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Math 250 - 2005 Practice Exam #1B 1. ( )( ) (a). 2i + 3j ! 4k " 3i ! 2j ! 5k = _______________ Solution: 20 (b). 2i + 3j + 6k = ____________________ Solution: 7 2. (a). Suppose that a, b and c are mutually perpendicular vectors all of length 3, suppose that v = 3a + 4b + 5c . Then v ! b = ______. Solution: 36 3x 2 y 2 does not exist. ( x, y )!(0,0) x 4 + 3y 4 Consider what happens as we approach along each of the lines y = 0 and y = x . Along y = 0 , all the functional values are 0, while along y = x , all the functional 3 values are . Thus since the limits along different lines have different values, the 4 limit of the function cannot exist. (b). Show that the limit lim 3. Find the equation of the line of intersection of the planes below. 2x + 3y ! z = 5 and 6x + 5y + z = 11 . Express your answer in symmetric form. Solution: A point that lies on the line of intersection is (0, 2, 1). i j k A vector parallel to the line is ( 2,!3,!!1) " (6,!5,!1) = 2 3 !1 = 8i ! 8 j ! 8k . 6 5 1 x y ! 2 z !1 Thus, the equation of the line is = or equivalently = 8 !8 !8 x y ! 2 z !1 . = = 1 !1 !1 4. The lines L1 , L2 below intersect. Find the point of intersection and also determine the equation of the plane that contains the two lines. x ! 2 y +1 z +1 x!2 y!7 z!9 = = L2 : = = . 3 2 4 !1 2 2 Solution: These lines interest at (5, 1, 3). A vector perpendicular to the plane is i j k ( 3,!2,!4 ) ! ( "1,!2,!2 ) = 3 2 4 = "4i " 10 j + 8k . Hence, the equation of the "1 2 2 plane is !4x ! 10y + 8z = !6 . L1 : 5. (a). Find the equation of the line that passes through the point (1, 4, 2) and is perpendicular to the plane 3x + 2 y ! z = 8 . Express your answer in parametric form. Solution: The line is parallel to the vector (3, 2, -1) and so its equation is x(t) = 1 + 3t,!!y(t) = 4 + 2t,!!z(t) = 2 ! t . (b). Find the equation of the plane that contains the line L1 : x = 2t + 1,!y = 3t ! 2,!z = 4t ! 1 and is parallel to the line L2 : x = 3t ! 1,!y = t ! 2,!z = t + 1 . Solution: A vector normal to the plane is n = ( 2i + 3j + 4k ) ! ( 3i + j + k ) = "i + 10 j " k . A point on the plane is (1,!!2,!!1) . And so the equation of the plane is !x + 10y ! z = !19 " x ! 10y + z = 19 . 6. (a). For v = 2i + 3j ! k, u = 6i + 5j ! k , projv u = ______________________. " % v!u 28 Solution: projv u = $ 2 ' v = v = 2v = 4i + 6j ( 2k . 14 $# v '& (b). Given that a, b are vectors with a = 3, b = 4 , and the angle between them is 60° , then proja b = _____________________. Solution: Since a ! b = a ! b cos" = 3 # 4 # a !b 6 1 = =2 = 6 , we get proja b = 3 2 a 7. Suppose that v and u are the vectors along two incident sides of a parallelogram. Show that the area of the parallelogram is v ! u . Solution: See your notes. ! 1 2 3$ 8. (a). Given that # 2 5 3& # & #" 1 0 8&% '1 ! '40 a 9 $ = # 13 '5 '3& , find a, b. # & #" 5 '2 b &% ! 1 2 3$ ! '40 a 9 $ ! 1 0 0$ Solution: We must have # 2 5 3& # 13 '5 '3& = # 0 1 0& . # &# & # & #" 1 0 8&% #" 5 '2 b &% #" 0 0 1&% Now carrying out the multiplications, we see that a ! 10 ! 6 = 0 " a = 16 , and 9 + 0 + 8b = 1 ! 8b = "8 ! b = "1 . 0 3 (b). 2 !1 2 1 3 2 0 4 4 3 0 3 Solution: 2 !1 0 5 = __________________ 1 2 2 1 3 2 0 4 4 3 0 3 4 5 5 = !2 2 4 1 = !2 " 45 = !90 . 1 !1 3 2 2 ( ) x2 y2 z2 + ! = 1. 9. (a). 9 4 1 (i). Sketch the level curve z = 4 . Solution: This is an ellipse. (ii). Sketch the intersection of the plane y = 1 with this surface. Solution: This is a hyperbola. (iii). Describe this surface. Solution: This is a Hyperboloid of one sheet with its major axis along the z-axis. x2 y2 + (b). z = 2 4 (i). What is the intersection of this surface with the xy-plane? Solution: Just the origin. (ii). What is the intersection of this surface with the xz-plane? Solution: This is a parabola. (iii). Describe this surface. Solution: This is a Elliptic Paraboloid with its major axis along the z-axis. (c). Describe this surface z = 4 ! x 2 ! y 2 . Solution: This is the upper hemisphere with center at the origin and radius 2.