richyw
  • richyw
Show that the line of intersection of the planes \[x+2y-z=2\]and\[3x+2y+2z=7\] is parallel to the line \[\langle x,y,x\rangle =\langle 1+6t,3-5t,2-4t\rangle\] Find the equation of the plane determined by these two lines.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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richyw
  • richyw
I can do the first part, the trouble I am having is finding the plane. How can I do this?
helder_edwin
  • helder_edwin
first, solve the system \[ \large \begin{cases} x+2y-z=2\\ 3x+2y+2z=7 \end{cases} \]
TuringTest
  • TuringTest
I would do it a bit differently I would cross the normal vectors of the two planes. The resulting vector should point in the same direction as the line.

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richyw
  • richyw
they are parallel lines
TuringTest
  • TuringTest
what are parallel line?
TuringTest
  • TuringTest
lines*
richyw
  • richyw
I need the plane defined by the two lines.
TuringTest
  • TuringTest
you are given the planes
TuringTest
  • TuringTest
and you are only given one line
richyw
  • richyw
right and I have already found the direction of the line defined by the intersection of the two planes, and shown that it is perpendicular to the given line.
richyw
  • richyw
now what I need to do is find the plane defined by the given line, and the line of intersection of the two planes.
TuringTest
  • TuringTest
it's supposed to be parallel to the line, I am getting confused
richyw
  • richyw
it is. but two parallel lines can still define a plane.
helder_edwin
  • helder_edwin
doing what @TuringTest sugested u get the vector \[ \large (6,-5,-4) \] which the direction of the given line.
TuringTest
  • TuringTest
I'll let @helder_edwin take it then, I am a bit confused as to what you're asking
TuringTest
  • TuringTest
^right, that's what the Q seems to be asking
richyw
  • richyw
so what I did was find a point on the line of intersection which I did by setting z=0 and got the point (5/2,-1/4, 0)
helder_edwin
  • helder_edwin
for the last part. two parallel lines CANNOT define a plane. u need two independent directions (vectors)
TuringTest
  • TuringTest
indeed
richyw
  • richyw
why can they not? I see no reason why they can't.
helder_edwin
  • helder_edwin
a plane is two-dimensional whareas a line is one-dimensional
TuringTest
  • TuringTest
you need a point and a vector to define a plane
helder_edwin
  • helder_edwin
but the point cannot belong to the line.
TuringTest
  • TuringTest
right, I mean a point in the plane and a normal vector|dw:1349383884636:dw|
richyw
  • richyw
but I have two lines. so I could get a displacment vector from one line and a point on the other line?
TuringTest
  • TuringTest
vectors can be moved, so having two does not define any point
richyw
  • richyw
but these are not vectors. These are lines!
richyw
  • richyw
I have one line, and If I could find the other, then I would have TWO LINES. I cannot visualize how there is more than one unique plane determined by these two lines
TuringTest
  • TuringTest
but you can't get another line because these vectors are linearly dependent
richyw
  • richyw
ok. what I need to do is find the equation of the line formed by the intersection of the two planes. This line has the direction (6,-5,-4), now I just need a point along that line. So I can just solve for where those planes equal eachother right?
richyw
  • richyw
doing that I find that the point (5/2,-1/4,0) is on the intersection of the two planes.
TuringTest
  • TuringTest
you just need one point of intersection, then use the vector you get from crossing the two normal vectors of the plane to write the equation of the line
richyw
  • richyw
now I have a point and a direction, meaning I have a line. How do I find this line?
TuringTest
  • TuringTest
use r(t)=P+tv where P is a point of intersection of the planes and v is the vector that points in that direction
richyw
  • richyw
ok I have done this and get \[\vec{r}(t)=\langle 5/2+6t, -1/4-5t,-4t \rangle\]
TuringTest
  • TuringTest
looks reasonable enough
TuringTest
  • TuringTest
oh I see, so now you want the plane between this line and the given one, eh?
richyw
  • richyw
ok now I have another line givin in the initial question.\[\langle x,y,x\rangle =\langle 1+6t,3-5t,2-4t\rangle\]
richyw
  • richyw
so two lines. need this plane!
TuringTest
  • TuringTest
take the given point P1=(1,3,2) and the point you just found P2=(5/2,-1/4,0) and make a vector u=P1-P2 then cross that with the vector for the line of intersection uXv=n (n is the normal vector of the plane you are looking for) then you can use n(dot)P2=n(dot)P where P=(x,y,z) to find the equation of that plane
richyw
  • richyw
alright cool thanks a lot! I got super confused when helder_edwin said "two parallel lines CANNOT define a plane. u need two independent directions (vectors)". So just to confirm I can get a second direction vector by subtracting a point on line 1 from a point on line two, correct?
helder_edwin
  • helder_edwin
yes. and by doing that u got the second direction u needed.
TuringTest
  • TuringTest
yes @richyw that is right. Sorry for the laps in communication earlier, I was thinking we were just talking about two parallel vectors
TuringTest
  • TuringTest
... and you're welcome :)

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