BRIEF ANSWERS FOR TEST 1

• T or F?
  • T, since v_1 dot v_2 = |v_1|*|v_2|*cos(theta), and in this problem the vectors v_1 and v_2 are unit vectors.
  • F. Let u = (1,0) and v = B = (1,1). Then proj_u B = (1,0) and proj_v B = (1,1), so B = (1,1) is not equal to proj_u B + proj_v B = (2,1).
  • T, both quantities equal the triple product of A, B and C.
  • T, since this is a polynomial equation of degree at most 2.
• The unit tangent vector must be plus or minus (1/sqrt[5], -2/sqrt[5]).
• B = proj_A B + (B - proj_A B) = ((3/2)i + (3/2)j) + ((-3/2)i + (3/2)j + 4k). The first vector is parallel to A = i + j and the second is orthogonal to A.
• Let v_1 and v_2 be the direction vectors of the two lines and let P_1 = (x_1,y_1,z_1), P_2 = (x_2,y_2,z_2) be points lying on each line. One idea for finding the distance between the lines is the following. If N = v_1 x v_2, then N is a normal vector to the plane containing v_1 and v_2. We can construct the plane through P_1 which has normal vector N. (Formula: N dot (x-x_1,y-y_1,z-z_1) = 0.) This plane contains the first line and is parallel to the second line (since v_2 is orthogonal to N). Therefore, the distance between the two lines is the same as the distance between the plane and the second line. The distance we seek agrees with the distance from P_2 and the plane. Thus we have `reduced' the problem to the problem of determining the distance from the point P_2 to the plane: N dot (x-x_1,y-y_1,z-z_1) = 0. This can be solved by the formula given in class: d = |vec[P_1 - P_2] x N|/|N| = |vec[P_1 - P_2] x (v_1 x v_2)|/|(v_1 x v_2)|.
• First we calculate normals to adjacent (planar) faces: N_1 = (0,1,1) x (1,0,1) = (1,1,-1), N_2 = (1,1,0) x (1,0,1) = (1,-1,-1). Next, the angles between planar faces equals the angle between normal vectors, so we use the dot product: cos(theta) = (N_1 dot N_2)/|N_1||N_2| = 1/3. Thus, theta = cos^{-1}(1/3).