Question

can someone help me with calc 3
i dont understand whats going on for this problem. ik it has to do something with scalar or vector projections

Answer 1

Oh girl hold on lemme get help lol

Answer 2

@Ultrilliam @Angelica @Warriorz13

Answer 3

thanks lol

Answer 4

I dont even know what Calc 3 is pft XD

Answer 5

I'm still in precalc @vocaloid

Answer 6

I take Geometry

Answer 7

D: i need help with this subject.. im so lost and her study guide looks like another mathematical language too me

Answer 8

I would help if I could but idk how

Answer 9

you know I should know vectors because of the c++ class I was in... but yea... one sec while I look

Answer 10

a) position 1
b) position 2



If I have that right...
orthogonal means the product of the 2 positions = 0 I believe

Answer 11

although I don't know the terminology to identify the positions in the vectors... RIP

Answer 12

looks right but im like which formula do we use

Answer 13

I just used intuition lol

Answer 14

i found this one for number 3 that is similar but idk http://math.arizona.edu/~mgilbert/HW1_Solutions_223_Fall2015.pdf

Answer 15

you know the formula for finding one vector component parallel to other vector ?

Answer 16

this might help you

Answer 17

do i have to draw the diagram ?

Answer 18

its your own choice but why so?
you have a formula ,just use it
diagram only simplifies your imagination that's all
but i recommend you to draw the diagram

Answer 19

idk how to draw the diagram. i have a hard time visualizing stuff

Answer 20

umm i'm here ,i'll try to guide you

Answer 21

now take the component of v along the direction of b
that gives you the vector component parallel to b
and that is vcos theta

Answer 22

am i right?

Answer 23

yeah

Answer 24

okay! so we get

do i write it as well ?

Answer 25

that formula is for parallel component right ?

Answer 26

yep

Answer 27

\[\theta \cos \theta | <1 ,2 , -1 >/ < 1 , 2 , -1 > \]

Answer 28

wait. i think i Salsaed up lol

Answer 29

are you there lool

Answer 30

i was wondering that what made you think like that lol

Answer 31

and still its difficult to understand you xD

Answer 32

i followed the formula e.e

Answer 33

okay so where we are

Answer 34

yes im having a hard time getting it lol

Answer 35

okay so we have a formula

Answer 36

can you find

Answer 37

we get | b | = sq root 6 right ?

Answer 38

aye dont leave me here lol

Answer 39

back

Answer 40

wb

Answer 41

thank you :D

Answer 42

yeah modulus of b is correct

Answer 43

find unit vector of b

Answer 44

okay! so i gotta go now
i'll be back tomorrow
see ya

Answer 45

no lool

Answer 46

B)
\[\large\rm Proj_b v =\frac{ v \cdot b }{ \left| v \right|^2 } v=\color{red}{answer}\]
\[\large\rm v-\color{red}{answer}= final~ answer\]

Answer 47

later gtg

Answer 48

wait which one is part a and b for that formula?

Answer 49

for the perpendicular

Answer 50

B)

Answer 51

B)
\[\large\rm Proj_b v =\frac{ v \cdot b }{ \left| v \right|^2 } v=\color{red}{answer}\]
\[\large\rm v-\color{red}{answer}= final~ answer\]
\[\large\rm \color{red}{v=<1,0,-1>}~~~~~\color{blue}{b=<1,2,-1>}\]
first we need to find the projection of v on b
use the above formula\[\large\rm Proj_\color{blue}{b} \color{red}{v}= \frac{ \color{red}{v}\cdot \color{blue}{b} }{ \left| \color{red}{v} \right|^2 } \color{Red}{v}\] \[\large\rm Proj_\color{blue}{b} \color{red}{v}= \frac{ \color{red}{<1,0,1>}\cdot \color{blue}{<1,2,-1>} }{ \left| \color{red}{<1,0,1>} \right|^2 } \color{Red}{<1,0,1>}\]
Lets do the denominator first
the magnitude of v square
but when we take the magnitude it will be square root of v and that will cancel out with the square \[\left| v \right|^2 = (\sqrt{v})^2=v\]
so
\[\large\rm Proj_\color{blue}{b} \color{red}{v}= \frac{ \color{red}{<1,0,1>}\cdot \color{blue}{<1,2,-1>} }{ \color{red}{ 1^2+0^2+1^2}} \color{Red}{<1,0,1>}\]
for numerator apply the dot product

<1,0,1> dot <1,2,-1> = ???

Answer 52

i don't think it asking if its orthogonal or not it just says find the orthogonal vector ?

Answer 53

ohhh lol

okay so i got < 1 ,0,-1> ?

Answer 54

yes :=))

Answer 55

yayy ^_^ so thats for part b

and what about part a ?

Answer 56

oh wait question
my answer sheet doesnt match the answer D:
she got

Answer 57

nnesha D:

Answer 58

i see what i did wrong i forgot the negative sign

Answer 59

https://prnt.sc/gk1q6v

Answer 60

but still i didn't get that answer hmmmm le me see...

Answer 61

ahhh projection of v on `b`
so replace all b with v
sorry

Answer 62

yeah or unless she did a mistake

Answer 63

okay hmm lets try again..

Answer 64

okie

Answer 65

\[\large\rm Proj_\color{blue}{b} \color{red}{v}= \frac{ \color{blue}{b}\cdot \color{red}{v} }{ \left| \color{blue}{b} \right|^2 } \color{blue}{b}\] \[\large\rm Proj_\color{blue}{b} \color{red}{v}= \frac{ \color{blue}{<1,2,-1>}\cdot \color{red}{<1,0,-1>} }{ \left| \color{blue}{<1,2,-1>} \right|^2 } \color{blue}{<1,2,-1>}\] \[=\frac{ 1+0+1 }{ 1^2 +2^2+(-1)^2} \times <1,2,-1>\]\[=\frac{ 2 }{ 6 }<1,2,-1>\]

Answer 66

one sec im writing this down lol

Answer 67

okay so then

Answer 68

that's the projection now subtract it from vector v

Answer 69

\[\frac{1}{3}<1,2,-1>=<\frac{1}{3},\frac{2}{3},-\frac{1}{3}>\]

Answer 70

so then i got < 2/3 , -2/3 , -2/3>

Answer 71

yeah :=))

Answer 72

yay :D

Answer 73

can you check my work for number 5 , bc her solution says they are parallel , but i think they orthogonal

Answer 74

can u do 4a ?? parallel one ?

Answer 75

yeah sure hurry up

Answer 76

this is what i did
https://docs.google.com/document/d/1Kz1qTwPADPJ5d-75l7Ik0Y7qU_agosgbaqhvTOInGmM/edit

Answer 77

if the dot product =0