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Yeah, Wholly or Fully
is there some theorem we can use?
can you give me a point on vector r
That's all it gives.
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
Why does t=0?
isn't r = 2i+4j+k supposed to be position vector passing through 0,0,0 ?
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?
No. It's negative... I thought it would be too... so that it could be 2=2?
if the line is wholly in the plane, then wouldn't all the points for all t be on the plane?
@dumbcow isn't r supposed to be position vector passing through the origin??
yeah i guess..oh i am interpreting it like its a line equation
well would you agree, you could say x = 2-4t y=4+4t z=1-5t
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
i think we should find this point P, since you can find direction cosines easily
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?
order, i don't think the line is wholly on the plane they seem to going in different directions
yes that is just like this question
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
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
ok, sorry i am not great at 3-d geometry or position vectors :|
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
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
Yes... So how does that work?
<(2-4t, 4+4t, 1-5t), (3, -2, 4)> = 0
it means the line only intersects with the plane at t=0
no, the line is parallel to plane at this value of t
since you are supposed to be verifying...im guessing there's a typo somewhere
i am sure ... this (3, -2, 4 would give the direction cosine of normal of the plane
sorry experimentX i don't follow, but based on the textbook example it doesn't work
right thats the normal vector to the plane
the dot product of this normal vector and our R must be 0 .. for R to be parallel to plane
yes, i beleve that R is <-4,4,-5> ?
but dot product is not zero
this is our R (2-4t, 4+4t, 1-5t)
and this is for normal vector (3, -2, 4)
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
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
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
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
our r is given by r1+txr2 |dw:1333360797528:dw| so we can adjust R
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...
go ahead, yeah thats my feeling too
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...
yeah .. but how do you find the points on r if you don't know where r starts from??
wait.. isn't r a line? so it extends infinitely so that it has no end or beginning
no ...not exactly ... it's a vector.
Verify that the LINE with equation r=2i +4j +k +t(-4i+4j-5k)
i think i was mis-informed .. my mistake.
i think i understand .. it now .. from this analogy. i thought it was just a vector plus r times vector
I still find a sense it what you say the the line is a sum of vectors or a result of that
@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
at least i can say from this picture http://www.ies.co.jp/math/java/vector/chok3D/intro.gif
Oh, Ok. that makes sense :)
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.
@order hope it helps you
Not much, but thanks for trying anyway :D
that's the easyest way to see it