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Math Help - Length and direction of vector

  1. #1
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    Length and direction of vector

    I just want to make sure my answer looks fine, but there is there a way to break it down.

    See attachmentMath1.rtf
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  2. #2
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    Quote Originally Posted by sderosa518 View Post
    I just want to make sure my answer looks fine, but there is there a way to break it down.

    See attachmentMath1.rtf
    Dear sderosa518,

    If you have a vector \underline{a} the direction is represented by the unit vector along that direction. That is, \displaystyle\hat{n}=\frac{\underline{a}}{\mid\und  erline{a}\mid} where \mid\underline{a}\mid is the magnitude of the vector.

    Therefore, \displaystyle\underline{a}=\mid\underline{a}\mid\f  rac{\underline{a}}{\mid\underline{a}\mid}

    In your case, \underline{a}=\frac{3}{5}i-\frac{12}{5}j+\frac{4}{5}k

    \displaystyle\hat{n}=\frac{\frac{3}{5}i-\frac{12}{5}j+\frac{4}{5}k}{\sqrt{\left(\frac{3}{5  }\right)^2+\left(\frac{12}{5}\right)^2+\left(\frac  {4}{5}\right)^2}}=\frac{3}{13}i-\frac{12}{13}j+\frac{4}{13}k

    \mid\underline{a}\mid=2.6

    Therefore, \underline{a}=2.6\left(\frac{3}{13}i-\frac{12}{13}j+\frac{4}{13}k\right)
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