1. ## Vector planes..

Prove that the line x = 1 + 2t, y = -1 + 3t, z = 2 + 4t is parallel to the plane x - 2y + z = 6.
What is the distance between the line and the plane?

2. Originally Posted by jozzo1000
Prove that the line x = 1 + 2t, y = -1 + 3t, z = 2 + 4t is parallel to the plane x - 2y + z = 6.
What is the distance between the line and the plane?
i do not think this line is parallel to this plane

3. Originally Posted by jozzo1000
Prove that the line x = 1 + 2t, y = -1 + 3t, z = 2 + 4t is parallel to the plane x - 2y + z = 6.
What is the distance between the line and the plane?
not that the vector in the direction of the line is <2,3,4>. we want this to be perpendicular to the normal to the plane, which is <1,-2,1>. as you can see, that does not happen

if you want to find the distance between a plane and a line parallel to it, use the formula:

$\displaystyle D = \frac {|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$

this gives the distance between a point $\displaystyle (x_0, y_0, z_0)$ and the plane $\displaystyle ax + by + cz + d = 0$, just make sure $\displaystyle (x_0, y_0, z_0)$ is a point on the line

4. Yes the line is parallel to the plane: the dot product of the direction vector and the normal vector is zero. To find the distance just take any point in the line and find its distance from the plane.