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Thread: Precalculus Homework Help

  1. #1
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    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.
    Last edited by skeeter; Dec 27th 2017 at 09:58 AM.
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  2. #2
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    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$
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    Re: Precalculus Homework Help

    Quote Originally Posted by HallsofIvy View Post
    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.
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