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deeps

Distance of a plane from the origin

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Hi, the equation for my plane is: 4x - 8y - 8z = 16 now, to find the distance to the origin, do i simply do this 16 / sqrt ( (4^2)+(8^2) + (8^2)) i have the equation of the plane given above, as well as 3 points on the plane as far as i know, the above method is how to solve it, or the answer is 16..but that seems too easy :S thanx for ur help

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The distance is 16. Remember the plane equation:

ax + by + cz - d = 0

a,b and c are the three components of the plane normal and d is the distance from the origin (to be exact, from the plane with the same normal that includes the origin).

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Guest Anonymous Poster
The distance is NOT 16! deeps, your original expression was correct: you need to divide d by the magnitude of the normal. Assuming you really want the *distance you additionally need to take the absolute value of the result, otherwise you are computing the *signed distance*.

In summary, the d term of the plane equation is only the ''distance'' so far as being the *signed* distance of the plane from the origin, *in units of the length of n*, where n is the plane normal.

Christer Ericson
Sony Computer Entertainment, Santa Monica

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ok, now i''m lost...crap, i was hoping this wouldn'' happen :S

sorry,i didn''t quiet understand "signed" distances,..are you talking about vectors?

i need to find the distance of the plane from the origin

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You (Christer) are right, Im an idiot --> i was assuming the normal was 'normalized', when of course it wasnt... sorry.

[edited by - psamty10 on April 22, 2004 2:26:38 PM]

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A normalized vector is a vector of unit length ( |V|(magnitude of vector V) = 1 )

In this case the magnitude of your normal vector is 12, which is why I was stupid to suggest the answer was 16.



[edited by - psamty10 on April 22, 2004 2:30:43 PM]

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