Posted by j on Thursday, May 24, 2012 at 10:52pm.
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.

Calculus  MathMate, Friday, May 25, 2012 at 7:43am
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.

Calculus  j, Friday, May 25, 2012 at 12:14pm
t(t+2) + 4(1t)  (2t+1) = 0
t = 1,3
like that?

Calculus  MathMate, Friday, May 25, 2012 at 1:26pm
Correct!

Calculus  j, Saturday, May 26, 2012 at 1:22am
thanks
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