http://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Tue, 01 Jul 2014 19:23:02 -0500http://calc3.askbot.com/question/40/finding-a-value-so-that-it-is-perpendicular-to-a-vector/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? Tue, 01 Jul 2014 13:09:26 -0500http://calc3.askbot.com/question/40/finding-a-value-so-that-it-is-perpendicular-to-a-vector/http://calc3.askbot.com/question/40/finding-a-value-so-that-it-is-perpendicular-to-a-vector/?answer=44#post-id-44**Christina:** Normally I would leave this as a comment, but perhaps this means that there are no vectors that satisfy our requirements? In other words, maybe there is no solution to this. *Comment*: I tend to agree.Tue, 01 Jul 2014 19:23:02 -0500http://calc3.askbot.com/question/40/finding-a-value-so-that-it-is-perpendicular-to-a-vector/?answer=44#post-id-44http://calc3.askbot.com/question/40/finding-a-value-so-that-it-is-perpendicular-to-a-vector/?answer=43#post-id-43I don't exactly consider this an answer, but just wanted to share some thoughts and pose some more questions. I approached this problem exactly as asmith14 did, setting the dot product of the two equal to zero. I also came up with the same answers as him/her: $$t = \frac{-1\pm\sqrt{-3}}{2}$$ Couldn't this also be written as: $$t = \frac{-1\pm i \sqrt{3}}{2}$$ Does this mean that there are 2 vectors that are perpendicular to $\langle1,1,1\rangle$? Does this also mean that the perpendicular vectors would be in the complex plane? I don't remember enough about complex numbers to know if this is right... **Comment** Justin, I tend to agree with you, a vector in the complex plain is one thing to consider, but in this case it makes a value of t a complex number...and correct me if I am wrong but isn't the complex plane just a way to represent the complex numbers? I don't know, it is a lot to consider...Thanks askbot!Tue, 01 Jul 2014 18:57:19 -0500http://calc3.askbot.com/question/40/finding-a-value-so-that-it-is-perpendicular-to-a-vector/?answer=43#post-id-43