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Math Help - General question about vectors. (point to plane)

  1. #1
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    General question about vectors. (point to plane)

    Ok, this is more a general question...There's a picture if that helps for me to explain my question .

    1/ Find the perpendicular distance of the point P(4, 5, 6) and the
    plane x + 2y - 2z = 0.
    This is the example in my text book. So i was reading through how the derive the formula

    PM =  \frac{n . p}{| n | }
    Now what i don't understand with this...n is the same vector as PM...
    So the modulus of n should be the distance of PM. But then if thats the case, the formula doesn't make sense because it would be saying
    |x| = \frac{x.y}{|x|}
    and we need to know |x| to work out |x|...therefore the formula is pointless...
    I know this mustn't be the case and its just me not understanding.

    Any help please ?
    Thanks
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  2. #2
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    I think that you have not really understood the proof.
    The usual formula for that distance is, using your diagram, \frac{{\left| {\overrightarrow {PT}  \cdot N} \right|}}{{\left\| N \right\|}}.
    But in that the N is the normal to the plane.
    That formula is derived by using the cosine of the angle between \overrightarrow {PT}~\&~N.
    The distance is the length of one side of a right triangle with hypotenuse \overrightarrow {PT}.
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  3. #3
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    Quote Originally Posted by Plato View Post
    I think that you have not really understood the proof.
    The usual formula for that distance is, using your diagram, \frac{{\left| {\overrightarrow {PT}  \cdot N} \right|}}{{\left\| N \right\|}}.
    But in that the N is the normal to the plane.
    That formula is derived by using the cosine of the angle between \overrightarrow {PT}~\&~N.
    The distance is the length of one side of a right triangle with hypotenuse \overrightarrow {PT}.
    Hi, thanks for the reply. Much appreciated
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