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Show that the line of intersection of the planes \[x+2yz=2\]and\[3x+2y+2z=7\] is parallel to the line \[\langle x,y,x\rangle =\langle 1+6t,35t,24t\rangle\] Find the equation of the plane determined by these two lines.
 one year ago
 one year ago
Show that the line of intersection of the planes \[x+2yz=2\]and\[3x+2y+2z=7\] is parallel to the line \[\langle x,y,x\rangle =\langle 1+6t,35t,24t\rangle\] Find the equation of the plane determined by these two lines.
 one year ago
 one year ago

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richywBest ResponseYou've already chosen the best response.0
I can do the first part, the trouble I am having is finding the plane. How can I do this?
 one year ago

helder_edwinBest ResponseYou've already chosen the best response.1
first, solve the system \[ \large \begin{cases} x+2yz=2\\ 3x+2y+2z=7 \end{cases} \]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
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.
 one year ago

richywBest ResponseYou've already chosen the best response.0
they are parallel lines
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
what are parallel line?
 one year ago

richywBest ResponseYou've already chosen the best response.0
I need the plane defined by the two lines.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
you are given the planes
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
and you are only given one line
 one year ago

richywBest ResponseYou've already chosen the best response.0
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.
 one year ago

richywBest ResponseYou've already chosen the best response.0
now what I need to do is find the plane defined by the given line, and the line of intersection of the two planes.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
it's supposed to be parallel to the line, I am getting confused
 one year ago

richywBest ResponseYou've already chosen the best response.0
it is. but two parallel lines can still define a plane.
 one year ago

helder_edwinBest ResponseYou've already chosen the best response.1
doing what @TuringTest sugested u get the vector \[ \large (6,5,4) \] which the direction of the given line.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
I'll let @helder_edwin take it then, I am a bit confused as to what you're asking
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
^right, that's what the Q seems to be asking
 one year ago

richywBest ResponseYou've already chosen the best response.0
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)
 one year ago

helder_edwinBest ResponseYou've already chosen the best response.1
for the last part. two parallel lines CANNOT define a plane. u need two independent directions (vectors)
 one year ago

richywBest ResponseYou've already chosen the best response.0
why can they not? I see no reason why they can't.
 one year ago

helder_edwinBest ResponseYou've already chosen the best response.1
a plane is twodimensional whareas a line is onedimensional
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
you need a point and a vector to define a plane
 one year ago

helder_edwinBest ResponseYou've already chosen the best response.1
but the point cannot belong to the line.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
right, I mean a point in the plane and a normal vectordw:1349383884636:dw
 one year ago

richywBest ResponseYou've already chosen the best response.0
but I have two lines. so I could get a displacment vector from one line and a point on the other line?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
vectors can be moved, so having two does not define any point
 one year ago

richywBest ResponseYou've already chosen the best response.0
but these are not vectors. These are lines!
 one year ago

richywBest ResponseYou've already chosen the best response.0
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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
but you can't get another line because these vectors are linearly dependent
 one year ago

richywBest ResponseYou've already chosen the best response.0
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?
 one year ago

richywBest ResponseYou've already chosen the best response.0
doing that I find that the point (5/2,1/4,0) is on the intersection of the two planes.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
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
 one year ago

richywBest ResponseYou've already chosen the best response.0
now I have a point and a direction, meaning I have a line. How do I find this line?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
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
 one year ago

richywBest ResponseYou've already chosen the best response.0
ok I have done this and get \[\vec{r}(t)=\langle 5/2+6t, 1/45t,4t \rangle\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
looks reasonable enough
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
oh I see, so now you want the plane between this line and the given one, eh?
 one year ago

richywBest ResponseYou've already chosen the best response.0
ok now I have another line givin in the initial question.\[\langle x,y,x\rangle =\langle 1+6t,35t,24t\rangle\]
 one year ago

richywBest ResponseYou've already chosen the best response.0
so two lines. need this plane!
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
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=P1P2 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
 one year ago

richywBest ResponseYou've already chosen the best response.0
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?
 one year ago

helder_edwinBest ResponseYou've already chosen the best response.1
yes. and by doing that u got the second direction u needed.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
yes @richyw that is right. Sorry for the laps in communication earlier, I was thinking we were just talking about two parallel vectors
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
... and you're welcome :)
 one year ago
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