how to get from that to one involving the dot product.

mag(u)2 = u * u

Now I don't know if I can cancel some u's. I don't know what properties to apply here, my book doesn't mention anything about cancelling vectors like someone would cancel regular variables in algebra.
 
  • #8
Okay, great, now you have everything you need! That's a good instinct you have to be careful about cancelling; indeed we can't cancel vectors as we can number. But remember: the dot product of two vectors is just a number, so after taking the dot product of two vectors, we can do anything to the result that we can do to any other number.
 
  • #9
Rasalhague said:
Okay, great, now you have everything you need! That's a good instinct you have to be careful about cancelling; indeed we can't cancel vectors as we can number. But remember: the dot product of two vectors is just a number, so after taking the dot product of two vectors, we can do anything to the result that we can do to any other number.

Honestly am still confused. I see that the top of the fraction is five but I don't know what u * u is on the bottom. And I still don't understand how to incorporate the other u outside the fraction. Am missing something here... you say that I can start cancelling once I get numbers but I only got one number.
 
  • #10
Don't worry, Justhanging, I've spent plenty of time confused myself! I'm sure you'll understand it very soon. You know that

[tex]\textbf{u} \cdot (\textbf{v} \times \textbf{w}) = 5[/tex]

and that

[tex]\text{Proj}_\textbf{u}(\textbf{v} \times \textbf{w}) = \frac{\textbf{u} \cdot (\textbf{v} \times \textbf{w})}{\textbf{u}\cdot \textbf{u}} \; \textbf{u}.[/tex]

So

[tex]\text{Proj}_\textbf{u}(\textbf{v} \times \textbf{w}) \cdot \textbf{u} = \frac{\textbf{u} \cdot (\textbf{v} \times \textbf{w})}{\textbf{u}\cdot \textbf{u}} \; \textbf{u} \cdot \textbf{u}[/tex]

[tex]= \textbf{u} \cdot (\textbf{v} \times \textbf{w}) = 5.[/tex]

The number I canceled there was [itex]\textbf{u} \cdot \textbf{u}[/itex]. This is allowed, so long as [itex]\textbf{u}[/itex] is not the zero vector, because whatever it's length, when we divide that squared length by itself, the result will be 1.

[tex]\frac{\textbf{u} \cdot \textbf{u}}{\textbf{u} \cdot \textbf{u}} = \frac{\left \| \textbf{u} \right \|^2}{\left \| \textbf{u} \right \|^2}=1.[/tex]

Is that any clearer?
 
  • #11
Rasalhague said:
Don't worry, Justhanging, I've spent plenty of time confused myself! I'm sure you'll understand it very soon. You know that

[tex]\textbf{u} \cdot (\textbf{v} \times \textbf{w}) = 5[/tex]

and that

[tex]\text{Proj}_\textbf{u}(\textbf{v} \times \textbf{w}) = \frac{\textbf{u} \cdot (\textbf{v} \times \textbf{w})}{\textbf{u}\cdot \textbf{u}} \; \textbf{u}.[/tex]

So

[tex]\text{Proj}_\textbf{u}(\textbf{v} \times \textbf{w}) \cdot \textbf{u} = \frac{\textbf{u} \cdot (\textbf{v} \times \textbf{w})}{\textbf{u}\cdot \textbf{u}} \; \textbf{u} \cdot \textbf{u}[/tex]

[tex]= \textbf{u} \cdot (\textbf{v} \times \textbf{w}) = 5.[/tex]

The number I canceled there was [itex]\textbf{u} \cdot \textbf{u}[/itex]. This is allowed, so long as [itex]\textbf{u}[/itex] is not the zero vector, because whatever it's length, when we divide that squared length by itself, the result will be 1.

[tex]\frac{\textbf{u} \cdot \textbf{u}}{\textbf{u} \cdot \textbf{u}} = \frac{\left \| \textbf{u} \right \|^2}{\left \| \textbf{u} \right \|^2}=1.[/tex]

Is that any clearer?

Ahh yes I see now. One last thing to clarify, the fraction in this equation is a scaler so really this equation is saying that it is a scaler multplied by a vector. Is this correct?[tex]\text{Proj}_\textbf{u}(\textbf{v} \times \textbf{w}) = \frac{\textbf{u} \cdot (\textbf{v} \times \textbf{w})}{\textbf{u}\cdot \textbf{u}} \; \textbf{u}.[/tex]
 
  • #12
Yes, that's right. The u on the far right is a vector, the rest is a scalar.
 

Related to Dot and cross product properties?

1. What is the difference between the dot product and cross product?

The dot product is a scalar value obtained by multiplying the corresponding components of two vectors, while the cross product is a vector perpendicular to both of the original vectors.

2. What are the properties of the dot product?

The dot product is commutative, distributive, and follows the law of cosines. It is also equal to the product of the magnitude of the two vectors and the cosine of the angle between them.

3. What are the properties of the cross product?

The cross product is anti-commutative, distributive, and follows the right-hand rule. It is also equal to the product of the magnitudes of the two vectors and the sine of the angle between them.

4. How do I calculate the dot and cross product of two vectors?

The dot product can be calculated by multiplying the corresponding components of the two vectors and then adding them together. The cross product can be calculated using a determinant or by breaking it down into its component form and using the laws of determinants.

5. What are the applications of dot and cross products in science?

Dot and cross products are commonly used in physics and engineering to calculate work, torque, and magnetic fields. They are also used in computer graphics to determine lighting and shading in 3D models.

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