Dot product / vector magnitudes

[pearss]

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)

 

[pearss]

sorry, reviewing my post i found my mistake.

 

[fzero]

If |\mathbf{V}|=2, then \mathbf{V}\cdot \mathbf{V}=4, with a similar correction to \mathbf{W}\cdot \mathbf{W}=4. Also, |2\mathbf{V}+\mathbf{W}| will always be a positive number.

 

[VietDao29]

...

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:
\| \mathbf{v} \| ^ 2 = \mathbf{v} \cdot \mathbf{v}

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

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

You seem to have forgotten to square it.

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