Calculus
posted by j on .
3. Determine that the vectors u=[t, 4, 2t+1] and v=[t+2, 1t, 1] are perpendicular, find the possible values of the contant, t.

Same as previous question,
http://www.jiskha.com/display.cgi?id=1337914331
Except that the dotproduct equated to zero results in a quadratic equation.
So solve for possible values of t by solving the quadratic. 
t(t+2) + 4(1t)  (2t+1) = 0
t = 1,3
like that? 
Correct!

thanks