Vector distance

  • Thread starter batballbat
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  • #1
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Let P and Q be two points and N a vector space in 3-space. Let P' be the point of intersection of the line through P, in the direction of N, and the plane through Q perpendicular to N. Prove that the distance between the plane and the point P is
[tex]\frac{|(Q-P) \cdot N|}{\|N\|}[/tex]
 
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Answers and Replies

  • #2
HallsofIvy
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I think you mean that N is a vector, not a vector space.

Here's an obvious, not very subtle method: Write P as [itex](p_1, p_2, p_3)[/itex] and Q as [itex](q_1, q_2, q_3)[/itex]. Write the vector N as <A, B, C>. The "the line through P in the direction of N" is given by parametric equations:
[itex]x= At+ p_1[/itex], [itex]y= Bt+ p_2[/itex], [itex]z= Ct+ p_3[/itex]
and the plane containing Q and perpendicular to N by
[itex]A(x- q_1)+ B(y- q_2)+ C(z- q_3)= 0[/itex]

So replace x, y, and z in the equation of the plane by their expressions in the parametric equations and solve for t, thus finding the point at which that line intersects the plane, then find the distance between the points, in terms of [itex]p_1, p_2, p_3, q_1, q_2, q_3[/itex] and A, B, C. Then, using those same values, work out the formula given showing that you get the same result.
 

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