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Test 1, MATH 550
Instructions: Write your name legibly. Name: There are 100 points. Show all work to get partial credit.
(1) Decompose 6i − 5 j + 2k into a vector parallel, plus a vector perpendicular, to i − 2 j + k. Solution: Let A = i − 2 j + k and B = 6i − 5 j + 2k. Need A·(B−tA) = (i − 2 j + k)·((6 − t)i + (−5 + 2t)j + (2 − t)k) = 0, i.e. 18 − 6 t − 0. Hence t = 3. Thus the parallel part is B|| = 3i − 6 j + 3k and the perpendicular part is B⊥ = 3i + j − k.
(2) Find all vectors of length 2, parallel to the plane x+6y +2z = 5 and perpendicular to the vector 23 i + j − 23 k. Let A = i + 6j + 2k denote the normal vector to the plane and let B = 23 i = j − 23 k. Then A × B = − 6 i + 2j − 3 k is parallel to the plane and perpendicular to B, Now the length of A × B is
36 + 4 + 9 = 7, so the answer to the problem is ±^27 (− 6 i + 2j − 3 k).
(3) a. Find the volume of the parallelepiped whose coterminal edges are arrows representing the vectors A = i − j + k, B = 2i + 4j − 3 k, C = −i + 4j + 4k. Solution: The volume is |[A, B, C]| = 28 + 5 + 12 = 45.
b. Find the area of the base parallelogram spanned by A and B. Solution: First compute A × B = −i + 5j + 6k. Now the area is |A × B| =
c. Find the altitude of the parallelepiped, if the base is taken to be the parallelogram determined by A and B. Solution: The altitude is the volume divided by the area of the base, so equal to √^4562.
(4) Let R(t) = (sin t − t cos t)i + (cos t + t sin t)j + 12 t^2 k be the position vector of a moving particle. a. Find the velocity and speed of the particle.
Solution: v = t sin ti + t cos tj + tk, so v = |v| = t
b. Determine the acceleration of the particle. Solution: a = (sin t + t cos t)i + (cos t − t sin t)j + k.
