Please help me with this question(URGENT):
Find the equation of the plane through the line
and parallel to the line
help me please...
Call it (p).
The line r=(1,-3,4)+t(2,1,1) is parallel to the vector (2,1,1). Since (p) contains r, the vector (2,1,1) is parallel to (p) also.
Now (p) is also parallel to r=s(1,2,3), so it is parallel to (1,2,3).
------To find its parametric equation, we use the fact that planes are generated by two linearly independent parallel vectors:
r=a(2,1,1)+b(1,2,3), for a,b real.
-----To find its equation in (x,y,z), we need a vector perpendicular to (p). For a perpendicular vector, we find the cross product of the two vectors parallel to (p).
So, a vector normal to the surface is (2,1,1)x(1,2,3)=(1,-5,1).
A point on (p) can be found by assigning a random value for t in the equation of the line r=(1,-3,4)+t(2,1,1), say t=0: r=(1,-3,4).
If (x,y,z) is a point (p), then (x,y,z)-(1,-3,4) is a vector parallel to the plane, and must be normal to (1,-5,1). We now have the equation of the plane: (1,-5,1) (x-1, y+3,z-4)=0 or (x-1)-5(y+3)+(z-4)=0.
call the plane p. Since p contains the given line the point A(1, -3, 4) belongs to the plane too. Now you need two linear independent vectors to span(?) the plane. These vectors are known to you. If is the position vector of/to every point of the plane the equation of the plane is: