12-1 - 0 32 If this were in the xy-plane it would be a...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Solutions to Homework Assignment 1 MATH 249-01 and -02 Section 12.1, Page 797 1-9, 11, 15-18, 23-33 odd, 35-38, 41 2. -10 -10 -10 -5 -5 -5 0 00 5 5 5 10 10 10 4. The length of the diagonal is radicalbig (2 - 0) 2 + (3 - 0) 2 + (5 - 0) 2 = 38 units. 5. We will see in Section 12.5 that this is an equation of a plane, as it appears to be. In fact, it is a vertical plane. 4 y -4 2 -4 -2 -2 00 0 z 2 x 2 -2 4 4 -4 7. PQ = radicalbig ( - 2 - 1) 2 + (4 - 2) 2 + (0 - ( - 1) 2 = 14 . PR = radicalbig ( - 2 - ( - 1)) 2 + (4 - 1) 2 + (0 - 2) 2 = 14 . QR = radicalbig (1 - ( - 1)) 2 + (2 - 1) 2 + ( - 1 - 2) 2 = 14 . Since all three sides have the same length, the triangle is equilateral. 9. (a) The points are collinear if and only if the distances add up correctly. We have AB = radicalbig (5 - 7) 2 + (1 - 9) 2 + (3 - ( - 1)) 2 = 84 = 2 21 . AC = radicalbig (5 - 1) 2 + (1 - ( - 15) 2 ) + (3 - 11) 2 = 336 = 4 21 . BC = radicalbig (7 - 1) 2 + (9 - ( - 15)) 2 + ( - 1 - 11) 2 = 756 = 6 21 . Since AB + AC = BC, the three points are collinear (and A is between B and C.
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
16. x 2 + y 2 + z 2 = 4 x - 2 y x 2 - 4 x + y 2 + 2 y + z 2 = 0 ( x 2 - 4 x + 4) + ( y 2 + 2 y + 1) + z 2 = 4 + 1 ( x - 2) 2 + ( y + 1) 2 + z 2 = 5 . The center is thus (2 , - 1 , 0) and the radius is
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: , , 0) . 32. If this were in the xy-plane, it would be a circle of radius 1 centered at the origin. Since it is in R 3 (and z can have any value), it is instead an inFnite cylinder of radius 1 with the z-axis as its axis. 36. 0 ≤ x ≤ 1 , ≤ y ≤ 2 , ≤ z ≤ 3 . 41. In order for a point ( x,y,z ) to be equidistant from both (-1 , 5 , 3) and (6 , 2 ,-2) , it must satisfy the equation r ( x + 1) 2 + ( y-5) 2 + ( z-3) 2 = r ( x-6) 2 + ( y-2) 2 + ( z + 2) 2 . This leads to ( x + 1) 2 + ( y-5) 2 + ( z-3) 2 = ( x-6) 2 + ( y-2) 2 + ( z + 2) 2 ( x 2 + 2 x + 1) + ( y 2-10 y + 25) + ( z 2-6 z + 9) = ( x 2-12 x + 36) + ( y 2-4 y + 4) + ( z 2 + 4 z + 4) 2 x-10 y-6 z + 35 =-12 x-4 y + 4 z + 44 14 x-6 y-10 z = 9 . This is a plane....
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern