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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 find a value of t so that \rangle{1},t,\langle{t^2} is perpendicular to \rangle {1},1,\langle{1}

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.

\rangle{1},t,\langle {t^2} \bullet \rangle {1},1,\langle{1} and I got 1+t+t^2=0

then I put those numbers into the quadratic formula -1\pm\surd{1-4(1)(1)} all over 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?

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 Exam 1 says to find a value of t so that \rangle{1},t,\langle{t^2} $ \langle 1,t, t^2 \rangle$ is perpendicular to $\langle 1,1,1 \rangle {1},1,\langle{1}$.

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.

\rangle{1},t,\langle {t^2} \bullet $$ \langle 1,t, t^2 \rangle {1},1,\langle{1} and I got 1+t+t^2=0\cdot \langle 1, 1, 1 \rangle = 0 $$ $$ 1+t+t^2 = 0 $$

then Then I put those numbers into the quadratic formula -1\pm\surd{1-4(1)(1)} all over 2formula: $$ t = \frac{-1\pm\sqrt{1-4(1)(1)}}{(2)(1)} = \frac{-1\pm\sqrt{-3}}{2} $$

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?