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 2 years 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.
 2 years 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.

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richyw
 2 years ago
Best ResponseYou've already chosen the best response.0I can do the first part, the trouble I am having is finding the plane. How can I do this?

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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2I 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.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2what are parallel line?

richyw
 2 years ago
Best ResponseYou've already chosen the best response.0I need the plane defined by the two lines.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2you are given the planes

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2and you are only given one line

richyw
 2 years ago
Best ResponseYou've already chosen the best response.0right 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
 2 years ago
Best ResponseYou've already chosen the best response.0now what I need to do is find the plane defined by the given line, and the line of intersection of the two planes.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2it's supposed to be parallel to the line, I am getting confused

richyw
 2 years ago
Best ResponseYou've already chosen the best response.0it is. but two parallel lines can still define a plane.

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

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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2^right, that's what the Q seems to be asking

richyw
 2 years ago
Best ResponseYou've already chosen the best response.0so 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
 2 years ago
Best ResponseYou've already chosen the best response.1for the last part. two parallel lines CANNOT define a plane. u need two independent directions (vectors)

richyw
 2 years ago
Best ResponseYou've already chosen the best response.0why can they not? I see no reason why they can't.

helder_edwin
 2 years ago
Best ResponseYou've already chosen the best response.1a plane is twodimensional whareas a line is onedimensional

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2you need a point and a vector to define a plane

helder_edwin
 2 years ago
Best ResponseYou've already chosen the best response.1but the point cannot belong to the line.

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

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

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

richyw
 2 years ago
Best ResponseYou've already chosen the best response.0but these are not vectors. These are lines!

richyw
 2 years ago
Best ResponseYou've already chosen the best response.0I 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
 2 years ago
Best ResponseYou've already chosen the best response.2but you can't get another line because these vectors are linearly dependent

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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2you 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
 2 years ago
Best ResponseYou've already chosen the best response.0now I have a point and a direction, meaning I have a line. How do I find this line?

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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2looks reasonable enough

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

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

richyw
 2 years ago
Best ResponseYou've already chosen the best response.0so two lines. need this plane!

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2take 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

richyw
 2 years ago
Best ResponseYou've already chosen the best response.0alright 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
 2 years ago
Best ResponseYou've already chosen the best response.1yes. and by doing that u got the second direction u needed.

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

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2... and you're welcome :)
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