Help on vector question please!?

Question

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?

Answer 1

It has been a while since I've worked with parametric equations and vector spaces. However, to prove the line is parallel to the plane, all you need do is show that there is no point of intersection between the line and the plane.

To do that, try to find where both line and plane contain the same point. For each variable in the equation of the plane, substitute the equivalent parametric value. That is, where x is in the plane equation, substitute (1 + 2t), and so forth.

Your equation should simplify to an impossibility, like 2 = 3 . Clearly this is impossible, so the line and the plane do not intersect.

To find the distance, you (should) know that the normal vector to the plane, is also normal to the line. Thus you need only to choose one point on the line (such as where t = 0), and use that as the reference point on a line from that point normal to the plane.

If I recall correctly, the normal vector is composed of the coefficients of the plane equations [1, -2, 1]. And your reference point on the line is (1, -1, 2). Constructing the equation for the normal line should yield:

x = n - 1; y = -n + 2; and z = n - 2

Now plug these into the equation of the plane, substituting n-1 for x, etc., to find the n-value where the normal line intersects the plane. (I believe this should be 13/4 ).

Substitute this n-value into the parametric equation for the normal line to get the coordinates (x,y,z) where the line meets the plane.

Then use the distance formula from this plane-normal point, to the initial line reference point, and you should have the distance between the line and the plane.

Answer 2

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

To prove this it is sufficient to show that the directional vector of the line lies in the plane. The directional vector v, of the line is:
v = <2, 3, 4>

The normal vector n, of the plane is:
n = <1, -2, 1>

The normal vector of the plane is perpendicular to every vector the lies in the plane. Therefore the dot product of v and n should equal zero if v lies in the plane.

v • n = <2, 3, 4> • <1, -2, 1> = 2 - 6 + 4 = 0

The dot product is zero. Therefore the vectors are perpendicular and v lies in the plane. Therefore the line and plane are parallel.
_____________

What is the distance between the line and the plane?

Since the line and plane are parallel, every point on the line should be the same distance from the plane. Let t = 0 to find a point P, on the line.

P(1, -1, 2)

Use the distance formula to find the distance from a point to the plane.

distance = | 1*1 - 2*(-1) + 1*2 - 6 | / √[1² + (-2)² + 1²]
distance = |1 + 2 + 2 - 6 | / √(1 + 4 + 1)
distance = 1 / √6