Question

find the eq of the plane that passes through the points:
Show that the lines
L1: x=-1+4t Y=+3+t z=1
L2: x=-13+12t y=1+6t z=2+3t
intersect and find an equation of the plane they determine.

Answer 1

@amistre64

Answer 2

if dot product of two vectors is not zero then these vectors are not intersecting ?
@amistre64

Answer 3

so the plane will look something like

Answer 4

if dot product of two vectors is not zero then these vectors are not intersecting ?

Answer 5

if the dot product of two vectors is zero, then the vectors are perpindicular

Answer 6

yes that i know

Answer 7

the thing i am asking is not correct ?

Answer 8

if dot product of two vectors is not zero then these vectors are not intersecting ?

Answer 9

if the dot product is non zero, we only know that they are not perpindicular. they could be intersecting or they could be parallel

Answer 10

heres an easier approach, we know that a plane is determined by three (non collinear) points

Answer 11

so pick two points from the first line, and then a point from your third line, and you have three collinear points

Answer 12

your graph is not correct
read the statement carefully....written that "find an equation of the plane they determine."
its means that we have to find an eq of such plane which already contains the these two given lines

Answer 13

that was just a generic picture of two intersecting lines, and a plane that contains them

Answer 14

hey, it looks like youre missing the first part of the question
find the eq of the plane that passes through the points:

Show that the lines
L1: x=-1+4t Y=+3+t z=1
L2: x=-13+12t y=1+6t z=2+3t intersect and find an equation of the plane they determine.

Answer 15

two show two lines intersect, you set the lines equal to each other and solve

Answer 16

first we will find the point of intersecting by these two lines then we check weather obtained point satisfy the given eq if yes or not.... we will state that after that we will take P2-P1 thus we will have a vector then we will cross product of this vector and any direction vector of given lines

Answer 17

alternatively, one could also show that L1 is not a multiple of L2

Answer 18

then we will have our required eq of the plane

Answer 19

m i right?

Answer 20

sorry im having trouble understanding, and the website is a laggy

Answer 21

you want to take the vector cross product?

Answer 22

thanks man.... we will help each other next time thank a lot

Answer 23

if v x w = 0, then v and w are parallel

Answer 24

so use that fact

Answer 25

For two lines to be parallel, the vectors defining their slopes have to be parallel, which means their vectors have to be constant multiples of one another.

Answer 26

thanks brother i solved it already

Answer 27

oh ok

Answer 28

is this from a book?

Answer 29

thanks for your help

Answer 30

yes calculus

Answer 31

https://answers.yahoo.com/question/index?qid=20071028224751AA98eZG

Answer 32

i did my question with same concept as explained in yahoo link

Answer 33

just to keep in practice :)

L1:
x=-1+4t
y= 3+1t
z= 1+0t

L2:
x=-13+12t
y= 1+ 6t
z= 2+ 3t

IF these lines are in teh same plane, they the cross product of their direction vectors will give us the normal to that plane.

(4,1,0) x (4,2,1) = (1,-4,4)

using either anchor point we can develop the plane equation that contains at least on of the lines: ill use (-1,3,1)

plane: 1(x+1)-4(y-3)+4(z-1)=0
------------------------------------

now, we know 2 things. the normal vector is perpendicular to both lines, and all the point of L1 is contained in the plane. if any point in L2 is in the plane, then L2 is also in the plane. Use the anchor point for L2 (-13,1,2) to see if its valid.

1(-13+1)-4(1-3)+4(2-1)

-12+8+4 = 0 .... both lines fit in the same plane.
-------------------------

do they intersect? well, since they exist are in the same plane, and they do not have the same unit direction vectors (they have different slopes) they have to intersect.

L1:L2
x:-1+4t =-13+12s
y: 3+1t = 1+ 6s
z: 1+0t = 2 + 3s <--- s=-1/3
------------------------------

x:-1+4t =-17
y: 3+1t = -1 <-- t = -4
z: 1 = 1
-----------------------------

x:-1-16 =-17
y: 3- 4 = -1
z: 1 = 1


they cross when t = -4 and s = -1/3