Question

Need pre-calculus help will give owlbucks

Answer 1

Let A (2, -1,) B (3, 1,) C (-4, 2,) and D (1, -5). Find the component form and magnitude of the vector. ---- ---- CD - AB?

Answer 2

hmmm

I assume you mean \(\bf C\cdot D-A\cdot B\impliedby \textit{as in dot product}\)

Answer 3

No they have line segments above them. and for CD I GOT -4,10 and AB I got 6,-1

Answer 4

hmm ohh hmm

Answer 5

\(\bf \overline{CD}-\overline{AB}\) right?

Answer 6

maybe A,B,C,D are points given
and we want to find the standard form for both vectors CD and AB.
and then find the difference

Answer 7

Sorry i was doing something and yes thats right

Answer 8

i will take a screen shot

Answer 9

hmm have you done 8 and 9 yet?
so I assume you've got those values already?

I mean, 10 is the same pretty much :)

Answer 10

\[\text{ a vector with initial point } (a,b) \text{ and terminal point } (x,y) \text{ has standard form } \\ (x-a,y-b)\]

Answer 11

No, i have not I just chose one of the hardest ones that look hardest so i didn't do the addition ones just did the subtraction one

Answer 12

heheh anyhow... notice above @myininaya 's line
to get teh vector component form from the given points

Answer 13

ok so i n ab and xy i would enter CD and AB?

Answer 14

Yes

Answer 15

so... CD will be
(-4, 2) and (1, -5)
thus

< 1-(-4) , -5 - 2 >

Answer 16

oh thats it ?

Answer 17

that isn't the final answer if that is what you mean

Answer 18

so i would need this formula

Answer 19

yes you do that same thing for other vector in your difference

Answer 20

Oh ok i expected more

Answer 21

i mean i do expect more

Answer 22

also I guess to be less confusing we could have used < > instead of ( ) for our vector notation thing

Answer 23

so 5 and -7?

Answer 24

oops but @jdoe0001 did that :p

Answer 25

for the vector component form, yes

and you'd do the same with AB

2, -1 and 3, 1
thus

AB wiill be
< 3-2, 1-(-1) >

notice, you have \(\bf \overline{CD}-\overline{AB}\implies <CD>-<AB>\)

which is pretty much, just an addition

Answer 26

oh so this is almsot like dot product but for each line segment

Answer 27

Ok so these two would be the component form <5,-7>VECTOR and <1,1> INITIAL points?

Answer 28

be careful 1-(-1) is 1+1 which is 2 not 1

Answer 29

o crap that got me

Answer 30

and then to do something like
<m,n>-<r,s>
=
<m-r,n-s>

Answer 31

thanks

Answer 32

so
<5,-7>-<1,2>
=
?

Answer 33

4,-8

Answer 34

o wait

Answer 35

your first component is right

Answer 36

your second one needs work

Answer 37

-9?

Answer 38

yes
<4,-9> is right if there wasn't some arithmetic error I made or I missed

Answer 39

i will most check this again cus i have test tommorow and am studying , but kinda learning at the same time

Answer 40

i apprieciate your help, so this is just the componenet vector?

Answer 41

wait i can review it

Answer 42

C(-4,2) D (1,-5) (X-A,Y-B)

Answer 43

yes
or you could write it as 4i-9j

Answer 44

(-4-1, 2--5) (-5,7)

Answer 45

where i and j are unit vectors
i=<1,0>
j=<0,1>

Answer 46

Ok i think its right, so how will i find magnitude , would i have to find the component vector first

Answer 47

Oh ok, so this I and J this is the terminal point or initial point?

