Given three points p=(2,1,3), q=(1,0,1) and r=(2,-1,1), find the Cartesian equation of the plane containing those points.
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Let the equation of the plane be ax+by+cz+d=0. Substitute your 3 points.Now you have 3 homogeneous equations in 4 unknowns. – P. Lawrence Mar 05 '20 at 23:51
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2X-2Y+2Z+8=0, that's the answer i got. – Antonius Mateus I Mar 05 '20 at 23:54
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The three given points don't satisfy the equation you got. (I guess there's at least one sign error in your computation, because changing $+8$ to $-8$ would make $p$ and $r$ satisfy the equation. But $q$ still doesn't, so there's another error involved.) – Andreas Blass Mar 06 '20 at 00:00
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By inspection, $x+y-z=0$ – J. W. Tanner Mar 06 '20 at 00:01
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Evaluate the direction vectors $\;\vec v:=\vec{pq}=q-p\;,\;\;\vec u=\vec{pr}=r-p\;$, then evaluate the vectorial product $\;(a,b,c)=\vec v\times\vec u\;$ , and this last is a normal vector to your plane, so your plane is $\;ax+by+cz+d=0\;\;$ . Finally, can you see how to evaluate $\;d\;$ ?
DonAntonio
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@AntoniusMateusI That's wrong: the plane passes through the origin. Using the notation of my answer, I get $;\vec v=(-1,-1,-2);,;;\vec u=(0,-2,-2);$ , and from here $;\vec v\times\vec u=(-2,-2,2);$ , so your plane is $;-2x-2y+2z+d=0;$ , and substituting into this $;q;$ , for example, gives $;d=0;$ , so the plane is $;x+y-z=0;$ (I divided the above by $;-2;$ to make it nicer) – DonAntonio Mar 06 '20 at 00:42
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Usually easier and much quicker than solving a $;3\times3;$ linear system of equations... – DonAntonio Mar 06 '20 at 00:47