The vector[13,-15] is a linear combination of the vectors [1,5] and [3,c]. Find c.
should be [1,5] + k [3,c] = [13,-15]
5+c=-15
find the value of k by comparison, from first coordinate. and compare second coordinate and find c
c=-20
i got k=10, c=-65 0.0"
solve this equations: 5+kc =-15 and 1+3k = 13
k= 4, c = -5
why can't i put k to the other two? i got different numbers
damn how could mertz get a different no??
no thinking, just typing:)
mertz wanted to say 5k+c=-15
i guess, but he continued with this misstype
i set k[1,5]+[3,c]=[13,-15] .... c=-65 and [1,5]+[3,c]=k[13,-15] .....c=125/13
i think k is 4 and c is -5
bouth are correct i think...
i got same as myko
loook at my other two settings
why can't i do that?
good question...
if you look at it this way, maybe it's more clear: (13,-15)-(1,5) =k(3,c)
=v=" what i concluded is [3,c] vector is the only vector that can be change since other two vectors are constant.
give me Medal =v="
not realy, becouse it says linear combo, so multiples of this vectors. Just there would be no way solution if we try to find it changing bouth at the same time. So we fix one and find the other... The question is way choosing different vector to keep it fixed different solutions come out
i only put one k on each equaltion i made. look at them.
a(1,5)+b(3,c) = (13,-15) so... a/b(1,5) +(3,c) =(13,-15)/b or (1,5) +b/a(3,c) = (13,-15)/a so if you take a =1 everything is ok, ho ho ho
3k + 1 = 13 -> k = 12/3 = 4 5 + 4c = -15 => c = -20/4 = -5
fine~
c= -20
two vector addition u + v= <1,5> + < 3,c> = <13, -15>...solve fr c.
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