How can I show that 2 lines lie in the same plane?
How can I get the equation of that plane?
if the two lines are parallel, then they lie in the same plane. if they are not parallel, they lie in the same plane only if they intersect. you can find whether they intersect simply by setting their equations equal to each other and attempting to solve.
as for finding the equation of the plane, i think a nice way of writing the equation of plane is as a set of vectors perpendicular to the normal vector to the plane, although this is only valid in R3. in general, you can always write the plane as the span of two vectors in the plane.
if the two lines intersect, then use the vectors along each line for the equation. if the two lines are parallel, use the vector along one line for the first, and choose any vector pointing from that line to the second line as your second vector.
Just a little more detail about Lethe's response:
If the two lines are given in parametric form:
l1: x= at+ b, y= ct+ d, z= et+ f
l2: x= ms+ n, y= ps+ q, z= rs+ u
You have 6 equations for 5 unknowns (x, y, z, s, t) which, in general cannot be solved (because,in general, two lines do not intersect).
You can obviously reduce these to at+ b= ms+ n, ct+ d= ps+q, and
et+ f= rs+ u, 3 equations for 2 unknowns. If you can solve these then the two lines intersect so the lines are in the same plane. If you can't, then look at at the vectors in the direction of the lines: ai+ cj+ ek and mi+ pj+ rk. If they are parallel (a= im, c= ip, and e= ir for the same number i) then they lie in the same plane. In any other situation, they do not.
Of course, the "typical" situation for two lines in 3 space is that they do not lie in the same plane.
If the two lines do lie on the same plane then you can find the equation of that plane just as I am sure you have done before:
Take the cross product of the two vectors ai+ cj+ ek and mi+ pj+ rk to find a vector, Ai+ Bj+ Ck, normal to the plane and take either t=0 or s= 0 in the equations of one of the lines to find a point,(x0, y0, z0), on the plane.
You should already know, if you are doing problems like this, that the equation of the plane with normal vector Ai+ Bj+ Ck containing point (x0,y0,z0) is
A(x-x0)+ B(y-y0)+ C(z-z0)= 0.
A nice way to find the equation of a plane in R3 is:
v1 = (d,e,f) - direction of one of the lines
v2 = (u,v,t) - direction of the other line
p = (a,b,c) - One known point
Calculate this determinant equaling 0
| (x - a) u d |
| (y - b) v e | = 0
| (z - c) t f |
Then you get the equation.
(sorry if some of the words are wrong, i've just studied linear algebra in swedish, so im not sure about all the words)
Until you said so, I would never have thought you a non-native speaker of English. You express yourself better than many who are.
Heckuva lot better than my Swedish!
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