Please help me with this question(URGENT):
Find the equation of the plane through the line
r=(1,-3,4)+t(2,1,1)
and parallel to the line
r=s(1,2,3)
help me please...
Welcome to Math Help Forum - Click here to Register
Welcome to the largest Math Help Forum, a free community dedicated to math help and math discussions.
We welcome everyone and the community is free to join so register today and become part of our math family!
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.
Hello,
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. Ifis the position vector of/to every point of the plane the equation of the plane is:
![]()