Math 53

Colleges and Universities A-Z University of California - Berkeley
1

<p>So that Math 53 midterm with Neu... probably one of the most disorganised midterms I've seen, but whatever. Anways, what did you all get for 3 and 4, I thought those were the most difficult, so if anyone here knows how it worked/what you did, feel free to explain. Also, 5 part C... </p>

<p>Also, does Neu's grading policy mean that the higher of our midterm 1 and 2 exams counts for 20%, while the lower does not? Also, past students: have you seen that he curves his grades? </p>

<p>Overall, I thought it was pretty decent though, maybe a little more difficult than what I had expected but nothing unreasonable. But I ramble, and have to go to a chem 3a midterm now... =(</p>

2

<p>Well…since I’m procrastinating, why not? hahaha I’m pretty sure this is how you do them…5c…yeah…idk what I did. I got like…I don’t even know…but whatever. </p>

<p>Sooooo I’ll separate each problem into a post…</p>

<p>3) The line contains the point (0,0,0), and has direction <1,1,1>. </p>

<p>Diagram: Draw a random line with a point (0,0,0) and vector b <1,1,1> going off that point, then draw a point (1,2,3) under the line. Label <1,2,3> (point on line (0,0,0) to the given point) as vector a. Finally, draw a line perpendicular to the given parametric line connecting the line to the given point]</p>

<p>Now, notice that you need the VECTOR projection of a onto b in order to find the vector representing (0,0,0) to where the perpendicular hits the line. Soooo after you get the vector projection, add that to (0,0,0), and I think the answer may have been (2,2,2) or something? idk…but that’s how I did it, not sure if right…I’m known for making a buttload of ■■■■■■■■ algebra mistakes.</p>

3

<p>4a) So, your point on the plane is (0,0,0). Now you need the normal to write your equation. </p>

<p>I was unsure as to what the point on the line was, but I took it as (1,0,-1) since it was i-k…but no parenthesis, so idk haha. The direction vector was <1,-2,1>. </p>

<p>Diagram: Again, draw a random line, and put the point (0,0,0) under it. On the line, label the point (1,0,-1), and let vector b be the direction <1,-2,1>. Draw vector a connecting (0,0,0) to (1,0,-1), which is <1,0,-1>. </p>

<p>Now we have two vectors on the plane! (since the line and the origin are on the plane). When you cross these, you should get your normal.</p>

<p>Since the origin is a point in the plane, it’s just your <normal vector=""> <em>dot</em> <x,y,z>.</x,y,z></normal></p>

4

<p>4b) Diagram : [Draw a random plane that kinda looks like the x-y plane in the x-y-z coordinate system. Draw point (1,0,0) above that plane. </p>

<p>From the previous part, you have the normal vector. Let’s call it vector b. You can move this to be anywhere in the plane, so place it where it will hit the point (1,0,0). The line segment that is connecting them is the distance you need. </p>

<p>From the previous problem, you know that (0,0,0) is on the plane, so put it somewhere on the plane. Draw a vector connecting (1,0,0) that’s hovering over the plane to (0,0,0). Make this vector a <-1,0,0>.]</p>

<p>Notice that what you need to find the distance is the SCALAR projection of a onto b. (if you didn’t catch that [our GSI went over it, so I knew what to do], label the angle between vectors a and b @(1,0,0) as theta. You now have a right triangle. The distance from (1,0,0) to the plane is the adjacent side, and the magnitude of vector a is the hypotenuse. This means the distance is |a| cos (theta). Since a<em>dot</em>b = |a||b| cos (theta), |a|cost(theta) is (a*b)/|b| >>>the scalar projection)</p>

5

<p>Wow…I got way too into that…but…yeah… correct me if I’m wrong…but wow…just spent 20 minutes on that hahaha</p>

<p>I may (probably) have signs wrong, but you should get the general idea. The nice thing about vectors though is that you can position them anywhere, meaning you can put them wherever is easier to look at :)</p>

<p>Hope that helps!</p>