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Math Help - Position vector of point of intersection of line and plane

  1. #1
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    Position vector of point of intersection of line and plane

    The line r=a+tb and the plane r.n=p. b.n is not equal to zero. Prove that the position vector of the point of intersection of the line and plane is a+\frac{(p-a.n)b}{b.n}

    I have no idea how to start this.
    Thanks
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  2. #2
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    Quote Originally Posted by arze View Post
    The line r=a+tb and the plane r.n=p. b.n is not equal to zero. Prove that the position vector of the point of intersection of the line and plane is a+\frac{(p-a.n)b}{b.n}
    Use the equation  \left( {a + tb} \right) \cdot n = p\, \Rightarrow \,a \cdot n + tb \cdot n = p\, \Rightarrow \,t = \frac{{p - a \cdot n}}{{b \cdot n}}.
    That is the t-value of the point of intersection.
    Now substitute it into the line.
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