Show that the 4 points all lie on the same plane

sigma

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Feb 19, 2006
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The question reads as:
Show that the 4 points P = (2,1,0), Q = (1,2,1), R = (3,1,1) and S = (4,1,2) all lie on the same plane. Give the standard form equation for this plane.

I'm not sure how to show that they all lie on the same plane. To find the standard form equation, I know that with 3 points you find the vectors with the 3 points for example: PQ and PR then you do the cross product of the 2 vectors then use the point normal form equation to find the standard equation for the plane. But how does this work with 4 points? You would have 3 vectors PQ, PR and PS but where do you go from there? How are you suppose to do the cross product of 3 vectors? So I just need help with that, and also how to show that all the points lie on the same plane.
 
If \(\displaystyle \vec{PS} \cdot \left( {\vec{PR} \times \vec{PQ} } \right) = 0\) then P, Q, R, and S are co-planar.

Use \(\displaystyle {\vec{PR} \times \vec{PQ} }\) as the normal to the plane.
 
Why wouldn't you use \(\displaystyle {\vec{PS}}\) as part of the normal to the plane?
 
sigma said:
Why wouldn't you use \(\displaystyle {\vec{PS}}\) as part of the normal to the plane?
Because basically there is no need for it.
If A, B and C are three non-collinear points we can use any of,
\(\displaystyle {\vec{AB} \times \vec{AC} }\), \(\displaystyle {\vec{BC} \times \vec{BA} }\) or \(\displaystyle {\vec{CB} \times \vec{CA} }\) as the normal.
 
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