Dot product / vector magnitudes

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Homework Help Overview

The discussion revolves around calculating the magnitude of the vector expression ||2V + W||, given the magnitudes of vectors V and W, the angle between them, and their dot product. The problem is situated within the context of vector mathematics, specifically focusing on vector magnitudes and dot products.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to find ||2V + W|| by first calculating ||2V + W||^2, leading to an expression involving dot products. Some participants question the correctness of the steps taken, particularly regarding the application of norms and the calculations of dot products.

Discussion Status

Participants are actively engaging with the problem, with some providing corrections to the original poster's calculations. There is a recognition of errors in the approach, and a few participants are clarifying the relationships between vector magnitudes and their dot products. However, there is no explicit consensus on the final outcome yet.

Contextual Notes

There are indications of confusion regarding the calculations of dot products and the application of norms, which may stem from the original poster's approach to the problem. The discussion highlights the importance of careful notation and understanding of vector properties.

pearss
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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)
 
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sorry, reviewing my post i found my mistake.
 
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.
 
pearss said:
...

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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