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[Dot Product] Vector Proection

by Highway
Tags: product, proection, vector
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Highway
#1
Nov3-11, 12:05 PM
P: 349
1. The problem statement, all variables and given/known data



2. Relevant equations





3. The attempt at a solution

I am not sure what to do here -- I know that the projection of u onto a "dotted" with w = 0 by definition, but I don't know how to show this.





added this second part after plugging in for the definition of the projection we derived in class, then simplified. . .
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Highway
#2
Nov3-11, 03:43 PM
P: 349
anyone?
I like Serena
#3
Nov3-11, 04:25 PM
HW Helper
I like Serena's Avatar
P: 6,189
Hi Highway!

You have an expression for the projection.
Can you substitute that (and only that) in the formula you have for w?

To show that 2 vectors are orthogonal, you need to show that their dot product is zero. That is, that [itex]\vec a \cdot \vec w = 0[/itex].

What you need to know, is that there are calculation rules for dot products.
For instance [itex]\vec a \cdot (\vec b+\vec c) = \vec a \cdot \vec b + \vec a \cdot \vec c[/itex].

Can you simplify the expression for [itex]\vec a \cdot \vec w = 0[/itex]?

Highway
#4
Nov3-11, 04:45 PM
P: 349
[Dot Product] Vector Proection

Quote Quote by I like Serena View Post
Hi Highway!

You have an expression for the projection.
Can you substitute that (and only that) in the formula you have for w?

To show that 2 vectors are orthogonal, you need to show that their dot product is zero. That is, that [itex]\vec a \cdot \vec w = 0[/itex].

What you need to know, is that there are calculation rules for dot products.
For instance [itex]\vec a \cdot (\vec b+\vec c) = \vec a \cdot \vec b + \vec a \cdot \vec c[/itex].

Can you simplify the expression for [itex]\vec a \cdot \vec w = 0[/itex]?
Thanks!! I got it figured out :P

I like Serena
#5
Nov3-11, 04:58 PM
HW Helper
I like Serena's Avatar
P: 6,189
Congrats!


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