projection of u along a

[hayood]

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?

 

[Aladdin]

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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 . . .

 

[hayood]

yup that's it..find the proj of u along a

 

[soroban]

Hello, hayood!

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

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

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

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

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

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

Can you finish it now?

 

[hayood]

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]

i got @ as in theta = 39.8
so 5|cos 39.8|?

 

[mmm4444bot]

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: \(\displaystyle \frac{x}{4}\)

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.

 

[mmm4444bot]

5 |cos 39.8°|

Yes, that looks good (for an approximation).

I confirmed it, using the following.

\(\displaystyle |\text{proj}_{\vec a}\vec u| \:=\:\text{comp}_{\vec a}\vec u \;=\; \frac{\vec u\cdot \vec a}{|\vec a|}\)

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

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

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

\(\displaystyle |\text{proj}_{\vec a}\vec u| \:=\: \frac{\vec u\cdot \vec a}{|\vec a|} \;=\; \frac{18}{\sqrt{22}} \; \approx \; 3.8376\)

 

[mmm4444bot]

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).