how to get, two vectors of length 26 units and slope 5/12?

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- Jul 6th 2010, 09:48 AMrabih2011vectors
how to get, two vectors of length 26 units and slope 5/12?

- Jul 6th 2010, 09:58 AMearboth
- Jul 6th 2010, 12:38 PMHallsofIvy
Call the vector <x, y>. Saying it has length 26 means that $\displaystyle \sqrt{x^2+ y^2}= 26$. Saying that it has slope 5/12 means that x/y= 12. That gives you two equations to solve for x and y. From x/y= 12, x= 12y so the first equation becomes $\displaystyle \sqrt{144y^2+ y^2}= \sqrt{145}y= 26$. You may assume that x and y are positive: if x and y are both positive or both negative, the sign disappears in both formulas. That is, if you find <x, y> which satisfies those equations, so does <-x, -y>. (I don't think it is necessary to "assume" your vectors are in $\displaystyle \mathbb{R}^2$- if they were not they would not have "slope".)

- Jul 6th 2010, 01:10 PMSoroban
Hello, rabih2011!

Quote:

Find two vectors of length 26 units and slope $\displaystyle \frac{5}{12}$

Let the vector be $\displaystyle \langle x,y\rangle$

Its length is 26: .$\displaystyle \sqrt{x^2+y^2} \:=\:26 \quad\Rightarrow\quad x^2 + y^2 \:=\:676$ .[1]

Its slope is $\displaystyle \frac{5}{12}:\;\;\frac{y}{x} \:=\:\frac{5}{12} \quad\Rightarrow\quad y \:=\:\frac{5}{12}x$ .[2]

Substitute [2] into [1]: .$\displaystyle x^2 + \left(\frac{5}{12}x\right)^2 \:=\:676 \quad\Rightarrow\quad x^2 + \frac{25}{144}x^2 \:=\:676 $

. . $\displaystyle \frac{169}{144}x^2 \:=\:676 \quad\Rightarrow\quad x^2 \:=\:576 \quad\Rightarrow\quad x \:=\:\pm24$

Substitute into [1]: .$\displaystyle y \:=\:\frac{5}{12}(\pm24) \:=\:\pm10$

The two vectors are: . $\displaystyle \begin{array}{c}\langle 24,\;10\rangle \\ \langle \text{-}24,\text{-}10\rangle \end{array}$

- Jul 6th 2010, 04:42 PMHallsofIvy
Arghh! I forgot to

**square**the 5/12! Thanks, Soroban.