Spring 2006 - Gross' Class - Exam 1

# Spring 2006 - Gross' Class - Exam 1 - Midterm #1 Solutions...

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Unformatted text preview: Midterm #1 Solutions Professor Gross Spring 2006 May 2, 2006 1 Green Version 1. Let r = i + 4 j + 3 k s = 2 i + j- k Compute: (a) 3 r- 2 s. 3 r- 2 s = 3( i + 4 j + 3 k )- 2(2 i + j- k ) = 3 i + 12 j + 9 k- 4 i- 2 j + 2 k =- i + 10 j + 11 j (b) r s. r s = (1)(2) + (4)(1) + (3)(- 1) = 2 + 4- 3 = 3 (c) cos , where is the angle between r and s. r s = | r || s | cos 3 = p (1) 2 + (4) 2 + (3) 2 p (2) 2 + (1) 2 + (- 1) 2 cos 3 = 1 + 16 + 9 4 + 1 + 1 cos 3 = 26 6 cos 3 = 156 cos 3 = 2 39 cos cos = 3 2 39 cos = 3 39 2 39 cos = 39 26 1 (d) r s. r s = fl fl fl fl fl fl i j k 1 4 3 2 1- 1 fl fl fl fl fl fl = fl fl fl fl 4 3 1- 1 fl fl fl fl i- fl fl fl fl 1 3 2- 1 fl fl fl fl j + fl fl fl fl 1 4 2 1 fl fl fl fl k =- 7 i + 7 j- 7 k 2. (a) Give a parametric equation for the line passing through the two points P = (1 , ,- 1) , Q = (2 , 1 , 3) . i. We may use the formula r ( t ) = r (1- t ) + r f t , where r is the initial point and r f is the final point. This will give us: r ( t ) = h 1 , ,- 1 i (1- t ) + h 2 , 1 , 3 i t = h 1(1- t ) + 2 t, 0(1- t ) + 1 t,- 1(1- t ) + 3 t i = h 1 + t, t,- 1 + 4 t i , so our parametric equations are: x = 1 + t y = t z =- 1 + 4 t . ii. Another method is to use P as the initial point and use- P Q as the direction vector for the line:- P Q = h 2- 1 , 1- , 3- (- 1) i = h 1 , 1 , 4 i , So we get the vector equation r ( t ) = r + t v = h 1 , ,- 1 i + t h 1 , 1 , 4 i , which again gives the parametric equations x = 1 + t y = t z =...
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## This note was uploaded on 04/29/2008 for the course MATH 20C taught by Professor Helton during the Spring '08 term at UCSD.

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Spring 2006 - Gross' Class - Exam 1 - Midterm #1 Solutions...

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