Solving Vectors Questions: Cosine & Sine Laws

[stunner5000pt]

Homework Statement

As in the diagram.

Homework Equations

Cosine law
sine law

The Attempt at a Solution

if we do attempt the magnitudes of u and w and the angle between them 60 degrees, I can solve for the opposite side where v is supposed to be.
I calculated
|k \vec{v}|^2 = |u|^2 + |w|^2 - 2|u||w| \cos 60
when I calculate the above I got the value of the right side as square root 52
\sqrt{52} = \frac{\sqrt{52}}{7} |\vec{v}|

Would this mean that to get w as a linear combination of u and v, we simply can write

\vec{w} = \vec{u} - \frac{\sqrt{52}}{7} \vec{v}

is that correct?

Thanks for your help.

 

[verty]

Think of this as a trig question: what are the component vectors of $\vec{w}$? Go from there.

 

[stunner5000pt]

Think of this as a trig question: what are the component vectors of $\vec{w}$? Go from there.

w_{x} = 8 \cos 60 \hat{i}
w_{y} = 8 \sin 60 \hat{j}

So of course x along the horizontal so
w_{x} = |u| - |v| \cos \theta
w_{y} = - |v| \sin \theta

But this will not yield exact answers when solved

 

[stunner5000pt]

w_{x} = 8 \cos 60 \hat{i}
w_{y} = 8 \sin 60 \hat{j}

So of course x along the horizontal so
w_{x} = |u| - |v| \cos \theta
w_{y} = - |v| \sin \theta

But this will not yield exact answers when solved

PS are we to assume that u and v are perpendicular? That would remove the need for the above

 

[haruspex]

The diagram is a little rough, but it looks as though it is supposed to be a closed triangle. If so, the given data (angle and magnitudes) are irrelevant.

 

[verty]

w_{x} = 8 \cos 60 \hat{i}
w_{y} = 8 \sin 60 \hat{j}

So of course x along the horizontal so
w_{x} = |u| - |v| \cos \theta
w_{y} = - |v| \sin \theta

But this will not yield exact answers when solved

You have assumed that $\vec{u}$ is parallel to $\hat{j}$ and has the same orientation, but you shouldn't make these assumptions. However, the assumption that $\vec{u}$ and $\vec{v}$ are perpendicular is necessary to make any sense of the question, I think.