projection of u along a

hayood

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Feb 16, 2010
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In each part, find ||projau||:
u=(3,0,4) a=(2,3,3)

||projau||=||u|| |cos@| --> pretend @ is angle

so what i did is i got ||u|| which = 5 and now i have 5 |cos @|

what should i do to get |cos @| the angle?
 
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Please, be more specific.

Your question is : Find the projection of vector U along a . . .

Sorry I didn't got you . . .
 
Hello, hayood!

Find projau:    u=3,0,4,  a=2,3,3\displaystyle \text{Find }|\text{proj}_{\vec a}\vec u|:\;\;\vec u\:=\:\langle3,0,4\rangle,\;\vec a\:=\:\langle 2,3,3\rangle

. . . . projau=ucosθ\displaystyle |\text{proj}_{\vec a}\vec u| \:=\:|\vec u| |\cos\theta|

I got: u=5 and now i have: 5cosθ\displaystyle \text{I got: }\:|\vec u|= 5\:\text{ and now i have: }\: 5|\cos\theta|

What should i do to get cosθ?\displaystyle \text{What should i do to get }|\cos \theta|\,?

Strange . . . you should have been given this formula, too:

The angle θ between two vectors u and v is given by:     cosθ  =  uvuv\displaystyle \text{The angle }\theta\text{ between two vectors }\vec u \text{ and }\vec v\text{ is given by: }\;\;\cos\theta \;=\;\frac{\vec u\cdot \vec v}{|\vec u||\vec v|}

Can you finish it now?

 
where do you get this better looking version of the question lool.. i will try solving it now and let you know if i get it ..
 
hayood said:
where do you get this better looking version of the question

The software for mathematical typesetting is called LaTex.

A simple example. Typing this (without the spaces inside the brackets): [ tex ] \frac{x}{4} [ /tex ]

yields this: x4\displaystyle \frac{x}{4}

To see the actual typing that generates any post at this site, click on the post's
button, in the upper-right corner of the post.

To read more about LaTex, use Google to find sites like THIS ONE.
 
hayood said:
5 |cos 39.8°|

Yes, that looks good (for an approximation).

I confirmed it, using the following.

projau=compau  =  uaa\displaystyle |\text{proj}_{\vec a}\vec u| \:=\:\text{comp}_{\vec a}\vec u \;=\; \frac{\vec u\cdot \vec a}{|\vec a|}

In other words, the magnitude of the projection of u\displaystyle \vec u onto a\displaystyle \vec a is simply the component of u\displaystyle \vec u along a\displaystyle \vec a.

compau\displaystyle \text{comp}_{\vec a}\vec u is also known as the "scalar projection" of u\displaystyle \vec u onto a\displaystyle \vec a.

As you see, the formula above does not require knowing the angle.

projau=uaa  =  1822    3.8376\displaystyle |\text{proj}_{\vec a}\vec u| \:=\: \frac{\vec u\cdot \vec a}{|\vec a|} \;=\; \frac{18}{\sqrt{22}} \; \approx \; 3.8376
 
hayood said:
yup that's it..find the proj of u along a Technically, no.

You asked for the magnitude of the projection of u onto a.

In other words, |v| is not the same as v.

If you actually DO want the projection of u onto a, instead of it's length, then you need to report the components of this projected vector, instead of 18/sqrt(22).
 
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