Question

Help with Vectors question?



Consider the following lines:

L1: [x, y] = [3, -2] + t[4, -5]
L2: [x, y] = [1 , 1] + s[7 , k]

a) For what value of k are the lines parallel?
b) For what value of k are the lines perpendicular?

Please show all work. Thanks :)

Answer 1

Hey Callisto :)

Answer 2

Hmm.... what is t and s?!

Answer 3

I dont know :P

Answer 4

That doesn't look like something I've learnt before :(
What I've learn is that

For two vectors
\[m_1i + n_1 j \ and \ m_2i+n_2j\]
(i) if they are //
\[\frac{m_1}{m_2} = \frac{n_1}{n_2}\]

(ii) if they are perpendicular
a.b =0
ie \(m_1m_2+n_1n_2=0\)

That's all I know :(

Answer 5

hmm... maybe they are variables?

Answer 6

s and t just mark the vectors, they scale the vectors to raech all the points on the line

Answer 7

stretch the first to equal the second and then you will be able to solve for k

Answer 8

Can you take me through this, Im new to this concept :P

Answer 9

the t resembles distribution; just push it thru the grouping symbols

Answer 10

[4t , -5t] = [7 , k]

solve for t with the first component and youll know what k has to be

Answer 11

So, 4t = 7?

Answer 12

yes

Answer 13

t=7/4

Answer 14

t=1.75

Answer 15

7/4 is fine; you can decimate it afterwards if it makes you feel better

Answer 16

1.75x-5=k?

Answer 17

k=-8.75

Answer 18

-5t = k

Answer 19

i read your times as an x lol

Answer 20

but t is 7/4 right so -5 times 1.75 right?

Answer 21

Oh sorry about that :P

Answer 22

yes, -5 times 1.75 = k for parallel

Answer 23

How would we do perpendicular?

Answer 24

callisto had that one called

multiply the vectors componentwise; then add the results and equate them to 0

Answer 25

4,-5
7, k
-----
28 -5k = 0

Answer 26

its called a dot product if you wanna get technical :)

Answer 27

So in hat last line we solve for k, and then are we done with the question?

Answer 28

**that

Answer 29

yes, unless you have something hiding underneath it :)

Answer 30

LOL, nope, I dont :P

Answer 31

THANKS FOR YOUR HELP amistre64! You are a life saver!

Answer 32

callisto did well too

Answer 33

I did nothing..... I'm still trying to understand the t and s :S

Answer 34

LOL

Answer 35

think of the t or s, they had to use 2 different scalars just becasue ... think of them as how far you would stretch a slinky to get it to cover 3 feet or 10 feet or whatever feetage

Answer 36

Ok

Answer 37

the vector itself gives us a direction to point in; the scalar allow us to hit all the points in that direction to form a line with

Answer 38

Can you tell me how to write the therefore statements?

Answer 39

therefore, yada yada ?? im no good with that part im afraid :)

Answer 40

LOL so just the lines are parallel when k = -8.75 right?

Answer 41

correct

Answer 42

Ok, THANKS AGAIN + CALLISTO!!

Answer 43

Why did you multiply the number in the bracket by t for t[4, -5], but did nothing for s[7 , k] ?

Answer 44

hmm.... yea I was wondering about that too, and why is the first part of each line left out?

Answer 45

@amistre64 Can you please elaborate :)

Answer 46

we simply were looking to get 4,-5 stretched to the same values as 7,k in order to have an identical vector .... having identical parts. there was no need to alter 7,k in any way

Answer 47

OHHH, I see noww...

Answer 48

Thanks :)

Answer 49

yep

Answer 50

and the things in front of t[.. , ..] and s[.. , ..] are not our concern?

Answer 51

parallel lines are parallel regardless of what points they are attached to

Answer 52

the direction of a line in space is determined by its vector part; not its point part. The point just anchors it to a specific spot

Answer 53

Oh... so those were points? :|

Answer 54

yes; read it as: at this point, anchor this vector, and stretch it to form a line

Answer 55

Wow! Thanks!!!! As I've said, I've never seen such thing before.... Thank you so much!!!

Answer 56

youre welcome