Unit Vector Question

**Joeda** · January 31st 2009, 09:18 PM

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.

**Chris L T521** · January 31st 2009, 09:30 PM

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?

**Joeda** · January 31st 2009, 09:42 PM

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

**Chris L T521** · January 31st 2009, 09:44 PM

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 $\sin\theta=\frac{b}{\sqrt{a^2+b^2}}$

**Joeda** · January 31st 2009, 10:00 PM

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?

**Chris L T521** · January 31st 2009, 10:03 PM

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.

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