Question

Let L be the line of intersection of the two planes:
3x-7y+2z=23
x-2y+3z=6

let L' be the line through the two points A=(2,-1,0) and B=(4,3,-1)

(i) Find in parametric form an expression for the general point of L.
(ii) Write down the vector AB and an expression for the general point of L'
(iii) Determine the point in which L' meets the plane x + y + z = 6.
(iv) Decide whether L meets L'
(v) Decide whether L 'meets the line of intersection M of the two planes x + y + z = 6, x + 2y − 3z = −3

I cant even do the first part i dont quite get how to go about doing it. Any help is gr8

Answer 1

every line in a plane is perpendicular to the plane's normal vector, yes ?

Answer 2

yes. I think ive done the first part

Answer 3

good, you just need direction vector and a point to write out the equation of line

Answer 4

part ii should be pretty straightforward ?

Answer 5

yeah i found the vector AB just by doing a-b...but then im not sure what it wanted by the general point of L'?

Answer 6

vector AB is B-A

Answer 7

sorry yes thats what i meant ><

Answer 8

just write out the equation of line

Answer 9

First you should set your functions as \(\sf z(x,y)= ...\)

Answer 10

Then for (i) your hint is: \(\sf \vec r(t) = (1-t)\vec r_0 +t\cdot \vec r_1\)

Answer 11

create parametric equations first, always always always.

Answer 12

So if you FIRST solve your equations in terms of z, then you can set them equal to each other to find the intersection :) That's what I think.

Answer 13

i did that, i have my equations in terms of z

Answer 14

this looks like what I just learned about line integrals and Green's Theorem...all that yummy stuff.

Answer 15

its ii), i found AB but im not sure how to find the general point of L'

Answer 16

are you done with part i ?

Answer 17

whats the equation of L ?

Answer 18

i believe so, i got an answer at least

Answer 19

Well i solved it in terms of Z, so i have x=7z+11 and y=-7-5z

Answer 20

that wont work

Answer 21

you need equation of form :

L = `point` + t( `direction vector`)

Answer 22

i reduced using REF to eliminate Z

Answer 23

And so when you set them equal to eachother you'll find the intersection? this is my guess <.<

Answer 24

What is REF...

Answer 25

Row echelon form

Answer 26

3x-7y+2z=23
x-2y+3z=6

cross product of (3, -7, 2) and (1, -2, 3) gives you the direction vector of line

Answer 27

because (3,-7,2) and (1, -2, 3) are the normal vectors of both the planes.

Answer 28

Taking their cross product gives you a direction vector.... :D

Answer 29

yes start by finding the cross product jbn

Answer 30

(-25,-7,1)

Answer 31

try again

Answer 32

(-17,-7,1)

Answer 33

\[\sf<-21 +4 ~, -(9-2)~, -6+7>\]

Answer 34

right, so we have direction vector of the line :

L = `point` + t `<-17, -7, 1>`

Answer 35

just find a `point` on the line and you're done with part i

Answer 36

any ideas on how to find a point on the line ?

Answer 37

set them equal and solve for where they intersect?

Answer 38

that would be tedious

Answer 39

you just need some point

Answer 40

so a better idea is to plugin z = 0 and solve the system :

3x-7y+2(0)=23
x-2y+3(0)=6

Answer 41

http://www.wolframalpha.com/input/?i=solve+3x-7y%3D23%2Cx-2y%3D6

so a point on the line is `(-4, -5, 0)`

Answer 42

Or rather, set two variables = 0 and solve 1 at a time,

i.e x=y=0, solve for z, etc.

Answer 43

parametric form of line would be :

L = `<-4, -5, 0>` + t `<-17, -7, 1>`

Answer 44

That was very simple, thank you

Answer 45

Similar method for ii i presume?

Answer 46

yes u just need a `point` and `direction vector` for writing out parametric form of line

Answer 47

find them

Answer 48

part ii

`let L' be the line through the two points A=(2,-1,0) and B=(4,3,-1)`

direction vector = AB = `<2, 4, -1>` yes ?

Answer 49

you have two points already, just pick one and write out the equation

Answer 50

yes thats what i got

Answer 51

use a different parameter to L like s yes?

Answer 52

you can use ur fav letter for parameter, but using t is also okay if you're aware that it is not the same as the t in L

Answer 53

okay gotcha, onto iii i go

Answer 54

so whats ur equation for L'

we need to use it in part iii

Answer 55

(2,-1,0)+s(2,4,-1)

Answer 56

looks good, write it in point form for part iii

Answer 57

L' = (2,-1,0)+s(2,4,-1)


which is same as

L' = (2+2s, -1+4s, -s)

yes ?

Answer 58

yes, and then sub the x y and z values into the equation of the plane?

Answer 59

plug this point in the given plane ` x + y + z = 6`

Answer 60

yes sub in and find the value of s at which the line and plane meet

Answer 61

i get s to be 2

Answer 62

so then sub this back into L' to find the point

Answer 63

try again

Answer 64

im getting s=1

Answer 65

yes s=6

Answer 66

sorry, 1

Answer 67

so now sub the correct s( 1 not 2) back into the L'

Answer 68

yes

Answer 69

L'=(14,23,-5)

Answer 70

L' = (2+2s, -1+4s, -s)

sub in s=1 for the intersection point of line L' and the plane `x+y+z=6`

Answer 71

i keep putting wrong values in ><

Answer 72

L'=(4,3,-5)

Answer 73

try again

Answer 74

(4,3,-1)

Answer 75

looks good

Answer 76

so now to see if L' and L meet, i would set them equal and solve for each parameter, and if they do meet when i sub them back in i would get the same answer. yes?

Answer 77

yes


L = (-4-17t, -5-7t, t)

L' = (2+2s, -1+4s, -s)

set the components equal to each other, you will get 3 equations with 2 unknowns. if the solution exists, the lines meet. otherwise they are skew

Answer 78

Im getting that they dont meet

Answer 79

thats right, the lines done meet and are not parallel. so they are skew

http://www.wolframalpha.com/input/?i=solve+-4-17t%3D2%2B2s%2C-5-7t%3D-1%2B4s%2Ct%3D-s

Answer 80

So for the last part, i need to make another parametric for the intersection of M

Answer 81

yes find the parametric form of M first

Answer 82

then change both M and L' to point forms and set their components equal

Answer 83

solve them.. same story