Through the point 111 and perpendicular to the plane

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
through the point (1,1,1) and perpendicular to the plane with vector equation The parameter value t for the point of intersection satisfies the equation Substituting this into the equation for the line reveals that the point has coordinates ( x , y , z ) = (5/3,7/3,7/3)
Image of page 1

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

View Full Document Right Arrow Icon
TEST1/MAC2313 Page 2 of 5 ______________________________________________________________________ 5. (5 pts.) What point (x 0 ,y 0 ) is four-fifths of the way from P = (-2,-3) to Q = (48,7) ?? The vector v with initial point P and terminal point Q has coordinates <50,10> in standard position. Consequently, the terminal point of <x 0 ,y 0 > = <-2,-3> + (4/5)<50,10> = <38,8> in standard position provides us with the coordinates for (x 0 ,y 0 ). Vector magic!! ______________________________________________________________________ 6. (5 pts.) Suppose v = <-3,-2, 1> and w = <-1,1,1>. Then v w = (-3)(-1) + (-2)(1) + (1)(1) = 3 - 2 + 1 = 2 ______________________________________________________________________ 7. (5 pts.) Suppose v = <-3,-2, 1> and w = <-1,1,1>. Then v × w = < -3, -(2), -5 > = < -3, 2, -5 > ______________________________________________________________________ 8. (5 pts.) Suppose v = <-3,-2, 1> and w = <-1,1,1>. Then proj w ( v ) = < -2/3, 2/3, 2/3 > , and the component of v perpendicular to w is w 2 = < -7/3, -8/3, 1/3 > . ______________________________________________________________________ The yz-plane part of Problem 20. (5 pts.) Do the three 2-space sketches of the traces in each of the coordinate planes of the surface defined by . z 1 x 2 9 y 2 4
Image of page 2
TEST1/MAC2313 Page 3 of 5 ______________________________________________________________________ 9. (5 pts.) Suppose v = <-3,-4, 5> and w = <-1,1,1>. If α , β , and γ are the direction angles of v , then cos( α ) = , cos( β ) = ,and cos( γ ) = . ______________________________________________________________________ 10. (5 pts.) Suppose v = <-3,-2, 1> and w = <-1,1,1>. What is the exact value of the angle θ between v and w ?? ______________________________________________________________________ 11. (5 pts.) Write a point-normal equation for the plane perpendicular to v = <-3,-2,1> and containing the point (-1,2,-3). -3(x - (-1)) - 2(y - 2) + (z - (-3)) = 0 ______________________________________________________________________ 12. (5 pts.) Which point on the line defined the vector equation <x,y,z> = <1,1,1> + t<2,-1,-1> is nearest the point, (0,-1,0)? Build the vector with initial point (0,-1,0) to an arbitrary point on the line with parameter t , The norm of this vector is the distance from the point (0,-1,0) to the point on the line with parameter value t . The norm will be smallest when the vector is perpendicular to the vector <2,-1,-1>. Consequently, provides us with the value of t needed for the closest point.
Image of page 3

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

View Full Document Right Arrow Icon
Image of page 4
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