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OpenStudy (anonymous):
@amistre64
OpenStudy (anonymous):
I'm not familiar with the notation. What does the down arrow between vectors represents? Cross product?
OpenStudy (anonymous):
its represents the projection of u onto v
OpenStudy (anonymous):
Wouldn't #3 imply that u and v are orthogonal or they are the same?
OpenStudy (anonymous):
whats orthogonal
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OpenStudy (isaiah.feynman):
Orthogonal means they are perpendicular.
OpenStudy (abb0t):
Just a fancy word for saying perpendicular, as the person has said above.
OpenStudy (anonymous):
oh okay so then number three implies they are equal
OpenStudy (isaiah.feynman):
Can you draw number three, because I don't know what "projection of v to u" means
OpenStudy (anonymous):
well thats the part on what i don't know how to do
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OpenStudy (isaiah.feynman):
Never heard of projection of one vector to another.
OpenStudy (isaiah.feynman):
Number 3 I'd say no.
OpenStudy (anonymous):
sorry what do you mean no?
OpenStudy (isaiah.feynman):
They are not equal
OpenStudy (anonymous):
ohh actually all of those are equal .. its just i have to prove they are equal
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OpenStudy (anonymous):
@ganeshie8
ganeshie8 (ganeshie8):
#3 or #4 ?
OpenStudy (phi):
The projection of vector u onto vector v will be a vector
It will point in the same direction as vector v
it will have a length | u | cos A where A is the angle between the two vectors
|dw:1385558127501:dw|
if we call "v hat" the vector v normalized to unit length, we can say
\[ u \downarrow v = \left(|u|\ \cos A\right) \hat{v}\]