http://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Tue, 01 Jul 2014 13:09:26 -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? asmith14Tue, 01 Jul 2014 13:09:26 -0500http://calc3.askbot.com/question/40/http://calc3.askbot.com/question/27/finding-the-cosine-of-the-angle-between-two-curves/So, I've been trying to work on the homework 13.2, and I'm stuck on both questions 6 and 7. But we'll start with question 6. "6. Find the cosine of the angle between the curves $\langle0,t^2,t\rangle$ and $\langle\cos(\frac{\pi t}{2}),\sin(\frac{\pi t}{2}),t\rangle$ where they intersect." Well I can tell that they intersect at t=1 And from the book on page 336 it states... " $ \cos(\theta) $ = $\mathbf {\frac{\vec r' * \vec s'}{|\vec r'||\vec s'|}}$ = $ \mathbf {\frac {\vec r'}{|\vec r'|} \cdot \frac{\vec s'}{|\vec s'|}} $ " So letting $\vec r$ = $\langle0,t^2,t\rangle$ and $\vec s$ = $\langle \cos(\frac{\pi t}{2}), \sin(\frac{\pi t}{2}),t\rangle$ I end up with $\vec r'$ = $\langle 0,2t,1\rangle$ and $\vec s' $ = $\langle (\frac{-1}{2} \pi \sin(\frac{\pi t}{2})), (\frac 12 \pi \cos(\frac{\pi t}{2})), 1\rangle $ using this to come up with | $ \vec r'$| I get |$\vec r'$| = $\sqrt{0^2 + (2t)^2 + 1^2} $ = $\sqrt{4t^2 +1} $ and |$\vec s'$|= $\sqrt{(\frac{-1}{2}\pi \sin(\frac {\pi t}{2}))^2 + (\frac 12 \pi \cos(\frac{\pi t}{2}))^2 + 1^2) } $ Now when we plug this into the $\cos(\theta)$ equation, do we substitute t=1 for the t in all of the equations to find the answer, or how exactly are we supposed to approach this problem? I end up with this ridiculous answer with a whole bunch of randomness in it that doesn't match up with the back of the book, so I'm beyond confused on this problem, and feel like there is a much easier way than how i'm trying to figure it out. HELLLPPPP! :) *Comment*: When you take the norm of $\vec{s}'$, note that you get a $\sin^2+\cos^2$. That yields a significant simplificatoin that might help.TiffanyMon, 30 Jun 2014 14:42:31 -0500http://calc3.askbot.com/question/27/