My Math Forum geometric to component form - am i right?

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 April 26th, 2012, 07:08 AM #1 Member   Joined: Mar 2012 Posts: 44 Thanks: 0 geometric to component form - am i right? Hi all, I just stuck understanding this question, if someone could help explain, that would be great thanks Three forces (and no others) act upon an object, which remains at rest. The forces are represented by the vectors P, Q and R, where P = ?4i ? 5j and Q = ?2i + 9j. give the magnitude and direction of the vector R, to one decimal place. How am I suppose to do this when I'm not given the component form for R? all I can think of would be P + Q + R = 0 so Q = - P - Q so Q would be ( 6 i - 4 j) ? so its magnitude would be 7.2 and direction 33.7 am i right ?
April 26th, 2012, 07:50 AM   #2
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Re: geometric to component form - am i right?

Hello, mathslog!

Your arithmetic is off . . .

Quote:
 $\text{Three forces act upon an object, which remains at rest.}$ $\text{The forces are represented by: }\:\begin{Bmatrix} \vec p=&\langle -5,-4\rangle \\ \\ \vec q=&\langle 9,-2\rangle \\ \vec r=&\langle a,\,b\rangle \end{Bmatrix}=$ $\text{Find the magnitude and direction of }\vec r\text{ to one decimal place.}$

$\text{Since the object is at rest: }\:\vec p \,+\,\vec q \,+\,\vec r \:=\:\vec 0$

[color=beige]. . . . . . . . . . . [/color]$\langle-5,-4\rangle\,+\,\langle9,-2\rangle\,+\,\langle a,\,b\rangle \;=\;\langle 0,\,0\rangle$

[color=beige]. . . . . . . . . . . . . . . . . . . .[/color]$\langle a\,+\,4,\,b\,-\,6\rangle \;=\;\langle 0,\,0\rangle$

[color=beige]. . . . . . [/color]$\begin{Bmatrix}a\,+\,4 &=&0 &\;\;&\Rightarrow &\;\;& a &=& -4 \\ \\ \\
b\,-\,6&=& 0 && \Rightarrow && b &=& 6 \end{Bmatrix} \;\;\;\Rightarrow\;\;\; \vec r \:=\:\langle -4,\,6\rangle$

$\text{Magnitude: }\:|\vec r| \:=\:\sqrt{(-4)^2\,+\,6^2} \;=\;\sqrt{52} \;\;\;\Rightarrow\;\;\;\fbox{|\vec r| \;\approx\;7.2}$

$\text{Direction: }\:\tan\theta \:=\:\frac{6}{-4} \:=\:-1.5 \;\;\;\Rightarrow\;\;\;\theta \:\approx\:-56.3^o$

$\text{A sketch shows that }\vec r\text{ is in Quadrant 2: }\:\fbox{\theta \:=\:123.7^o}$

 April 26th, 2012, 08:30 AM #3 Member   Joined: Mar 2012 Posts: 44 Thanks: 0 Re: geometric to component form - am i right? Hello soroban Thanks for the quick reply, Your explanation helped out alot thank-you! made it much clearer to see

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