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anonymous
 4 years ago
the symbol\[A _{b}\] stands for the projection of vector A onto vector B. In other words, \[A _{b}\] represents the component of A that is parallel to B. Derive an expression for \[A _{b}\] in terms of the vectors A and B.
anonymous
 4 years ago
the symbol\[A _{b}\] stands for the projection of vector A onto vector B. In other words, \[A _{b}\] represents the component of A that is parallel to B. Derive an expression for \[A _{b}\] in terms of the vectors A and B.

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amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[\frac{a.b}{a^2}a\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wow u did that really quickly

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1just got out of calc3 that just went over it :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in the numerator...is that a times b?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0a dot b divided by magnitude of a squared all multiplied by a?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1yes, the left side is a scalar; and the right side is vector a scaled to that length

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0would it be a bother to run me through how u got to this expression?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1327527260794:dw

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1b cos(t) = the length of Ab, if we use your notation for proj{a} b

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[bcos(t) =b \frac{a.b}{ab}=\frac{a.b}{a}\] good so far?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i'm also trying to decode your picture :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is that the x coordinate down there at the end?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1now, we need to scale that to a unit vector of a; since a is a long, lets divide it by a to get it to a length of 1

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[\frac{a.b}{a}\ \frac{a}{a}=; unit\ a, scaled\ by\ needed\ length\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1and since a a = a^2 i just condensed it alittle bit

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1327527605851:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can u redraw that picture from above?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1lol, i cant draw on this thing very well period :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay....so that's it for the equation now?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1to find the the vector that is the proj of b onto a; yes; reduce a to ints unit equivalent; and scale it up by the magnitude of b*cos(t)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and what u first initally gave me is that correct?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1yes, seeing how that is what we ended up with in the end i would say that its correct :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0u are an absolute life saver
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