Mathematics
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OpenStudy (cutiecomittee123):
The projection of v=(6,4) onto w=(1,1) is _____
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OpenStudy (loser66):
you have formula to do it. what is it?
OpenStudy (cutiecomittee123):
p=(v(w))/(absolute value of w)
OpenStudy (cutiecomittee123):
I get (10)/(absolute value of (1,1))
OpenStudy (loser66):
\(proj_a u =\dfrac{u\bullet a}{||a||^2}a\)
OpenStudy (loser66):
your a is w, your u is v. Apply it. show your work, please
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OpenStudy (xapproachesinfinity):
Proj of v onto w=v.w/||w|| times w
OpenStudy (loser66):
hey kid, the denominator is power of 2.
OpenStudy (xapproachesinfinity):
|dw:1452551821376:dw|
OpenStudy (xapproachesinfinity):
yes old man just forget it :)
OpenStudy (loser66):
I am busy now, but you must teach me how to do that grid later, kid?
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OpenStudy (xapproachesinfinity):
|dw:1452551952191:dw|
OpenStudy (xapproachesinfinity):
just a visualization of what is going on
OpenStudy (xapproachesinfinity):
at any rate do you know how to use the formula
OpenStudy (cutiecomittee123):
sorta I am confused on how to calculate the absolute value of a point though
OpenStudy (cutiecomittee123):
@xapproachesinfinity
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OpenStudy (cutiecomittee123):
So then wouldnt it be 10?
because after plugging in the numbers I get 10/1 which is just 10
OpenStudy (cutiecomittee123):
@Loser66
OpenStudy (phi):
|a| can be thought of as the length of a vector from (0,0) to the point a
use pythagoras. e.g. if a=(x,y) then |a| = \(\sqrt{x^2+y^2}\)
OpenStudy (cutiecomittee123):
Okay
OpenStudy (cutiecomittee123):
explain more because that kinda confused me