Question

vectors ʕ·ᴥ·ʔ

Answer 1

Determine whether the vectors u and v are parallel, orthogonal, or neither.

u = <6, -2>, v = <8, 24>

Answer 2

you answered my koala :D

Answer 3

dot product again

Answer 4

it was too cute, couldn't help it
find the dot product
if it is zero then they are "orthogonal"

Answer 5

\[6\times 8+(-2)\times 24=0\] i think
so they are "orthogonal" i .e. perpendicular

Answer 6

more gibberish...... i.e. things i dont understand

Answer 7

you know this because
\[\cos^{-1}(0)=\frac{\pi}{2}\] they make a right angle

Answer 8

we did the dot product the other day, remember? it is the number you get like i wrote above

Answer 9

soooo 48+-48 so its 0?

Answer 10

yes

Answer 11

oh i ment orthogonal is gibberish

Answer 12

oh c: so its orthogonal

Answer 13

.-. they should have just said perpendicular rather than create a different word for it all together :P

Answer 14

yes, they change the names to annoy you

Answer 15

✧･ﾟ:*✧･ﾟ:* \(◕‿◕✿)/ *:･ﾟ✧*:･ﾟ✧ you get stars for answering my question because medals dont matter to you XD
by any chance do you know triagonometric stuffs?

Answer 16

you mean "tickonometry"? yes i know some

Answer 17

.-. No?

Answer 18

i like the starts though, very nice
bear is nice too

Answer 19

*stars

Answer 20

yes, sure, fire away

Answer 21

its a koala >:c

Answer 22

and okay c:
Express the complex number in trigonometric form.

-3i

Answer 23

ok a koala
\(-3i\) we need only two number, the absolute value, which is pretty clearly 3, and the angle, which, is \(-90\) in degrees or \(-\frac{\pi}{2}\) in radians
which one are you using?

Answer 24

if you are using radians (which you should be) it is
\[3\left(\cos(-\frac{\pi}{2})+i\sin(-\frac{\pi}{2})\right)\]

Answer 25

if the angle is positive you could write
\[3\left(\cos(\frac{3\pi}{2})+i\sin(\frac{3\pi}{2})\right)\]

Answer 26

clear how i found the angle?

Answer 27

so both angles are 270?

Answer 28

yes
they are both 270, and they are both always the same, no matter what the angle is
it is
\[a+bi=r(\cos(\theta)+i\sin(\theta))\] same \(\theta\) for both

Answer 29

★~(◕‿◕✿) yay heres a star c:

Answer 30

don't be confused by this, this is just an equality between numbers
since \(\cos(270)=0\) and \(\sin(270)=-1\) you get \(3\times -i=-3i\)

Answer 31

thanks

Answer 32

yay c:
Express the complex number in trigonometric form.

-3 + 3 square root of threei
this one harder .-.

Answer 33

yes we tried this recently
it is not really much harder
but remember that
\[\sqrt{3^2+(3\sqrt3)^2}=\sqrt{9+27}=\sqrt{36}=6\] recall?

Answer 34

wait so would it be 5pi/6?

Answer 35

so it is going to be
\[-3+3\sqrt3i=6(\cos(\theta)+i\sin(\theta))\] now we need \(\theta\)

Answer 36

let me check

\[\cos(\theta)=\frac{a}{r}=\frac{-3}{6}=-\frac{1}{2}\] and
\[\sin(\theta)=\frac{b}{r}=\frac{3\sqrt{3}}{6}=\frac{\sqrt3}{2}\]

Answer 37

which is 5pi/6? :D

Answer 38

you got them backwards i think
it is the point \((-\frac{1}{2},\frac{\sqrt3}{2})\) which is at \(\frac{2\pi}{3}\)

Answer 39

I think it has something to do with the latin word orthogon, which means rectangle.
Perpendicular would be from the combination latin words per+ pendere to hang or balance equally.

Answer 40

now it seems you are working in radians, not degrees
weird they switch
in any case it is
\[-3+3\sqrt3i=6(\cos(\frac{2\pi}{3})+i\sin(\frac{2\pi}{3}))\]

Answer 41

(ノಠ益ಠ)ノ彡┻━┻ noooo

Answer 42

??

Answer 43

nooooo?

Answer 44

i wanted to be right :c
okay 1 more question with a+bi form c:

Answer 45

oh i thought you meant it was wrong
ok go ahead, shoot

Answer 46

( •_•) ( •_•)>⌐■-■ (⌐■_■)
okay then
Write the complex number in the form a + bi.

five divided by two(cos 150° + i sin 150°)

Answer 47

this one is the easiest yet
evaluate the numbers

Answer 48

\[\cos(150)=-\frac{\sqrt3}{2},\sin(150)=\frac{1}{2}\]

Answer 49

\[\frac{5}{2}(-\frac{\sqrt3}{2}+\frac{1}{2}i)\]

Answer 50

i.e.
\[\frac{-5\sqrt3}{4}+\frac{5}{4}i\]

Answer 51

okay so -5sqrt3 over 4

Answer 52

yeah like that

Answer 53

੭•̀ω•́)੭̸*✩⁺˚ yeah!

Answer 54

how do you do that???

Answer 55

computer codes XD most of its Japanese symbols look up "kawaii text faces" on google

Answer 56

thank you for your help c:

Answer 57

̿ ̿ ̿'̿'\̵͇̿̿\з=(•_•)=ε/̵͇̿̿/'̿'̿ ̿

Answer 58

yw
gotta run, hope you are done