The vector

The unit vector is the vector in the same direction, but with unit length. So

To show that is perpendicular to , use the dot product. The dot product of two perpendicular vectors is 0.

Now this is just silly to me. Vectors can be moved all over the place without changing them, just as long as their length and direction aren't changed.

I think what the problem is trying to say is "Place the tails of the vectors and at the origin. Then let S be the midpoint..." Only in this way is the wording unambiguous.

So let's assume the rewritten problem is right. The heads of and are at (2, 3, 1) and (4, -5, 21), respectively. The midpoint S of these points is (3, -1, 11). So the vector is

The vector form of a line in 3-D is

where is the position vector of a point on the line, is the position vector to a point that the line passes through, and is a vector, and t is a scalar parameter generating the points along the line. (The line will be parallel to .)

So for L1 we want:

For c) consider this: What was the vector that forced L1 to be parallel to ? Similarly what is the vector that L2 is parallel to? Once you have these answers, the cross product of two parallel vectors is 0. Is this true for the L1 and L2 vectors?

For d) you know that the (x, y, z) coordinate points of L1 are (3 + 2t, -1 + 3t, 11 + t). For L2 they are (5 - 2s, 10 + 5s, 10 - 3s). How can you find if they have a point in common? (ie. for what value of s and t are the two points the same?)

-Dan