Results 1 to 2 of 2

Math Help - vectors

  1. #1
    Member
    Joined
    Aug 2007
    Posts
    239

    vectors

    1. Prove that  \alpha^{2} + \beta^{2} + \gamma^{2} = 1 (direction cosines).

    So  \alpha = \frac{A_x}{|\bold{A}|} ,  \beta = \frac{A_y}{|\bold{A}|} ,  \gamma = \frac{A_z}{|\bold{A}|} which implies  \alpha^{2} + \beta^{2} + \gamma^{2} = 1 .

    2. Show that if  |\bold{A} - \bold{B}| = |\bold{A}+ \bold{B}| then  \bold{A} is perpendicular to  \bold{B} .

    So  \sqrt{(A_x-B_x)^{2} + (A_y-B_y)^{2} + (A_z - B_z)^{2}}  = \sqrt{(A_x+B_x)^{2} + (A_y+B_y)^{2} + (A_z + B_z)^{2}} .

    Then  (A_x - B_x)^{2} = (A_x+B_x)^{2} or  A_{x}^{2} - 2A_{x}B_{x} + B_{x}^{2} = A_{x}^{2} + 2A_{x}B_{x} + B_{x}^{2} so that  A_{x}B_{x} = A_{y}B_{y} = A_{z}B_{z} = 0 which implies that  \bold{A} \cdot \bold{B} = 0 .

    Are these correct?
    Last edited by shilz222; November 26th 2007 at 11:47 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,824
    Thanks
    1717
    Awards
    1
    In # 2, Square both sides.
    \begin{array}{rcl}<br />
 \left\| {A + B} \right\|^2  & = & \left( {A + B} \right) \cdot \left( {A + B} \right) \\   & = & A \cdot A + 2A \cdot B + B \cdot B \\  \end{array}

    \begin{array}{rcl}<br />
 \left\| {A - B} \right\|^2  & = & \left( {A - B} \right) \cdot \left( {A - B} \right) \\ <br />
  & = & A \cdot A - 2A \cdot B + B \cdot B \\  \end{array}

    This implies that A \cdot B = 0
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 15th 2011, 05:10 PM
  2. Replies: 3
    Last Post: June 30th 2011, 08:05 PM
  3. Replies: 2
    Last Post: June 18th 2011, 10:31 AM
  4. [SOLVED] Vectors: Finding coefficients to scalars with given vectors.
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: January 23rd 2011, 12:47 AM
  5. Replies: 4
    Last Post: May 10th 2009, 06:03 PM

Search Tags


/mathhelpforum @mathhelpforum