http://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Thu, 24 Jul 2014 11:06:57 -0500http://calc3.askbot.com/question/157/quiz-1-review/I would also be posting quiz 2, but sadly it was lost by an anonymous individual named SpacemanSpiff. **1:** Prove that scalar multiplication of vectors is distributive over the addition of 2-D vectors. We can start by defining two arbitrary vectors, $\vec{u}=\langle u_1, u_2 \rangle$, and $\vec{v} = \langle v_1, v_2 \rangle$. Now let $a$ be a real number. Then we just add our arbitrary vectors together and multiply the resulting vector by $a$, and break apart the vectors into their components, then put them back together: $a(\vec{u} + \vec{v}) = a(\langle u_1, u_2 \rangle + \langle v_1, v_2 \rangle $ $= a\langle u_1+v_1, u_2 + v_2 \rangle $ $= \langle a(u_1+v_1), a(u_2+v_2) \rangle $ $= \langle au_1 + av_1, au_2+av_2 \rangle $ $= \langle au_1,au_2 \rangle + \langle av_1,av_2 \rangle $ $= a\langle u_1, u_2 \rangle + a \langle v_1, v_2 \rangle $ $= a\vec{u} + a\vec{v}$. And our proof is done. **2:** The intersection between the $xy$ plane and the plane $x+2y -3z = 6$ is a line. Find an $xy$ equation for that line. The key here is to see that the intersection occurs at $z=0$, we just take the equation for the plane and set $z=0$: $x+2y = 6$. This is our $xy$ equation. **3:** Find a parametrization of the line containing the point $(1,2,3)$ and perpendicular to the plane $x+2y-3z = 6$. The coefficients on $x$, $y$, and $z$ form a vector perpendicular to the plane, as we showed in class. So we just take this vector $\langle 1, 2, -3 \rangle $ as the direction vector for our line, and then use the point we want to be on the line as our $\vec{p_0}$: $\vec{l}(t) = \langle 1, 2, 3 \rangle + \langle 1, 2, -3 \rangle t $ **4:** Find all values of $t$ such that the vector $\vec{p}(t) = \langle t^2, t \rangle $ is perpendicular to the vector $\vec{q}(t) = \langle t, -1 \rangle $. Two vectors are perpendicular if and only if their dot product is equal to 0, so we can just set the dot product of these two vectors equal to 0: $\langle t^2, t \rangle \cdot \langle t, -1 \rangle = t^3 - t$ $ = t(t^2-1) = 0$ This gives us solutions where $t = 0$, $t= 1$ and $t=-1$. So for those values of $t$, the vectors are perpendicular. Wed, 23 Jul 2014 20:24:27 -0500http://calc3.askbot.com/question/157/quiz-1-review/http://calc3.askbot.com/question/157/quiz-1-review/?answer=162#post-id-162Are you asking a question about something on this, or just posting the answers that we went over in class? If it's a question, I'm confused as to what you're needing help on? This looks more like a review?Thu, 24 Jul 2014 10:08:00 -0500http://calc3.askbot.com/question/157/quiz-1-review/?answer=162#post-id-162http://calc3.askbot.com/question/157/quiz-1-review/?answer=166#post-id-166This is a part of the Database that SpacemanSpiff started here: http://calc3.askbot.com/question/143/final-exam-review-test-1/ Admittedly this quiz was not intensely difficult and I was thinking of not posting it because I didn't know if anyone really wanted to review it, but I realized when my second quiz got lost that having an online copy of the questions and work might be useful to someone (including myself). **Note: I am adding this as an answer because the comment system is still not working on my end.**Thu, 24 Jul 2014 11:06:57 -0500http://calc3.askbot.com/question/157/quiz-1-review/?answer=166#post-id-166