Results 1 to 2 of 2

Math Help - Vector/projection question

  1. #1
    VkL
    VkL is offline
    Member
    Joined
    Oct 2008
    Posts
    96

    Vector/projection question

    a.) Write v = 1i + 1j + 1k as the sum of two vectors, one parallel to and one perpendicular to the vector v=4i + 2j + 0k,

    i.e - find the projection of u on to v and v on to u.

    ??

    Any help would be appreciated. thank you
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,389
    Thanks
    1325
    Quote Originally Posted by VkL View Post
    a.) Write v = 1i + 1j + 1k as the sum of two vectors, one parallel to and one perpendicular to the vector v=4i + 2j + 0k,

    i.e - find the projection of u on to v and v on to u.

    ??

    Any help would be appreciated. thank you
    No, the "i.e." is NOT what the question is asking.

    Do you understand what "projection" means? Draw the two vectors, starting at the same point. Draw a line from the tip of v perpendicular to u. The vector from the base of u to the foot of that perpendicular is the "parallel projection of v on u" (typically what is meant by just "projection of v on u". The vector from the foot of that perpendicular to the tip of v is the "perpendicular projection of v on u". Those two vectors add to v.

    To find the projetion of u on v, you would draw the perpendicular from the tip of u, a completely different construction.

    There is probably a formula in your book but:

    You have a right triangle in which v is the hypotenuse, and the length of the projection of v on u is the near side, with angle \theta the angle between the two vectors so the length of the projection is |v|cos(\theta).

    You may also recall that the "dot product" of two vectors, u and v, can be defined by u\cdot v= |u||v|cos(\theta) so that |v|cos(\theta)= \frac{u\cdot v}{|u|}.

    To get a vector with that length in the direction of u, multiply by the unit vector in that direction, [tex]\frac{u}{|u|}.

    The (parallel) projection of v on u is \frac{u\cdot v}{|u|^2} u.

    The perpendicular projection is v minus that vector.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Vector projection
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 13th 2011, 09:13 AM
  2. Projection of a vector
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: November 28th 2010, 02:13 PM
  3. Vector projection question
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: September 28th 2010, 07:47 AM
  4. 3D vector projection
    Posted in the Algebra Forum
    Replies: 8
    Last Post: May 19th 2010, 06:28 PM
  5. vector projection
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 15th 2008, 12:11 PM

Search Tags


/mathhelpforum @mathhelpforum