Dot and cross product properties?

[Justhanging]

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.

 

[mathman]

I'm guessing what Proj_u means, but if it is the projection of the cross product onto u, then answer should be 5.

 

[Rasalhague]

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?

 

[Justhanging]

I'm guessing what Proj_u means, but if it is the projection of the cross product onto u, then answer should be 5.

Can you elaborate?

 

[Justhanging]

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.

 

[Rasalhague]

You're nearly there. The projection is the vector

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

rather than the scalar

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

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.

 

[Justhanging]

You're nearly there. The projection is the vector

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

rather than the scalar

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

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.

 

[Rasalhague]

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.

 

[Justhanging]

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.

 

[Rasalhague]

Don't worry, Justhanging, I've spent plenty of time confused myself! I'm sure you'll understand it very soon. You know that

\textbf{u} \cdot (\textbf{v} \times \textbf{w}) = 5

and that

\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}.

So

\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}

= \textbf{u} \cdot (\textbf{v} \times \textbf{w}) = 5.

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

\frac{\textbf{u} \cdot \textbf{u}}{\textbf{u} \cdot \textbf{u}} = \frac{\left \| \textbf{u} \right \|^2}{\left \| \textbf{u} \right \|^2}=1.

Is that any clearer?

 

[Justhanging]

Don't worry, Justhanging, I've spent plenty of time confused myself! I'm sure you'll understand it very soon. You know that

\textbf{u} \cdot (\textbf{v} \times \textbf{w}) = 5

and that

\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}.

So

\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}

= \textbf{u} \cdot (\textbf{v} \times \textbf{w}) = 5.

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

\frac{\textbf{u} \cdot \textbf{u}}{\textbf{u} \cdot \textbf{u}} = \frac{\left \| \textbf{u} \right \|^2}{\left \| \textbf{u} \right \|^2}=1.

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?\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}.

 

[Rasalhague]

Yes, that's right. The u on the far right is a vector, the rest is a scalar.