# How to find a vector A that has the same direction as ⟨−8,7,8⟩ but has length 3 ?

Oct 7, 2015

$\left(- \frac{24}{\sqrt{177}} , \frac{21}{\sqrt{177}} , \frac{24}{\sqrt{177}}\right)$

#### Explanation:

The idea is based on a concept of scaling and similarity.
Any vector that "has the same direction" as $\left(- 8 , 7 , 8\right)$ has all the coordinates proportional to this given vector and, therefore, can be described by coordinates $\left(- 8 f , 7 f , 8 f\right)$ where $f$ is a scaling factor.

All we need now is to find a scaling factor that leads to a vector with the length $3$.

The length of a vector with coordinates $\left(- 8 f , 7 f , 8 f\right)$ equals to
$\sqrt{64 {f}^{2} + 49 {f}^{2} + 64 {f}^{2}} = f \cdot \sqrt{177}$

So, if we want the length to be equal to $3$, we should choose
$f = \frac{3}{\sqrt{177}}$

The coordinates of a vector with the same direction as $\left(- 8 , 7 , 8\right)$ but with the length $3$ will be
$\left(- \frac{24}{\sqrt{177}} , \frac{21}{\sqrt{177}} , \frac{24}{\sqrt{177}}\right)$

Oct 7, 2015

${\overline{A}}_{3} \cong < - 1.80 , 1.578 , 1.80 >$

#### Explanation:

Let
$\textcolor{w h i t e}{\text{XXX}} {\overline{A}}_{o}$: be the original vector $< - 8 , 7 , 8 >$
$\textcolor{w h i t e}{\text{XXX}} {\overline{A}}_{1}$: be unit vector in same direction as ${\overline{A}}_{o}$
$\textcolor{w h i t e}{\text{XXX}} {\overline{A}}_{3}$: be vector in same direction as ${\overline{A}}_{o}$ with length of $3$

$| {\overline{A}}_{o} | = \sqrt{{\left(- 8\right)}^{2} + {7}^{2} + {8}^{2}} = \sqrt{177} \cong 13.304$

${\overline{A}}_{1} = \frac{{\overline{A}}_{o}}{|} {\overline{A}}_{o} | = < \frac{- 8}{\sqrt{177}} , \frac{7}{\sqrt{177}} , \frac{8}{\sqrt{177}} >$

${\overline{A}}_{3} = 3 \cdot {\overline{A}}_{1}$
$\textcolor{w h i t e}{\text{XXX}} = < \frac{- 24}{\sqrt{177}} , \frac{21}{\sqrt{177}} , \frac{24}{\sqrt{177}} >$

$\textcolor{w h i t e}{\text{XXX}} \cong < - 1.80 , 1.578 , 1.80 >$