Proving Distance Between Plane and Point P

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SUMMARY

The discussion focuses on proving the distance between a point P and a plane defined by point Q and a normal vector N in 3-dimensional space. The distance formula derived is \(\frac{|(Q-P) \cdot N|}{\|N\|}\). The method involves expressing points P and Q in Cartesian coordinates and utilizing parametric equations to find the intersection of the line through P in the direction of N with the plane through Q. The solution confirms the distance formula by substituting the coordinates into the plane equation and solving for the intersection point.

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  • Understanding of vector operations in 3D space
  • Familiarity with parametric equations
  • Knowledge of plane equations and their geometric interpretations
  • Proficiency in dot product and vector norms
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  • Learn about parametric equations and their applications in geometry
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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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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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