absolute values and magnitudes

Hi Dr. Taylor, I'm just wondering if on the last portion of the problem if
|w| is really supposed to be ||w|| since vectors don't technically have
absolute values

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We discussed this in class: technically yes, you're a little bit right.  It's a common usage to write |w| instead of ||w|| for the very reason you mention: there's no such thing as the absolute value of a vector, so what else could  |w| mean but ||w||.  There's really no room for confusion so no harm in this usage.

10.3#5

this:
I am not sure where i have made my mistake on this problem, i took the
cross product of the vectors given <1,1,0>x<0,1,-5>  which gave me the new
orthogonal vector <-6,5,1> but since this problem needs the orthogonal
vector to have an i component that is equal to 1, i divided the 5 and 1 by
-6 getting the resultant vector <1,-5/6,-1/6>. Web work is saying that this
answer is incorrect. Is my set up wrong? Is the method for finding the
answer different that what i am attempting?
and this:
in response to the previous email, my set up was right, i ended up making a
small computational error while doing the cross product, i should have
equalled -5 not -6, throwing my answer off. Ended up getting the right
answer.

*************
Glad you figured it out on your own, this is the best possible outcome, especially because you learn to catch your own algebraic and arithmetic errors.

10.4#8

Hello Dr. Taylor!

I got the answer correct by finding the cross product of
the two vectors, then finding the ((magnitude of the cross
product)/(distance between Q and R)). Could you explain why this works
though? I would've thought it would have something to do with projection.

Thanks!

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Oh. Yeah, I never thought about that, but you can do that.  Here's a picture of the points Q,R,P:

There is a dotted line passing through the two points Q, R, and a displacement vector from Q to P.  Notice that I've drawn a right triangle, one corner of which is at Q, the other corner at P, and the right angle on the line passing through Q and R.  I suppose that when you say "the two vectors" you mean the displacement vectors u=R-Q and v=P-Q.  Note that the hypotenuse of the triangle is ||v||. Then the part of v along the direction of  u  is (u.v)/(u.u) u, this is the displacement vector that goes from Q to the place where the blue line intersects the line QR, and has magnitude ||v|| Cos(θ) where θ is the angle RQP.  Of course a little trigonometry tells you that the opposite leg of the right triangle must have magnitude ||v||Sin(θ) = ||uxv||/||u||

10.3#11

Dear Dr. Taylor,
I am a little confused about what I am doing wrong.  I
have been using the equation W=|F||D|cos(Theta) and received the answer
(31sqrt(3))/2.

Thanks,

**************

You meant 3*1*sqrt(3)/2.  That kind of mistake will kill you on the exam though, so be careful.

10.2#10

Hi, I'm not sure if you can see my attempts for this problem but I've done
it a bunch of times and I can't figure out where I am going wrong with the
second part to this question. If you could give me a little guidance I
would really appreciate it thank you!

****************

Well, you got the speed right.  The angle is a mistake of course.   It's interesting though because, although it's close to the correct answer in an absolute sense, the relative error is about 10%.  I went through the usual list of mistakes people make in computing angles and they all gave smaller relative errors than that, so I really can't imagine how you got that answer, except maybe bad arithmetic or a typo. For instance using 13 instead of 14 would give about the answer you have. 

 

To reiterate a comment I made in an earlier post, when you submit a question like this you need to tell me what you tried so I can catch your mistake. 

Yes I can see how many times you tried to do the problem. So as far as guidance, just flailing and trying lots of guesses isn't going to help you. Stop doing that.  Stepping back and making sure you understand the math, and making sure you did the steps correctly will work better than that 

Lecture Notes 2/28/17 through 9/1/17 (complete!)

Lecture Notes 8/28/17

Lecture Notes 8/30/17

Lecture Notes 9/1/17

10.3#14

Mr. Taylor,

I just don't understand this question at all, and as you can

probably see I've tried it a ton of different times

Thank you for any help

**************************Well you clearly do understand something since you got the a) correct.  It *looks like maybe* you tried to do part b) correctly too, except you made a wee little mistake subtracting the one vector from the other.  I can't tell though, because *you didn't tell me what you did*.  May I suggest that you do that when send me a question?--it usually is a better teaching moment if I can tell you exactly where you made a mistake.   BTW, we used exactly the terminology this vector parallel and that vector perpendicular in the lecture notes from last Wednesday and/or Friday, and those lecture notes are already posted here in the blog, and this stuff is also discussed in section 10.3 in the textbook.  I didn't lecture on the stuff about work, but is also discussed in the textbook in glorious detail.  The upshot is that the work is defined  as the Force vector dot product with the displacement vector.