[Dot Product] Vector Proection

  • Thread starter Highway
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  • #1
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[Dot Product] Vector Projection

Homework Statement



OA1VY.jpg


Homework Equations



3Xrpu.jpg


b1Jy6.jpg


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.

X0lLw.jpg


JldYG.jpg


added this second part after plugging in for the definition of the projection we derived in class, then simplified. . .
 
Last edited:

Answers and Replies

  • #2
349
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anyone?
 
  • #3
I like Serena
Homework Helper
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Hi Highway! :smile:

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]?
 
  • #4
349
1
Hi Highway! :smile:

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

M0r1c.jpg
 
  • #5
I like Serena
Homework Helper
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Congrats! :wink:
 

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