Math 21a Practice Hourly 1 Answers(Fall 2000)Note: Problems 1-6 count 8 points each, while Problem 7 counts for 2 points.1. Answer :a) |v| = 21, |w| = 1, |v - 3w| = √6.b) This number is v•w = 4.c) (-2, 0, 1).d) b = -2 and c = 0.2. Answer :a) (3/2, 0, 0), (0, -3/2, 0) and (0, 0, 3). (There are infinitely many other possibilities)b) (2, -2, 1), or any non-zero multiplec) The distance is 1.d) Parametric: Send t → (3t, 3t, 3). Nonparametric: x = y & z = 3 - 2x + 2y. (There are infinitely many other possibilities.)3. Answer :a) (0, -3, 4 π).b) 2 √π (-3, 0, 4)c) t → (-3t, -3, 4 (π + t)).d) The velocity vector at general t is 2t (3 cos(t2), -3 sin(t2), 4) whose length is 10 t.Thus, the distance is the integral of this last function from 0 to √π which is 5 √π.4. Answer :a) p is on L and v is tangent to L.b) t → p + t v.c) (-4, -3, 0); the case of t = -1 in the preceding parameterization.d) d = |p × v|/|v| = 3435.5. Answer :a) The plane where x - 3z = 0. (There are infinitely many other possibilities.)b) w•v = 0 requires b = 16.c) b = -6 and c = -2.d) If s is parallel to v, then u × s will be perpendicular to v. For this, take c = -6.6. Answer :a) True . Indeed, since r•r = |r|2 and ddt(r•r) = 2 r•ddtr, which is zero, the distance of the particle to the origin stays constant. Thus, it moves on the surface of a sphere.b) False : If the particle motion is given by r(t) = t k, then it moves on a line and not a sphere.c) True : See the preceding answer. In fact, any r(t) of the form r(t) = r0 + t k will do.d) True : Write r(t) = (a(t), b(t), c(t)), so k × r = (-b(t), a(t), 0) and k × ddtr = (-ddtb(t), ddta(t), 0). As this is orthogonal to k × r, so addta + bddtb = 0, which says that2−1ddt(a2 + b2) = 0 so a2 + b2 is constant. Thus, the x and y coordinates of the particlemove on a circle while the z coordinate can do what it likes. This puts the motion on acircular cylinder.7. Answer : 0. The vector u × v is perpendicular to both u and
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