# Length and direction of vector

• March 16th 2011, 08:51 PM
sderosa518
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 attachmentAttachment 21168
• March 16th 2011, 11:21 PM
Sudharaka
Quote:

Originally Posted by sderosa518
I just want to make sure my answer looks fine, but there is there a way to break it down.

See attachmentAttachment 21168

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