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I get it up to the point m(vector a) is m a scalar then?
we can just say that Bv is <0,4> right? thats a vecotr of magnitude 4 :)
no we cant, we have to base it off of A... and 60 degree from it.... back to the draing pad :)
A<3,4> : ||A|| = 5
i found that (vector a) dot (vector b)= 10
by process of the math? or looking at the answers :)
process of math
can you type it out so I can see it :) that way I know what we both are looking at.... get me up to speed.
cos 60= 1/2 1/2= ((vector a) dot (vector b))/(magnitude of vector a * magnitude of vector b) 1/2= ((vector a) dot (vector b))/4*5 1/2= ((vector a) dot (vector b))/20 ((vector a) dot (vector b)) = 10
to keep things clean; we can use a to mean vector a and b likewise. magnitude is |a| works good. cos(60) = 1/2 1/2 = a*b / |a| |b| 1/2 = a*b / 4*5 1/2 = a*b /20 a*b = 10 ..good work :)
vectors that are perpendicular have a dot product of zero.
[m<3,4> + b] * b = 0
do we know what the vector parts for b are? I was working on that :)
no we dont
we can figure it out :) do we need to?
yeah, i think so,
b has 2 options up and to the left, or down and to the right.
the original angle of "a" is tan-1(4/3) = 53.1301
b has an angle of either -6.8699 or b has an angle of 113.1301
4sin(t) will give us the y value and 4cos(t) will give us the x values right?
the left vector option is <-1.57128, 3.67846> for "b" that gives us an angle betwwen them of 60 and a magnitude of 4
the other option for b is <3.97128, -.47846>
since we cant move a :) we need to work with these 2 possibilities and see what we get :)
Do you know if we should include the m in the vector like this?
+ < x , y > "b1"
ok i think i can take it from here
itd a been nice if they had given us a pretty angle to work with off of "a" lol
yeah i know, vector is very difficult to deal with
if you got it from here....good luck :)
you can pick whichever vector for b gets you that 10 ;)
thank you very much for your help