chapter_12 - January 27, 2005 11:53 L24-ch12 Sheet number 1...

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January 27, 2005 11:53 L24-ch12 Sheet number 1 Page number 524 black 524 CHAPTER 12 Three-Dimensional Space; Vectors EXERCISE SET 12.1 1. (a) (0 , 0 , 0) , (3 , 0 , 0) , (3 , 5 , 0) , (0 , 5 , 0) , (0 , 0 , 4) , (3 , 0 , 4) , (3 , 5 , 4) , (0 , 5 , 4) (b) (0 , 1 , 0) , (4 , 1 , 0) , (4 , 6 , 0) , (0 , 6 , 0) , (0 , 1 , 2) , (4 , 1 , 2) , (4 , 6 , 2) , (0 , 6 , 2) 2. corners: (2 , 2 , ± 2), (2 , 2 , ± 2), ( 2 , 2 , ± 2), ( 2 , 2 , ± 2) y x z (–2, –2, 2) (–2, 2, 2) (–2, 2, –2) (–2, –2, –2) (2, 2, –2) (2, –2, –2) (2, –2, 2) (2, 2, 2) 3. corners: (4 , 2 , 2), (4,2,1), (4,1,1), (4 , 1 , 2), ( 6 , 1 , 1), ( 6 , 2 , 1), ( 6 , 2 , 2), ( 6 , 1 , 2) (–6, 2, 1) (–6, 2, –2) y x z (–6, 1, –2) (4, 1, 1) (4, 1, –2) (4, 2, 1) 4. (a) ( x 2 ,y 1 ,z 1 ) , ( x 2 2 1 ) , ( x 1 2 1 )( x 1 1 2 ) , ( x 2 1 2 ) , ( x 1 2 2 ) (b) The midpoint of the diagonal has coordinates which are the coordinates of the midpoints of the edges. The midpoint of the edge ( x 1 1 1 ) and ( x 2 1 1 )i s µ 1 2 ( x 1 + x 2 ) 1 1 ; the midpoint of the edge ( x 2 1 1 ) and ( x 2 2 1 s µ x 2 , 1 2 ( y 1 + y 2 ) 1 ; the midpoint of the edge ( x 2 2 1 ) and ( x 2 2 2 )) is µ x 2 2 , 1 2 ( z 1 + z 2 ) . Thus the coordinates of the midpoint of the diagonal are µ 1 2 ( x 1 + x 2 ) , 1 2 ( y 1 + y 2 ) , 1 2 ( z 1 + z 2 ) . 5. (a) a single point on that line (b) a line in that plane (c) a plane in 3 space 6. (a) R (1 , 4 , 0) and Q lie on the same vertical line, and so does the side of the triangle which connects them. R (1 , 4 , 0) and P lie in the plane z = 0. Clearly the two sides are perpendicular, and the sum of the squares of the two sides is | RQ | 2 + | RP | 2 =4 2 +(2 2 +3 2 ) = 29, so the distance from P to Q is 29. y x z P (3, 1, 0) Q (1, 4, 4) R (1, 4, 0)
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January 27, 2005 11:53 L24-ch12 Sheet number 2 Page number 525 black Exercise Set 12.1 525 (b) S (3 , 4 , 0) and P lie in the plane z = 0, and so does SP . S (3 , 4 , 0) and Q lie in the plane y = 4, and so does SQ . Hence the two sides | | and | | are perpendicular, and | PQ | = p | PS | 2 + | QS | 2 =3 2 +(2 2 +4 2 ) = 29. y x z P (3, 1, 0) Q (1, 4, 4) S (3, 4, 0) (c) T (1 , 1 , 4) and Q lie on a line through (1 , 0 , 4) and is thus parallel to the y -axis, and TQ lies on this line. T and P lie in the same plane y = 1 which is perpen- dicular to any line which is parallel to the y -axis, thus TP , which lies on such a line, is perpendicular to . Thus | | 2 = | PT | 2 + | QT | 2 = (4 + 16) + 9 = 29. y x z P (3, 1, 0) Q (1, 4, 4) T (1, 1, 4) 7. The diameter is d = p (1 3) 2 +( 2 4) 2 +(4+12) 2 = 296, so the radius is 296 / 2= 74. The midpoint (2 , 1 , 4) of the endpoints of the diameter is the center of the sphere. 8. Each side has length 14 so the triangle is equilateral. 9. (a) The sides have lengths 7, 14, and 7 5; it is a right triangle because the sides satisfy the Pythagorean theorem, (7 5) 2 =7 2 +14 2 . (b) (2,1,6) is the vertex of the 90 angle because it is opposite the longest side (the hypotenuse). (c) area = (1/2)(altitude)(base) = (1 / 2)(7)(14) = 49 10. (a) 3 (b) 2 (c) 5 (d) p (2) 2 3) 2 = 13 (e) p ( 5) 2 3) 2 = 34 (f) p ( 5) 2 + (2) 2 = 29 11. (a) ( x 1) 2 + y 2 z +1) 2 =16 (b) r = p ( 1 0) 2 +(3 0) 2 0) 2 = 14, ( x 2 y 3) 2 z 2) 2 =14 (c) r = 1 2 p ( 1 0) 2 2) 2 +(1 3) 2 = 1 2 5, center ( 1 / 2 , 2 , 2), ( x +1 / 2) 2 y 2) 2 z 2) 2 =5 / 4 12. r = | [distance between (0,0,0) and (3 , 2 , 4)] ± 1 | = 29 ± 1 , x 2 + y 2 + z 2 = r 2 = ( 29 ± 1 ) 2 =30 ± 2 29 13. ( x 2) 2 y 2 z +3) 2 = r 2 , (a) r 2 2 =9 (b) r 2 =1 2 (c) r 2 =2 2 =4
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January 27, 2005 11:53 L24-ch12 Sheet number 3 Page number 526 black 526 Chapter 12 14. (a)
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This note was uploaded on 06/10/2010 for the course MATH 200-177 taught by Professor Richardwhite during the Spring '10 term at Drexel.

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chapter_12 - January 27, 2005 11:53 L24-ch12 Sheet number 1...

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