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Math Help - The perpendicular distance-vector

  1. #1
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    The perpendicular distance-vector

    Find the perpendicular distance of the point (p,q,r) from the plane ax+by+cz=d

    Explain concisely please
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  2. #2
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    Hello, kingkaisai2!

    You've been asked to derive this distance formula . . . messy!


    Find the perpendicular distance of the point (p,q,r) from the plane ax+by+cz\:=\:d

    We want a line from P_1(p,q,r) perpendicular to the plane.

    The plane has normal direction: \vec{n} = \langle a,b,c\rangle
    So our line has parametric equations: \begin{Bmatrix}x = p + at\\y = q + bt\\z = r + ct\end{Bmatrix}

    To find P_2, the intersection of the line and the plane:
    . . . a(p + at) + b(q + bt) + c(r + ct) \:= \:d

    . . and solve for t:\;\;t\;=\;\frac{d - ap - bq - cr}{a^2+b^2+c^2} . . . call this T

    Then P_2 is: . (p + aT,\:q + bT,\:r + cT)


    Let D be the desired distance.

    Then: P_1P_2)^2\:=\p + aT - p)^2 + (q + bT - q)^2 +" alt="D^2\:=\P_1P_2)^2\:=\p + aT - p)^2 + (q + bT - q)^2 +" />  (r + cT - r)^2

    . . . . D^2 \;= \;a^2T^2 + b^2T^2 + c^2T^2\;=\;T^2(a^2+b^2+c^2)


    Since T = \frac{d - ap - bq - cr}{a^2+b^2+c^2}

    . . we have: . D^2\;=\;\frac{(d - ap - bq - cr)^2}{(a^2+b^2+c^2)^2}(a^2+b^2+c^2) \;= \;\frac{(d - ap - bq - cr)^2}{a^2 + b^2 + c^2}


    Hence: . D \;= \;\sqrt{\frac{(d - ap - bq - cr)^2}{a^2 + b^2 + c^2}}

    Therefore: . \boxed{D \;= \;\frac{|ap + bq + cr - d|}{\sqrt{a^2+b^2+c^2}}}

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