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the vector would have to be a scalar of (4,-1) in order to be parallel to it
4 = 8t -1 = 2t
it can also be in the opposit direction and be parallel
-1/4 is the slope we want to obtain right?
what is it possible?
are you talking about the first one or second?
b/c the second one I got -1/4 and I can't square root it
the second one cant be parallel; there is no hope of a negative slope with it
and the first one I can't get a value that's works because one of them needs to be negative?
thats the way i see it as well; the cant hav the same direction as the original no matter what real values you plug in
okay so the second one doesn't work I can't explain it
I just can't completely find t when I tried
t^2 is always positive; no way to get a negative slope
ohh that's makes more sense thanks!
youre welcome :)
@amistre64: why has the vector be a scalar of (4,-1), it can be any scalar(4x,-1x) isn't it?
allsmiles: is the question asking that both the vectors (8t,2t) and (1,t^2) should be parallel simultaneously or independently?
The first one is always parallel, whatever the value of t be, the second one is parallel (or anti-parallel) when t = +- 1/2
a scalar of (4,-1) simply means that any scalar will do
t<8,2> != x<4,-1> no matter what you do to it right?
they are orthoganal vector
maybe not ortho; but no the same slope :)
sry. what does <> mean?
<> denotes a vector as opposed to () which is a point
<3,5> is a vector; (3,5) is a point
Oh, so for no value of t are any of the two vecs parallel, isnt it?
for \[t \in R\]
the question asked shows point notation; but is assumed to be vectors :)