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Practice 1 Solutions

# Practice 1 Solutions - Math E-21a Practice Exam#1 Solutions...

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1 Math E-21a – Practice Exam #1 Solutions – Fall 2009 1) True/False questions (circle one) – you do not need to show your reasoning: a) The vector 1,2, 4 v is perpendicular to the plane with equation 4 2 2 100 x y z . False : In fact, the vector 1,2, 4 v is parallel to this plane. [It’s perpendicular to the normal vector 4,2,2 n .] b) If the vectors u v and u v are perpendicular, then the vectors u and v have the same length. True : Calculate 2 2 ( ) ( ) 0 u v u v u v . Therefore u v . c) For any two vectors v and w , it is the case that 2 2 2 v w v w . False : 2 2 2 ( ) ( ) 2 v w v w v w v u v w . This will only equal 2 2 v w when either u 0 , v 0 , or u and v are perpendicular. d) For any three vectors u , v , w , the identity ( ) ( ) u v w w v u holds. False : ( ) ( ) ( )   u v w v w u w v u e) If 0 v w , then either 0 v or 0 w or v w . False : This will be the case whenever v and w are parallel. f) If u and v are perpendicular nonzero vectors, then ( ) u v u is parallel to v . True : Draw a picture. 2) Short Answer Questions – show your calculations and reasoning. a) Find the scalar projection (component) of the vector 5,1, 2 u in the direction of the vector 2,1,2   v . Solution : The scalar projection is given by 10 1 4 13 3 4 1 4   v u v . b) Find the vector projection of the vector 5,1, 2 u in the direction of the vector 2,1,2   v . Solution : The vector projection of u in the direction of v is 2 13 Proj 2,1,2 9   v v v u v u u v v v v . c) Find the angle between the vectors 5,1, 2 u and 2,1,2   v . Solution : For the angle between u and v, we use 1 13 cos 142.29 3 30 . d) Find the volume of the parallelepiped which contains the points (1,1,1) A , (2,3,4) B , ( 1,1,3) C   , and (3,3,2) D , and where AB , AC , and AD are adjacent edges.

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Practice 1 Solutions - Math E-21a Practice Exam#1 Solutions...

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