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Mathematics 15 Online
OpenStudy (lukecrayonz):

JUST #28. http://screensnapr.com/v/wmqb1W.png

OpenStudy (turingtest):

\[\vec u\cdot\vec v=|u||v|\cos\theta\]

OpenStudy (anonymous):

Why did you specify the absolute value for a dot product, Turing?

OpenStudy (lukecrayonz):

Thats not absolute value, its magnitude.

OpenStudy (turingtest):

normally it has two lines around it, but that looks messy on here

OpenStudy (lukecrayonz):

He could have put || but yeah :P

OpenStudy (anonymous):

Oh, alright.

OpenStudy (lukecrayonz):

so just 100*250(pi/6)

OpenStudy (turingtest):

no cosine!

OpenStudy (lukecrayonz):

no wait thats like 740000 degreees

OpenStudy (lukecrayonz):

HHAHAHA I WAS LIKE OMG THIS IS CRAZY

OpenStudy (turingtest):

gotta have that cosine or it's not worth the paper it's printed on...

OpenStudy (lukecrayonz):

I'm still pretty sure this answer is wrong.. 24998..

OpenStudy (turingtest):

let me see what I get...

OpenStudy (turingtest):

I got 21650...

OpenStudy (turingtest):

21651 rounded

OpenStudy (turingtest):

which is reasonable, they are big vectors

OpenStudy (lukecrayonz):

u=100, v=250, theta=(pi/6) 100*250=25000*cos(pi/6)=24998.

OpenStudy (lukecrayonz):

Calculator is 100% sure in degree too.

OpenStudy (turingtest):

perhaps you are in degree mode?

OpenStudy (lukecrayonz):

Oh......

OpenStudy (turingtest):

lol that's not a good thing

OpenStudy (turingtest):

if it's got pi in it you're almost certainly in radians

OpenStudy (lukecrayonz):

I was just doing angle between vectors which said make sure it's in degree mode :P

OpenStudy (turingtest):

cos(pi/6) is a known special trig value: \[\cos\frac\pi6=\frac{\sqrt3}2\]and I'm gonna guess the other excercise you did did not have pi degrees in it

OpenStudy (lukecrayonz):

Oh! a quick question, how do you find if two vectors are orthogonal or parallel? <15,45>

OpenStudy (lukecrayonz):

and <-5,12> sorry :P

OpenStudy (turingtest):

again, same formula if they are parallel what is the angle between them?

OpenStudy (lukecrayonz):

And no, it was a different section :)

OpenStudy (lukecrayonz):

Isnt it just 15*-5+45*12? 465.

OpenStudy (turingtest):

\[\vec u\cdot\vec v=|u||v|\cos\theta\]if u and v are parallel, what is theta? what about if they are perpendicular?

OpenStudy (lukecrayonz):

http://screensnapr.com/v/5gMbRA.png

OpenStudy (turingtest):

right, and how did we know those vectors in your example were orthogonal?

OpenStudy (lukecrayonz):

equals 0 then vectors are orthogonal

OpenStudy (lukecrayonz):

But does 465 mean anything?And how do you know if they are parallel?

OpenStudy (lukecrayonz):

besides graphing them of course

OpenStudy (turingtest):

can you see why from this formula?\[\vec u\cdot\vec v=|u||v|\cos\theta\]that will answer your question, just answer mine what is the angle between two orthogonal vectors?

OpenStudy (lukecrayonz):

Well they're perpindicular :P

OpenStudy (lukecrayonz):

perpendicular.

OpenStudy (turingtest):

and a perpendicular angle is how many degrees?

OpenStudy (lukecrayonz):

180 or 90.

OpenStudy (turingtest):

just 90... and what is the cosine of 90 degrees?

OpenStudy (lukecrayonz):

zero

OpenStudy (lukecrayonz):

OH! I get it :D

OpenStudy (lukecrayonz):

Not the parallel part but I get the reasoning of the 0.

OpenStudy (turingtest):

so when our formula becomes\[\vec u\cdot\vec v=|u||v|\cos\frac\pi2=0\]that's why that works yeah! so what about parallel? what is the angle there?

OpenStudy (turingtest):

pi/2=90 degrees btw...

OpenStudy (turingtest):

|dw:1329975086069:dw|what is theta between these two parallel vectors?

OpenStudy (lukecrayonz):

What..

OpenStudy (turingtest):

theta is the angle what is the angle between them?

OpenStudy (lukecrayonz):

Yes, I know, but I have no idea!

OpenStudy (turingtest):

you can always draw vectors on top of each other to compare the angle|dw:1329975259129:dw|it looks like the angle is 0 degrees (or 180 degrees) what is the cosine of zero degrees?

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