SolRev1_fall11 - a = ? NO. In fact for any vector a it is...

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MAC 2313 Review Exam 1 09/01/2011 Answers 1. Center has coordinates (1 , 1 , 1) so equation is ( x - 1) 2 + ( y - 1) 2 + ( z - 1) 2 = 1 2. a) 2 b) r ( t ) = (1 - t ) A + t B , where A and B are the position vectors of the points A and B . c) Midpoint= (5 / 2 , 1 / 2 , 3) d) AB = h 1 , - 1 , 0 i e) v = 3 C + 1 2 AB = h 8 , 1 , 9 i f) cos α = 5 / 62, cos β = 1 / 62 and cos γ = 6 / 62 3. 9 x 2 + y 2 + z 2 < π 2 4. a) Suppose a · a = 0, what can we say about a ANS: a = 0 (the zero vector) b) Suppose we know that a × b = 0 , what can we say about a and b ? ANS: Since we’ll have | a × b | = 0, the vectors are parallel. c) Suppose we know they are unit vectors and the angle between them is π 6 , what can we say about a × b and about a · b ANS: We know that | a × b | = sin π/ 6 = 1 / 2 and a · b = cos π/ 6 = 3 / 2 d) If a × a = 0 , does it mean that
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Unformatted text preview: a = ? NO. In fact for any vector a it is true that a × a = 5. | v × w | = √ 122 6. Find the area of the triangle with vertices (0 , , 0), (0 , , 1) and (0 , 1 , 2) ANS= 1 / 2 7. Find | triple product | = 0 Thus the vectors are coplanar. 8. a) h-3 / √ 35 , 5 / √ 35 ,-1 / √ 35 i b) Not parallel (clear). If we equate x’s and y’s we can get s = 0 and t = 0 to solve the system of 2 × 2 but these values of the parameters don’t make the z’s equal, so they don’t intersect. Thus L 1 and L 2 ARE skew....
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This note was uploaded on 12/15/2011 for the course MAC 2313 taught by Professor Keeran during the Spring '08 term at University of Florida.

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