prac1Csol

# prac1Csol - 18.02 Practice Exam 1 Solutions Problem 1. a) P...

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Unformatted text preview: 18.02 Practice Exam 1 Solutions Problem 1. a) P = (1 , , 0), Q = (0 , 2 , 0) and R = (0 , , 3). Therefore QP = 2 and QR = 2 + 3 k . b) cos = QP QR vextendsingle vextendsingle vextendsingle QP vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle QP vextendsingle vextendsingle vextendsingle = ( 1 , 2 , ) ( , 2 , 3 ) 1 2 + 2 2 2 2 + 3 2 = 4 65 Problem 2. a) PQ = ( 1 , 2 , ) , PR = ( 1 , , 3 ) . PQ PR = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle k 1 2 1 3 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = 6 + 3 + 2 k . Then area () = 1 2 vextendsingle vextendsingle vextendsingle PQ PR vextendsingle vextendsingle vextendsingle = 1 2 radicalbig 6 2 + 3 2 + 2 2 = 1 2...
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## This note was uploaded on 04/18/2008 for the course 18 18.02 taught by Professor Auroux during the Fall '08 term at MIT.

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