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PostPosted: Thu, 11 Sep 2008 02:25:44 UTC 
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Joined: Mon, 29 Mar 2004 04:23:45 UTC
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Location: Charlotte
Let P be a point not on the plane that passes through the points Q,R, and S. Show that the distance d from P to the plane is
d = magnitude of (a * (b x c)) / magnitude of (a x b)
where a = QR, b = QS, and c = QP.

Use the distance to find the distance from the point (1,2,3) to the line x = 2 + t,
y = 2 - 3t, z = 5t


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PostPosted: Thu, 11 Sep 2008 16:19:57 UTC 
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Joined: Mon, 19 May 2003 19:55:19 UTC
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Location: Lexington, MA
Hello, boadufk!

I have an "eyeball" approach to the formula . . .


Quote:
Let P be a point not on the plane that passes through the points Q, R, S.

Show that the distance d from P to the plane is: .$d \:= \:\frac{|\vec a\cdot(\vec b \times \vec c)|}{|\vec a\times \vec b|}
where: .\vec a = \overrightarrow{QR},\;\;\vec b = \overrightarrow{QS},\;\;\vec c = \overrightarrow{QP}.

[Sorry, you'll have make your own sketch.]


The vectors \vec a, \vec b, \vec c determine a parallelepiped.

Its volume is: .\text{(Volume)} \;=\;\text{(area of base)}\, \text{(height)} .[1]

. . The volume is given by: .|\vec a\cdot(\vec b \times \vec c)|

. . The area of the base is given by: .|\vec a \times \vec b|


Substitute into [1]: . |\vec a\cdot(\vec b \times \vec c)| \;=\;(|\vec a \times \vec b|)\,d

Therefore: . $d \;=\;\frac{|\vec a\cdot(\vec b\times\vec c)|}{|\vec a \times\vec b|}



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