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...
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)$\displaystyle \cdot$(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. If $\displaystyle \vec x$ is the position vector of/to every point of the plane the equation of the plane is:
$\displaystyle
p: \vec x = \left(\begin{array}{c}1 \\-3 \\ 4 \end{array} \right) + r \cdot \left(\begin{array}{c} 2 \\1 \\ 1 \end{array} \right) + s \cdot \left(\begin{array}{c} 1 \\ 2\\ 3 \end{array} \right)~, ~r,s~\in~\mathbb{R}$