Answer 48

\[v=ai+bj \\ \text{ had maginitude } \sqrt{(a)^2+(b)^2} \text{ which is denoted } |v| \\ \text{ I put ( ) around the } i \text{ and } j \text{ component } \\ \text{ because most people don't square the whole thing when there is a negative } \\ \text{ it normally is written } \sqrt{a^2+b^2}\]

Answer 49

oh ok

Answer 50

so for example if you have v=-i+2j
then |v|=sqrt((-1)^2+2^2)=sqrt(1+4)=sqrt(5)

and another example if you have v=3i-j
then |v|=sqrt(3^2+(-1)^2)=sqrt(9+1)=sqrt(10)

Answer 51

i and j are unit vectors

Answer 52

Wait it did not convert to symbols for me i have to refresh

Answer 53

I didn't use pretty symbols

Answer 54

Oh ok just making sure

Answer 55

sqrt( ) just means principal square root in thise case

Answer 56

I know just sometimes my openstudy glitches but i can write them down

Answer 57

\[v=-i+2j \implies |v|=\sqrt{(-1)^2+2^2}=\sqrt{1+4}=\sqrt{5} \\ v'=3i-j \implies |v'|=\sqrt{3^2+(-1)^2}=\sqrt{9+1}=\sqrt{10}\]

Answer 58

those were the two examples I used above

Answer 59

Oh ok thanks

Answer 60

i went ahead and renamed the other vector v' instead of v

Answer 61

I don't like two different vectors having the same name :p

Answer 62

Lol thats fine! Ive met you before your very good on mathematics especially on unique sets

Answer 63

So i would do this to the unit vectors i got before which were 4 and -9

Answer 64

?

Answer 65

you can write it as <4,-9>
or 4i-9j
but the form they are looking for is <4,-9>

Answer 66

Ok i got sqrt65

Answer 67

and remember i and j are just unit vectors
i=<1,0>
and
j=<0,1>
always!


4i-9j
4<1,0>-9<0,1>
<4,0>+<0,-9>
<4,-9>

the forms are equivalent though as shown here

Answer 68

Im sorry Myinana but i must go i have to go eat, i will be back and hoepfully youll be back!

Answer 69

\[|<4,-9>|=\sqrt{4^2+(-9)^2}=\sqrt{16+81}=...\]

Answer 70

O sorry 97 ill be back

Answer 71

k sqrt(97)
and okay
p.s. it looks like you didn't square both of components fully as I was talking about earlier
don't miss squaring the negative part of the component

Answer 72

see ya later

Answer 73

?

Answer 74

Hi

Answer 75

hi

Answer 76

Ok sorry so Sqrt-97 did i miss something

Answer 77

why you insert a negative

Answer 78

sorry i meant it as atachment

Answer 79

Sqrt97

Answer 80

\[\sqrt97\]

Answer 81

sqrt(97) is correct
where as sqrt(65) incorrect
the note earlier was refering to... what it appeared that you did
it appeared you did sqrt(-9^2+4^2)
instead of sqrt((-9)^2+4^2)

Answer 82

Yes

Answer 83

This is the magnitude?

Answer 84

yep

Answer 85

Wow thank you!

Answer 86

the magnitude is usually denoted as |vector goes here|

Answer 87

\[v=ai+bj \\ \text{ has magnitude } |v|=\sqrt{a^2+b^2}\]

Answer 88

Wait im sorry Myin what do i put when it says terminal point: and then Initial point: is that part of this problem? or is that the I , J you were talking about

Answer 89

if you want to find the vector in standard form (position)
and you are given the initial point of the vector is (a,b) and the terminal point (x,y)
then the vector in standard position is <x-a,y-b>

Answer 90

i and j are unit vectors

Answer 91

i is the vector <1,0>
where as j is the vector <0,1>

Answer 92

another way to write ai+bj is <a,b>

these two forms are equivlanet
ai+bj
a<1,0>+b<0,1>
<a,0>+<0,b>
<a,b>

Answer 93

Oh ok thank you

Answer 94

I have sent you owlbucks

Answer 95

awww thanks
you didn't have too
i would have helped anyways

but thank you kindly

Answer 96

Have a good night i guess, i will be making a new question if you are interested!