Results 1 to 7 of 7
Like Tree1Thanks
  • 1 Post By HallsofIvy

Math Help - Vector Proof

  1. #1
    Junior Member
    Joined
    Mar 2014
    From
    Canada
    Posts
    42

    Vector Proof

    Hi all,

    Im working on a vector proof as follows:

    u and v are vectors in R3.

    Prove that |u|v + |v|u bisects the angle between u and v.

    I have tried looking at the dot product formula cos(theta) = u.v/|v||u| but I am a bit stuck.

    I figure I need to be dividing through by the magnitudes of theses vectors to give me unit vectors but not quite sure how to go about things.

    Any help would be greatly appreciated, thanks

    Andy
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,973
    Thanks
    1638

    Re: Vector Proof

    I don't see any need to reduce to unit vectors, although that could be done. Note that the dot product of |u|v+ |v|u with itself is |u||v|(|u|+ |v|+ 2).
    Go ahead and use that formula to calculate the angle |u|v+ |v|u makes with u and the angle it makes with v and compare. You should find that a lot of things cancel.
    Thanks from andy000
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2014
    From
    Canada
    Posts
    42

    Re: Vector Proof

    Hey, sorry, I have probably been at this for too long but I feel like I am missing something.

    i have only been able to get to:

    |v||u| . |v||u| = |u|^2|v| + |u||v|^2 + 2|u||v|

    don't know if I have done that right or where to go next.

    Andy
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,054
    Thanks
    368
    Awards
    1

    Re: Vector Proof

    Quote Originally Posted by andy000 View Post
    Hey, sorry, I have probably been at this for too long but I feel like I am missing something.

    i have only been able to get to:

    |v||u| . |v||u| = |u|^2|v| + |u||v|^2 + 2|u||v|

    don't know if I have done that right or where to go next.

    Andy
    Are you adding or multiplying? |u| and |v| are scalars so all you need to do is multiply them. No fuss, no mess.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor ebaines's Avatar
    Joined
    Jun 2008
    From
    Illinois
    Posts
    1,109
    Thanks
    325

    Re: Vector Proof

    There's an easier way in my opinion, but unfortunately it involves geometry rather than vectors, so perhaps isn't what the assignment is asking for. But nevertheless, here goes:

    Note that |u|v and |v|u have the same magnitude. If you construct a triangle with sides (a) the origin to the tip of |u|v, (b) from there to the the tip of |u|v+|v|u, and (c) from there back to the origin it is an isosceles triangle. If the vector u makes an angle alpha with the horizontal, and the vector v angle theta, then the angle between them is theta-alpha. It's not hard to show that the angle of the vertex of the isosceles triangle is 180+alpha-theta , or 180 minus the angle between u and v; let's call this angle gamma. Since theh other two angles of the isosceles triangle must be equal, and the sum of angles of a triangle is always 180, those other two angles e must each be 1/2 of 180 minus gamma, or 1/2(theta - alpha). Hence the base of the isosceles triangle bisects the angle between u and v.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,792
    Thanks
    1687
    Awards
    1

    Re: Vector Proof

    Quote Originally Posted by andy000 View Post
    Prove that |u|v + |v|u bisects the angle between u and v.
    I have tried looking at the dot product formula cos(theta) = u.v/|v||u| but I am a bit stuck.
    Let \overrightarrow b  = \left\| {\overrightarrow u } \right\|{\overrightarrow v } + \left\| {\overrightarrow v } \right\|{\overrightarrow u } then \frac{{\overrightarrow b  \cdot \overrightarrow v }}{{\left\| {\overrightarrow v } \right\|}} = \frac{{\overrightarrow b  \cdot \overrightarrow u }}{{\left\| {\overrightarrow u } \right\|}}
    To see that note that \overrightarrow b  \cdot \overrightarrow v  = \left\| {\overrightarrow u } \right\|\overrightarrow v  \cdot \overrightarrow v  + \left\| {\overrightarrow v } \right\|\overrightarrow u  \cdot \overrightarrow v  = \left\| {\overrightarrow u } \right\|{\left\| {\overrightarrow v } \right\|^2} + \left\| {\overrightarrow v } \right\|\overrightarrow u  \cdot \overrightarrow v


    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Dec 2012
    From
    Athens, OH, USA
    Posts
    659
    Thanks
    271

    Re: Vector Proof

    Hi,
    Your problem is really two dimensional; work in the plane of vectors u and v. Here's a geometric solution:

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Vector proof
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: April 13th 2010, 05:41 AM
  2. Vector Proof
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 7th 2009, 06:32 PM
  3. Vector Proof
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: November 19th 2008, 07:33 AM
  4. Vector proof
    Posted in the Calculus Forum
    Replies: 3
    Last Post: July 30th 2008, 06:54 PM
  5. Vector proof
    Posted in the Geometry Forum
    Replies: 2
    Last Post: May 6th 2008, 10:00 PM

Search Tags


/mathhelpforum @mathhelpforum