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I found a linear algebra problem asking to determine if different structures form subspaces. In one of the examples, it gives a parametric equation of a plane. It has one constant vector followed by the two parameters, something like this: Plane: $(x,y,z)= (2,3,4) + t_1(1,1,1) + t_2(1,2,3)$

The answer at the back says that this is a subspace. This equation does satisfy the condition with $t_1=-1$ and $t_2=-1$, to contain the zero vector, which is the first condition of a subspace.

But I know a theorem or corollary that says that planes of the Cartesian form:

$Ax+By+Cz=0$, go through the origin and are subspaces.

So if I convert my above parametric equation to Cartesian form, I have doubt that it will be this form, due to the constant. (NOTE: I put one constraint on this problem, you can't use Cross Product to find the Normal vector.)

So I guess my question is, if a parametric equation of a plane has the constant vector $(2,3,4)$ does that imply that it has a constant $D$ in the Cartesian form: $Ax+By+Cz + D =0$. (which means it does not pass through the origin) But I can find parameters to make the zero vector.

So what is the relationship between $D$ and the constant vector, if there is any.

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But the plane does go through the origin! Take $t_1=t_2=-1$. Yes, $D=0$.

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  • $\begingroup$ Hi there, I figured it out just 1 minute ago.the constant vector is a linear combination of the basis vectors, so hence one can convert it to the form: Ax+By+Cz=0, so hence D=0. Thanks for your input Ted. $\endgroup$ – Palu Oct 7 '13 at 3:44
  • $\begingroup$ You're welcome. Best to accept the answer so that this doesn't stay on the "unanswered" stack! $\endgroup$ – Ted Shifrin Oct 7 '13 at 3:48

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