math192_hw2sp06 - HW2 Solutions 12.5 Lines and Planes in...

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

HW2 Solutions 12.5 Lines and Planes in Space 2. x = 1 - 2 t, y = 2 - 2 t, z = - 1 + 2 t . 6. x = 3 + 2 t, y = - 2 - t, z = 1 + 3 t . 22. 3 x + y + z = 5 31. n 1 × n 2 = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle i j k 2 1 - 1 1 2 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = 3 i - 3 j + 3 k is a vector in the direction of the line of intersection of the planes 3( x - 2) + ( - 3)( y - 1) + 3( z + 1) = 0 3 x - 3 y + 3 z = 0 x - y + z = 0 is the desired plane containing P 0 (2 , 1 , - 1). 36. d = radicalBig 14 3 is the distance from S to the line 44. d = 3 2 2 is the distance from S to the plane 56. 2 x - 3 z = 7 2( - 1 + 3 t ) - 3(5 t ) = 7 ⇒ - 9 t - 2 = 7 t = - 1 x = - 1 - 3 , y = - 2 , z = - 5 ( - 4 , - 2 , - 5) is the point
Image of page 1

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

13.1 Vector Functions 4. v = 6 j and a = - 4 i at t = 0 8. for t = - 1 , v ( - 1) = i - 2 j and a ( - 1) = 2 j for t = 0 , v (0) = i and a (0) = 2 j for t = 1 , v (0) = i + 2 j and a (1) = 2 j 12. v = (sec t tan t ) i +(sec 2 t ) j + 4 3 k and a = (sec t tan 2 t +sec 3 t ) i +(2sec 2 t tan t ) j ; Speed vextendsingle vextendsingle v ( π 6 )vextendsingle vextendsingle = 2; Direction: v ( π 6 ) | v ( π 6 ) | = 1 3 i + 2 3 j + 2
Image of page 2
This is the end of the preview. Sign up to access the rest of the 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