# Chapter 10 - SECTION 10.1(PAGE 542 R A ADAMS CALCULUS...

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SECTION 10.1 (PAGE 542) R. A. ADAMS: CALCULUS CHAPTER 10. VECTORS AND COORDI- NATE GEOMETRY IN 3-SPACE Section 10.1 Analytic Geometry in Three Dimensions (page 542) 1. The distance between ( 0 , 0 , 0 ) and ( 2 , 1 , 2 ) is ± 2 2 + ( 1 ) 2 + ( 2 ) 2 = 3 units. 2. The distance between ( 1 , 1 , 1 ) and ( 1 , 1 , 1 ) is ± ( 1 + 1 ) 2 + ( 1 + 1 ) 2 + ( 1 + 1 ) 2 = 2 3 units . 3. The distance between ( 1 , 1 , 0 ) and ( 0 , 2 , 2 ) is ± ( 0 1 ) 2 + ( 2 1 ) 2 + ( 2 0 ) 2 = 6 units . 4. The distance between ( 3 , 8 , 1 ) and ( 2 , 3 , 6 ) is ± ( 2 3 ) 2 + ( 3 8 ) 2 + ( 6 + 1 ) 2 = 5 3 units. 5. a) The shortest distance from ( x , y , z ) to the xy -plane is | z | units. b) The shortest distance from ( x , y , z ) to the x -axis is ± y 2 + z 2 units. 6. If A = ( 1 , 2 , 3 ) , B = ( 4 , 0 , 5 ) , and C = ( 3 , 6 , 4 ) , then | AB |= ± 3 2 + ( 2 ) 2 + 2 2 = 17 | AC ± 2 2 + 4 2 + 1 2 = 21 | BC ± ( 1 ) 2 + 6 2 + ( 1 ) 2 = 38 . Since | | 2 +| AC | 2 = 17 + 21 = 38 =| | 2 , the triangle ABC has a right angle at A . 7. If A = ( 2 , 1 , 1 ) , B = ( 0 , 1 , 2 ) , and C = ( 1 , 3 , 1 ) , then c ± ( 0 2 ) 2 + ( 1 + 1 ) 2 + ( 2 + 1 ) 2 = 3 b AC ± ( 1 2 ) 2 + ( 3 + 1 ) 2 + ( 1 + 1 ) 2 = 3 a ± ( 1 0 ) 2 + ( 3 1 ) 2 + ( 1 + 2 ) 2 = 26 . By the Cosine Law, a 2 = b 2 + c 2 2 bc cos ± A 26 = 9 + 9 18 cos ± A ± A = cos 1 26 18 18 116 . 4 . 8. If A = ( 1 , 2 , 3 ) , B = ( 1 , 3 , 4 ) , and C = ( 0 , 3 , 3 ) , then | ± ( 1 1 ) 2 + ( 3 2 ) 2 + ( 4 3 ) 2 = 2 | AC ± ( 0 1 ) 2 + ( 3 2 ) 2 + ( 3 3 ) 2 = 2 | ± ( 0 1 ) 2 + ( 3 3 ) 2 + ( 3 4 ) 2 = 2 . All three sides being equal, the triangle is equilateral. 9. If A = ( 1 , 1 , 0 ) , B = ( 1 , 0 , 1 ) , and C = ( 0 , 1 , 1 ) , then | |=| AC 2 . Thus the triangle is equilateral with sides 2. Its area is, therefore, 1 2 × 2 × ² 2 1 2 = 3 2 sq. units. 10. The distance from the origin to ( 1 , 1 , 1 ,..., 1 ) in n is ± 1 2 + 1 2 + 1 2 +···+ 1 = n units. 11. The point on the x 1 -axis closest to ( 1 , 1 , 1 1 ) is ( 1 , 0 , 0 0 ) . The distance between these points is ± 0 2 + 1 2 + 1 2 1 2 = n 1 units. 12. z = 2 is a plane, perpendicular to the z -axis at ( 0 , 0 , 2 ) . x y z z = 2 2 Fig. 10.1.12 13. y ≥− 1 is the half-space consisting of all points on the plane y =− 1 (which is perpendicular to the y -axis at ( 0 , 1 , 0 ) ) and all points on the same side of that plane as the origin. x y z y =− 1 1 Fig. 10.1.13 14. z = x is a plane containing the y -axis and making 45 angles with the positive directions of the x - and z -axes. 386

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INSTRUCTOR’S SOLUTIONS MANUAL SECTION 10.1 (PAGE 542) x y z z = x ( 1 , 0 , 1 ) Fig. 10.1.14 15. x + y = 1 is a vertical plane (parallel to the z -axis) passing through the points ( 1 , 0 , 0 ) and ( 0 , 1 , 0 ) . x y z x + y = 1 1 1 Fig. 10.1.15 16. x 2 + y 2 + z 2 = 4 is a sphere centred at the origin and having radius 2 (i.e., all points at distance 2 from the origin). 17. ( x 1 ) 2 + ( y + 2 ) 2 + ( z 3 ) 2 = 4 is a sphere of radius 2 with centre at the point ( 1 , 2 , 3 ) . 18. x 2 + y 2 + z 2 = 2 z can be rewritten x 2 + y 2 + ( z 1 ) 2 = 1 , and so it represents a sphere with radius 1 and centre at ( 0 , 0 , 1 ) . It is tangent to the xy -plane at the origin. x y z ( 0 , 0 , 1 ) x 2 + y 2 + z 2 = 2 z Fig. 10.1.18 19. y 2 + z 2 4 represents all points inside and on the circular cylinder of radius 2 with central axis along the x -axis (a solid cylinder).
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Chapter 10 - SECTION 10.1(PAGE 542 R A ADAMS CALCULUS...

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