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Dot product / vector magnitudes

  • Thread starter pearss
  • Start date
  • #1
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Homework Statement


||V|| = 2
||W|| = 3
angle between vectors = 120 degrees
V(dot)W = -3

Find ||2V+W||


Homework Equations


Cos120 = V(dot)W / ||V|| ||W||

V(dot)V = ||V|| ^2
W(dot)W = ||W|| ^2



The Attempt at a Solution




instead of solving for ||2V+W||

i instead try to solve for ||2V+W|| ^2

||(2V+W)(2V+W)||

= 4V(dot)V + 4V(dot)W + W(dot)W

=|| 4(2) + 4(-3) + 3||

i add these up and get -1

the book has an answer of 13^(1/2)
 

Answers and Replies

  • #2
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sorry, reviewing my post i found my mistake.
 
  • #3
fzero
Science Advisor
Homework Helper
Gold Member
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If [tex]|\mathbf{V}|=2[/tex], then [tex]\mathbf{V}\cdot \mathbf{V}=4[/tex], with a similar correction to [tex]\mathbf{W}\cdot \mathbf{W}=4[/tex]. Also, [tex]|2\mathbf{V}+\mathbf{W}|[/tex] will always be a positive number.
 
  • #4
VietDao29
Homework Helper
1,423
2
...

The Attempt at a Solution




instead of solving for ||2V+W||

i instead try to solve for ||2V+W|| ^2
So far, so good. :)

||(2V+W)(2V+W)||
This line is incorrect!!! There should be no norm sign there. You should note that:
[tex]\| \mathbf{v} \| ^ 2 = \mathbf{v} \cdot \mathbf{v}[/tex]

= 4V(dot)V + 4V(dot)W + W(dot)W

=|| 4(2) + 4(-3) + 3||
You seem to have forgotten to square it.

-------------------

Whoops... it seems that fzero beats me by 3 minutes. =.="
 

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