Verify that the line with equation
r=2i +4j +k +t(-4i+4j-5k) lies wholly in the plane with equation 3x-2y+4z=2

- anonymous

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- perl

wholly?

- anonymous

Yeah, Wholly or Fully

- perl

is there some theorem we can use?

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## More answers

- experimentX

can you give me a point on vector r

- anonymous

That's all it gives.

- dumbcow

the vector normal to plane is <3,-2,4>
if you can show that this is perpendicular to line vector then they are going same direction
finally you have to show they then go through the same point
look at t=0
r = 2i+4j+k, which is point (2,4,1)
the plane also goes through this point --> 3*2 -2*4+4*1 = 2

- anonymous

Why does t=0?

- experimentX

isn't r = 2i+4j+k supposed to be position vector passing through 0,0,0 ?

- dumbcow

to show 2 vectors are perpendicular, the dot product will be 0
<-4,4,-5> * <3,-2,4> = -40 ??
hmm is that maybe a positive 5?

- anonymous

No. It's negative... I thought it would be too... so that it could be 2=2?

- dumbcow

if the line is wholly in the plane, then wouldn't all the points for all t be on the plane?

- anonymous

yes

- experimentX

@dumbcow isn't r supposed to be position vector passing through the origin??

- dumbcow

yeah i guess..oh i am interpreting it like its a line equation

- dumbcow

well would you agree, you could say
x = 2-4t
y=4+4t
z=1-5t

- experimentX

if it is then ... the given plane never passes through the origin and position vector passes through the origin
moreover, when i studied 3d geometry, i studied line as
(x-x1)/l = (y-y1)/n = ... k

- experimentX

|dw:1333358695991:dw|

- experimentX

i think we should find this point P, since you can find direction cosines easily

- anonymous

An example in my textbook says Find whether or not the lines line in th eplane 2x-y-x=1
Would that be an example that could be used for this question too?

- dumbcow

order, i don't think the line is wholly on the plane
they seem to going in different directions

- dumbcow

yes that is just like this question

- experimentX

no, they might align ... since r = r1+r2 .. you are wholly ignoring r2 by putting t=0
my point is they could be parallel, but untill and unless you find the point origin of vector R, it can never be determined if it lies in the plane or not

- anonymous

It answers it like this:
(a) r= (1,2,-1) +s(4,3,5)
Writing r as (x,y,z) gives
x=1+4s
y=2+3s
z=-1+5s
Substituting for x,y,z in the equation of th eplane gives
2(1+4s)-(2+3s)-(-1+5s)=1
---> 1=1

- dumbcow

ok, sorry i am not great at 3-d geometry or position vectors :|

- dumbcow

here are the graphs
http://www.wolframalpha.com/input/?i=plot+x%3D2-4t%2C+y%3D4%2B4t%2C+z%3D1-5t
http://www.wolframalpha.com/input/?i=plot+3x-2y%2B4z%3D2
seems like they are going in opposite directions

- experimentX

|dw:1333359374980:dw|

- dumbcow

order, if you do that same process for this problem you end up with
--> -40t = 0,
the t doesn't cancel because dot product is not zero

- anonymous

Yes... So how does that work?

- experimentX

<(2-4t, 4+4t, 1-5t), (3, -2, 4)> = 0

- dumbcow

it means the line only intersects with the plane at t=0

- experimentX

no, the line is parallel to plane at this value of t

- dumbcow

since you are supposed to be verifying...im guessing there's a typo somewhere

- experimentX

i am sure ... this (3, -2, 4 would give the direction cosine of normal of the plane

- dumbcow

sorry experimentX i don't follow, but based on the textbook example it doesn't work

- dumbcow

right thats the normal vector to the plane

- experimentX

|dw:1333359910407:dw|

- experimentX

the dot product of this normal vector and our R must be 0 .. for R to be parallel to plane

- dumbcow

yes, i beleve that R is <-4,4,-5> ?

- dumbcow

but dot product is not zero

- experimentX

this is our R (2-4t, 4+4t, 1-5t)

- experimentX

and this is for normal vector
(3, -2, 4)

- dumbcow

same thing...
dot product:
3(2-4t) -2(4+4t)+4(1-5t)
=6-12t-8-8t+4-20t
=-40t+2
does not equal 0 for all t

- experimentX

of course it is not equal ... so we find the value of t such that dot product will be equal to zero ... with will give our parallel R

- dumbcow

hmm but then you could make any line parallel to any given plane, as long as you could solve for t
sorry i guess i just don't get what you mean

- dumbcow

experimentX, you shouldn't involve "t" in the dot product since the direction of the line is constant...r2 is the directional vector (like the slope), r1 is like the initial point

- experimentX

our r is given by r1+txr2
|dw:1333360797528:dw|
so we can adjust R

- anonymous

So, is there a typo in the question...? or? I need to put up another question... Do you mind if I give the medals and close it, so you can still try and work it out? I just have a feeling there's a typo...

- dumbcow

go ahead, yeah thats my feeling too

- anonymous

If line r will be wholly on the plane, then the points on r would be also a solution to the equation of our plane right? correct me if I'm wrong...

- experimentX

yeah .. but how do you find the points on r if you don't know where r starts from??

- anonymous

wait.. isn't r a line? so it extends infinitely so that it has no end or beginning

- experimentX

no ...not exactly ... it's a vector.

- anonymous

Verify that the LINE with equation
r=2i +4j +k +t(-4i+4j-5k)

- experimentX

i think i was mis-informed .. my mistake.

- experimentX

i think i understand .. it now .. from this analogy. i thought it was just a vector plus r times vector

- anonymous

I still find a sense it what you say the the line is a sum of vectors or a result of that

- experimentX

@order you still there?? your answer will be
if (2,4,1) satisfies this equation of line 3x-2y+4z=2
and <(3,-2,4),(-4. 5, -5)> = 0 then the line is contained in plane

- experimentX

at least i can say from this picture http://www.ies.co.jp/math/java/vector/chok3D/intro.gif

- anonymous

Oh, Ok. that makes sense :)

- anonymous

One way to define the plane is to take a vector that's perpendicular to it and one point of the plane. This way, all the vectors that lie on the plane would be perpedicular to this chousen vector. In the case of 3x-2y+4z=2 vector (3,-2,4) plays the roll of this perpendicular vector. The roll of the point, takes the 2 on the right side. It's the distance from the origin of coordinates. To find this point, aplie the (3,-2,4) at the origin and find modulus = 2. In the problem you facing, first you would have to check if the point (2,4,1) belongs to the plane. To do that, just plug in the values in the plane equation and check if it is matched. To check if the line is in the plane, is to check if its direction vector is perpendicular to (3,-2,4). You can use dot product, that's the esyest way, i guess. 3*(-4)+(-2)*4+4*(-5) =/ 0 so it is not on the plane.

- anonymous

@order hope it helps you

- anonymous

Not much, but thanks for trying anyway :D

- anonymous

that's the easyest way to see it

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