Vectors: The line l1 passes through the point A with position vector i - 2j - 2k and is parallel to the...?

Question

The line l1 passes through the point A with position vector i - 2j - 2k and is parallel to the vector 3i - 4j - 2k. The variable line l2 passes through the point (1 + 5 cos t)i - (1 + 5sin t)j - 14k, where 0 ≤ t ≤ 2π, and is parallel to the vector 15i + 8j - 3k. The points P and Q are on l1 and l2 respectively, and PQ is perpendicular to both l1 and l2.

Find the length of PQ in terms of t.

Oh, gosh. How on earth do I start? The cos and sin t look horribly menacing. Help'd be very much appreciated, and workings, too. My lecturer hasn't really gotten into such complicated questions, so this one came as quite a shock to me, and I've absolutely no idea how to go about answering this.

Answer 1

This is a complicated way of asking how to find the minimum distance between two skew lines.

L1 = A + ru
L2 = B + sv
where u and v are directional vectors for the lines and
where r and s are scalars that range over the real numbers

L1 = <1, -2, -2> + r<3, -4, -2>
L2 = <1 + 5cost, -1 - 5sint, -14> + s<15, 8, -3>
where r and s are scalars that range over the real numbers

The shortest distance between the two lines will be along a line segment PQ that is perpendicular to both lines. Take the cross product of the directional vectors of the two lines.

n = u X v = <28, -21, 84>

Any non-zero multiple of n is also a normal vector to the two lines. Divide by 7.

n = <4, -3, 12>

Take one point on each line by setting r and s equal to zero.

A(1, -2, -2) and B(1 + 5cost, -1 - 5sint, -14)

AB = B - A = <5cost, 1 - 5sint, -12>

The distance between the two lines is the projection of AB onto the unit vector n / || n ||.

| AB • n | / || n || = <5cost, 1 - 5sint, -12> • <4, -3, 12> / || n ||
= | 20cost - 3 + 15sint - 144 | / √[4² + (-3)² + 12²]
= | 20cost + 15sint - 147 | / √(16 + 9 + 144)

| AB • n | / || n || = | 20cost + 15sint - 147 | / 13

Answer 2

First determine the direction of PQ. Since it is perpendicular to the given directions, you can find it as their cross product:

PQ ~ (3, -4, -2) x (15, 8, -3)
... = (12 + 16, -30 + 9, 24 + 60)
... = (28, -21, 84) = 28i - 21j + 84k.
For simplicity, I reduce this vector by dividing by 7: PQ ~ n = (4, -3, 12).

Think of this vector as the normal (perpendicular) vector to two planes; plane I contains point A, line l1 and point P; and plane II contains the variable point (which, by the way, moves along a circle), line l2 and point Q. The distance between P and Q equals the distance between the points.

The equations for planes I and II can now be written as n . p = k1 or k2, that is,
... 4 x - 3 y + 12 z = k1 or k2;
k1 and k2 can be determined from the given points, A and the point on the circle.

As for point A,
... 4.1 - 3.(-2) + 12.(-2) = -14 = k1
and for the other point
... 4.(1 + 5 cos t) + 3.(1 + 5 sin t) + 12.(-14)
... ... = -161 + 20 cos t + 15 sin t = k2.

The distance between P and Q is now equal to the distance between the planes. A nice formula for it is

D = |k1 - k2| / |n| =
... = |-14 + 161 - 20 cos t + 15 sin t| / 13 =
... = (147 - 20 cos t + 15 sin t) / 13.

Note: |n| = 13 because |n|^2 = 3^2 + 4^2 + 12^2 = 169.
Moreover, we can drop the absolute value signs because |sin| and |cos| can't be greater than one, so 147 - 20 cos t + 15 sin t is at least 112.

Answer 3

see the link. a particular example is worked out at the bottom. hope it helps.