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Vector projection problem

  1. Feb 17, 2008 #1
    Let [tex]\vec{x}[/tex] and [tex]\vec{y}[/tex] be vectors in Rn and define

    [tex]p = \frac{x^Ty}{y^Ty}y[/tex]
    [tex]z = x - p[/tex]

    (a) Show that [tex]\vec{p}\bot\vec{z}[/tex]. Thus [tex]\vec{p}[/tex] is the vector projection of x onto y; that is [tex]\vec{x} = \vec{p} + \vec{z}[/tex], where [tex]\vec{p}[/tex] and [tex]\vec{z}[/tex] are orthogonal components of [tex]\vec{x}[/tex], and [tex]\vec{p}[/tex] is a scalar multiple of [tex]\vec{y}[/tex]

    (b) If [tex]||\vec{p}|| = 6[/tex] and [tex]||\vec{z}|| = 8[/tex], determine the value of [tex]||\vec{x}||[/tex]

    My problem:
    I understand the question, but have no idea how to approach it. Hints?
  2. jcsd
  3. Feb 18, 2008 #2
    Draw any arbitrary two vectors, draw the projection of one onto the other and stare at it until you realize that [tex] \sqrt{6^2 + 8^2} [/tex] is obvious.
  4. Feb 18, 2008 #3


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    Science Advisor

    And then think "3-4-5 right triangle" twice! :rofl:

    Your "[itex]x^T y[/itex] is a fancy way of writing the dot product. In my simpler mind, what you really want to prove is that
    [tex]\frac{\vec{x}\cdot\vec{y}}{||\vec{x}||} \vec{x}[/tex]
    is perpendicular to [itex]\vec{p}- \vec{x}[/itex].

    Okay, go ahead and take the dot product:
    [tex]\frac{\vec{x}\cdot\vec{y}}{||\vec{x}||} \vec{x}\cdot (\vec{x}- \frac{\vec{x}\dot\vec{y}}{||\vec{x}||} \vec{x})[/itex]
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