how to find the projection of a=(1,3) on b=(4,0,2) ??
a onto b eh... vector right? its the |a| cos(angle between)
really ? I know diff. formula!!
angle is not mentioned at all in the question
t = \(cos^{-1}(a.b/(|a||b|))\)
well, in order to know the length of the projection onto b you gotta know the angle between a and b
not the length ! it said just the vector that is proj.
vectors are length and direction...cant have vector without length right?
sure! but I'm using the dot product ,, (a.b/||b||^2) x b
ok...
<1,3,0> <4,0,2> ------- 4+0+0 =4
aha ,, then ?
sqrt(10).sqt(20) = sqrt(200)
cos_1(4/sqrt(200)) is what id try
73.57 degrees is the angle between them...
but we already have the cos of that: 4/sqrt(200) lol
4sqrt(10) -------------- = 4/sqrt(20) right? this is the length of a vector sqrt(10).sqrt(20) that is parallel to b
b/|b| = unit vector of b; then *that to get the vector projection
aha that the answer I've found but in diff. way
but the problem that it's multiple choice Qs. and I didn't find the answer !!
<4,0,2> 4 ------- * ----- :) sqrt(20) sqrt20
whats the choices?
i get: <16/20 , 0 ,8/20> if im right lol
the answers are : <1,1> , <2,1> , <2,-1> , <1,2>
that's the answer I've got ! but it seems that its wrong !!!
well, then we misunderstood what it was asking for then :)
the question which is up there is the exact same ques. !!!!!
yeah, perhaps its the vector that I drew in green?
thanks allot for trying ..
a.b --- * b |b|^2 4/20 = 1/5<4,0,2> = <4/5,0,2/5> cant really see it being nuthin else
youre welcome :) good luck
yes! that's my opinion two !
thank you ..
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