Question
Find the value(s) of t for which the vectors are parallel.
a = (t, t+2)
b = (2-3t, t)
I managed to work it down to t = 2k - 3t - 6, but I don't know what to after that.
a = (t, t+2)
b = (2-3t, t)
I managed to work it down to t = 2k - 3t - 6, but I don't know what to after that.
Answers
Reiny
recall that for any vector of the form <a,b> the slope of the vector is b/a.
So their slopes must be the same:
(t+2)/t = t/(2-3t)
t^2 = 2t -3t^2 + 4 -6t
4t^2 + 4t -4 = 0
t^2 + t - 1 = 0
solve using the quadratic formula.
So their slopes must be the same:
(t+2)/t = t/(2-3t)
t^2 = 2t -3t^2 + 4 -6t
4t^2 + 4t -4 = 0
t^2 + t - 1 = 0
solve using the quadratic formula.