Finding a value so that it is perpendicular to a vector
asked 2014-07-01 13:09:26 -0600
I thought that if you are looking for a value that is perpendicular to a vector you could use the dot product and just set it equal to zero. Number 7 on the review for Exam 1 says to find a value of t so that $ \langle 1,t, t^2 \rangle$ is perpendicular to $\langle 1,1,1 \rangle $.
When I did the dot product and set it equal to zero I used the quadratic formula and it will not work because there is a negative under the radical.
$$ \langle 1,t, t^2 \rangle \cdot \langle 1, 1, 1 \rangle = 0 $$ $$ 1+t+t^2 = 0 $$
Then I put those numbers into the quadratic formula: $$ t = \frac{-1\pm\sqrt{1-4(1)(1)}}{(2)(1)} = \frac{-1\pm\sqrt{-3}}{2} $$
So am I doing it wrong, or am I doing it right but my numbers are wrong, or what is going on because this answer does not make sense to me?
