Here's the question you clicked on:
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.
I can do the first part, the trouble I am having is finding the plane. How can I do this?
first, solve the system \[ \large \begin{cases} x+2y-z=2\\ 3x+2y+2z=7 \end{cases} \]
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.
what are parallel line?
I need the plane defined by the two lines.
you are given the planes
and you are only given one line
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.
now what I need to do is find the plane defined by the given line, and the line of intersection of the two planes.
it's supposed to be parallel to the line, I am getting confused
it is. but two parallel lines can still define a plane.
doing what @TuringTest sugested u get the vector \[ \large (6,-5,-4) \] which the direction of the given line.
I'll let @helder_edwin take it then, I am a bit confused as to what you're asking
^right, that's what the Q seems to be asking
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)
for the last part. two parallel lines CANNOT define a plane. u need two independent directions (vectors)
why can they not? I see no reason why they can't.
a plane is two-dimensional whareas a line is one-dimensional
you need a point and a vector to define a plane
but the point cannot belong to the line.
right, I mean a point in the plane and a normal vector|dw:1349383884636:dw|
but I have two lines. so I could get a displacment vector from one line and a point on the other line?
vectors can be moved, so having two does not define any point
but these are not vectors. These are lines!
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
but you can't get another line because these vectors are linearly dependent
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?
doing that I find that the point (5/2,-1/4,0) is on the intersection of the two planes.
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
now I have a point and a direction, meaning I have a line. How do I find this line?
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
ok I have done this and get \[\vec{r}(t)=\langle 5/2+6t, -1/4-5t,-4t \rangle\]
looks reasonable enough
oh I see, so now you want the plane between this line and the given one, eh?
ok now I have another line givin in the initial question.\[\langle x,y,x\rangle =\langle 1+6t,3-5t,2-4t\rangle\]
so two lines. need this plane!
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
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?
yes. and by doing that u got the second direction u needed.
yes @richyw that is right. Sorry for the laps in communication earlier, I was thinking we were just talking about two parallel vectors
... and you're welcome :)