Question:
Find the equations o the line that intersects the lines
2x + y - 1 = 0 = x - 2y + 3z (say 1)
3x - y + z - 2 + 2 = 0 = 4x + 5y - 2x - 3 (say 2)
and is parallel to the line x/1 = y/2 = z/3 (say 3)
=============================================
My work so far:
I noted that the RHS of (2) did not have a z coefficient, which I'm not sure is a typo, since this is how it was given in the question.
To solve this, I had an approach to the question as follows:
(1) Equations of general lines passing through (1) & (2) are given by,
2x + y - 1 +k(x - 2y + 3z)----- (4) and
3x - y + z - 2 + 2 + r(4x + 5y - 2x - 3) ----(5) for some real number k &r, respectively.
Let A & B be the points that the required line intersects (4) & (5).
Also, the required line is given to be parallel to the line x/1 = y/2 = z/3
So due to parallelism, the direction ratios of (3) & (4) and (3) & (5) should be proportional at the points of intersection A & B.
i.e. (2+k)/1 = (1-2k)/2 = 3k/3 ---(6) and
(3+2r)/1 = (5r-1)/2 = 1/3 ---(7), which gives,
2+k=k from (6), which gives 2=0(??????)
Am I taking a wrong approach here?
first of all what you state as:
2x + y - 1 = 0 = x - 2y + 3z (say 1)
to me looks like the equation of 2 planes , which are
2x + y -1 = 0 and
x - 2y + 3z = 0
equations of planes take the form ax + by + cz = k
equation of lines cannot be written like that.
The intersection of those two planes would be a line,
as would be the intersection of 3x - y + z - 2 + 2 = 0 = 4x + 5y - 2x - 3
I also noticed that
4x + 5y - 2x - 3 = 0 probably should say: 4x + 5y - 2z - 3 = 0
otherwise why not add up the x terms ??
Look over your typing carefully, and see if you have some typos.
(btw, it is just about midnight where I live, so I am cashing in right now.
Unless oobleck is still hanging around, you might not get an answer until
the morning)