# Thread: Precalculus Homework Help

1. ## Re: Precalculus Homework Help

for vector $\left<a,b\right>$, the unit vector is $\left< \dfrac{a}{\sqrt{a^2+b^2}} ,\dfrac{b}{\sqrt{a^2+b^2}} \right>$

oh, and refrain from including links to pay sites like "studydaddy", please.

2. ## Re: Precalculus Homework Help

I thought I had already responded to this.

First, "the unit vector to <a,b>" means the unit vector in the same direction as <a,b>.

Taking $\theta$ to be the angle the vector <a, b> makes with the positive x-axis, $tan(\theta)= \frac{b}{a}$

But also $\frac{\frac{b}{\sqrt{a^2+b^2}}}{\frac{a}{\sqrt{a^ 2+b^2}}}= \frac{b}{a}$

In order that this be a unit vector, its length must be 1.

$\sqrt{\left(\frac{a}{\sqrt{a^2+b^2}}\right)^2+ \left(\frac{b}{\sqrt{a^2+b^2}}\right)^2}$$= \sqrt{\frac{a^2}{a^2+b^2}+ \frac{b^2}{a^2+b^2}}= \sqrt{\frac{a^2+b^2}{a^2+b^2}}= 1$

3. ## Re: Precalculus Homework Help

Originally Posted by HallsofIvy
I thought I had already responded to this.
These folks don't care about the problems or answers. They just want you to click on study daddy . com.