# Thread: Unit Vector Question

1. ## Unit Vector Question

Going back over textbook problems and having a bit of trouble with this one.
If v=ai+bj, show that a/((a^2+b^2)^1/2)=Cos theta & b/((a^2+b^2)^1/2)=Sin theta, where theta is the direction of v.
Sorry if the question is not written the best. Newbie to the site. Thanks for your help in advance.

2. Originally Posted by Joeda
Going back over textbook problems and having a bit of trouble with this one.
If v=ai+bj, show that a/((a^2+b^2)^1/2)=Cos theta & b/((a^2+b^2)^1/2)=Sin theta, where theta is the direction of v.
Sorry if the question is not written the best. Newbie to the site. Thanks for your help in advance.
I would suggest that you first draw a diagram.

The vector $\mathbf{v}=a\mathbf{i}+b\mathbf{j}$ has a component of length $a$ in the x direction, and a component of length $b$ in the y direction. The length of the hypotenuse of the triangle (which is the length of the vector) is given by $\parallel\!\mathbf{v}\!\parallel=\sqrt{a^2+b^2}$.

Now, assign the angle between the vector and the positive x axis the value $\theta$. Can you try to visualize what I described?

You should then be able to determine $\cos\theta$ and $\sin\theta$.

Does this make sense?

3. That does make sense thanks.
So to answer the question would I just have to draw the diagram and label all vectors?

4. Originally Posted by Joeda
That does make sense thanks.
So to answer the question would I just have to draw the diagram and label all vectors?
That isn't the answer itself. It leads you to the answer, which is to show that $\cos\theta=\frac{a}{\sqrt{a^2+b^2}}$ and $\sin\theta=\frac{b}{\sqrt{a^2+b^2}}$

5. ok so i think ive got it. Im using {v} as my symbol for magnitude of v.

v=ai+bj
{v}=((a^2+b^2)^1/2)

and u=v/{v}
=a/((a^2+b^2)^1/2)

and we have this equatioin: u=(cos theta)i + (sin theta)j

so how do I put it all together to prove the question?

6. Originally Posted by Joeda
ok so i think ive got it. Im using {v} as my symbol for magnitude of v.

v=ai+bj
{v}=((a^2+b^2)^1/2)

and u=v/{v}
=(ai+bj)/((a^2+b^2)^1/2)

and we have this equatioin: u=(cos theta)i + (sin theta)j

so how do I put it all together to prove the question?
I inserted missing information in your post in red.

Now break it up into $\frac{a}{\sqrt{a^2+b^2}}\mathbf{i}+\frac{b}{\sqrt{ a^2+b^2}}\mathbf{j}$ and then the result follows.