#### Archived

This topic is now archived and is closed to further replies.

# Distance of a plane from the origin

This topic is 5232 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

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

##### Share on other sites
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).

##### Share on other sites
16 was my first answer...but it seemed too obvious... thankyou!

##### Share on other sites
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

##### Share on other sites
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

##### Share on other sites
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]

##### Share on other sites
so it''s my original thingi up the top... cool

wats a normalized normal?

##### Share on other sites
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]

##### Share on other sites
ah, is that wats also known as an unit vector?

1. 1
2. 2
3. 3
4. 4
frob
13
5. 5

• 16
• 13
• 20
• 12
• 19
• ### Forum Statistics

• Total Topics
632169
• Total Posts
3004545

×