Multivariable calculus vector problem?
For what values of s and t does the equality
<-(3t+3),2s+3t> = <3s-t,5t-1>
i know that i have to set the directions equal to each other but i cant figure out the algebra after that
what are the values of s and t?
Comments
-(3t+3) = 3s-t
2s+3t = 5t-1
-3t-3 = 3s-t (multiply negative through)
2s - 2t = -1 (subtract 5t from both sides)
-2t-3 = 3s (add t to each side)
-2t = -1 - 2s (subtract 2s from each side)
-2t-3 - (-2t) = 3s - (-1-2s) (subtract bottom equation)
-2t = -1 - 2s
-3 = 5s + 1 (simplify)
-2t = -1 - 2s
-4 = 5s (solving for s)
-2t = -1 - 2s
-4/5 = s (solved for s)
-2t = -1 - 2s
-4/5 = s
-2t = -1 - 2(-4/5) (plug in value for s)
-4/5 = s
-2t = -1 - (-8/5) (solving for t)
-4/5 = s
-2t = 3/5 (solving for t)
-4/5 = s
t = -3/10 (solved for t)
Two vectors are equal iff their components are equal, so we have the system:
(1) -(3t + 3) = 3s - t ==> -3t - 3 = 3s - t ==> 3s + 2t = -3
(2) 2s + 3t = 5t - 1 ==> 2s - 2t = -1.
Since the coefficients of t are equal in magnitude but opposite in sign, adding these two equations will cancel out t and allow us to solve for s. Doing this gives:
(3s + 2t) + (2s - 2t) = -3 + (-1) ==> 5s = -4
==> s = -4/5.
By plugging s = -4/5 back into either equation, t = -3/10.
Therefore, s = -4/5 and t = -3/10.
I hope this helps!