University, Harvard Math 20 Fall 2011, Instructor: Rachel Epstein
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Lines and Planes
September 16, 2011
Find the following line and planes, using either parametric form or an equation. Let P = (1, 2, 3, 4), Q = (2, 3, 4, 4), and R = (0, 0, 2, 0). 1. Find the line through the points P and Q. This line is the set: {Qt + P (1 - t) | t R}. This is the same as the set 1+t 2 + t |tR 3 + t 4 There are other right answers to this problem. To check your answer, see if you can choose t in such a way that you recover the points P and Q. For the answer above, choosing t = 0 and t = 1 works. 2. Find the plane through the points P , Q, and R. This is a bit harder, since we didn't specifically do an example like this in class. We must first convert it to a problem that says, "Find the plane through P , parallel to vectors v and w." But what should v and w be? We can use the vectors P Q and P R, or any other vector through two points on the plane. P Q = Q - P = (1, 1, 1, 0). P R = R - P = (-1, -2, -1, -4). So, the parametric form is 1 -1 1 -2 2 1 + t + u | t, u R , -1 1 3 4 0 -4 which is the same as 1+t-u 2 + t - | 2u t, u R . 3 + t - u 4 - 4u As in the previous problem, there are other right answers. See if you can recover P , Q, and R by choosing appropriate values of t and u. In the
Harvard University, Math 20 Fall 2011, Instructor: Rachel Epstein
2
above solution, when t = 0 = u, we get P , when t = 1 and u = 0, we get Q, and when t = 0 and u = 1, we get R. 3. the hyperplane through P , and perpendicular to the vector R = Find 0 0 . 2 0 In this case, the hyperplane is 3-dimensional. In general, a hyperplane can be any number of dimensions. The solution is the set of all vectors x such that (x - P ) R = 0. Since R has 3 zero coordinates, the only part that really matters is the third coordinate. We get the equation (x3 - 3)2 = 0, which is the same as 2x3 = 6, or simply x3 = 3. If we convert this problem to one in only three dimensions, instead of four, you get the same equation, although you may have written it as z = 3. This plane isn't too hard to picture. It is the plane parallel to the x and y axes at height 3. In four dimensions, it is the 3-dimensional surface parallel to the x1 , x2 , and x4 axes, with x3 = 3. This is not so easy to picture.