Distance of a plane from the origin
Author
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
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).
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).
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
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
Author
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
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
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]
[edited by - psamty10 on April 22, 2004 2:26:38 PM]
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]
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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