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.
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 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?)