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

See attachmentAttachment 21168

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- Mar 16th 2011, 07:51 PMsderosa518Length 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 attachmentAttachment 21168 - Mar 16th 2011, 10:21 PMSudharaka
Dear sderosa518,

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

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

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

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

$\displaystyle \mid\underline{a}\mid=2.6$

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