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I've been reading Linear Algebra by Jacob for self study and I'm wondering about the geometric interpretation of a unique solution to systems of equations in $\mathbb R^3$.

For example, the homogeneous system $$ \begin{eqnarray*} x+5y-2z &=&0, \\ x-3y+z &=& 0, \\ x+5y-z &=& 0 \end{eqnarray*} $$ has the unique solution $x=y=z=0$. From Calc III, I would think of this as being 3 planes and the various $(x,y,z)$ values of the line (vector) that forms the intersection as the solution to the system. But the solution here and in many other systems in $\mathbb R^3$ is a point and that doesn't seem to be possible.

Am I wrong in trying to understand this geometrically or is there something I'm missing?

Sorry if this is hard to read, I haven't learning latex yet and thanks to everyone who responded to my question about self study books for real analysis.

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Thanks to Srivatsan for cleaning up the mess! – CritChamp Oct 5 '11 at 1:30

You're correct in your interpretation it is the intersection of three planes. Three planes can indeed intersect at one point for example, the origin is the only point contained in $x = 0, y = 0, z = 0$.

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When you intersect three planes in $\mathbb{R}^3$, you can get a plane, a line, a point, or the empty set.

  • You get the empty set if two of the planes are parallel but distinct.

  • You get a plane if the three planes are identical.

In all other cases two of the planes are non-parallel, so they intersect in a straight line; call it $L$.

  • If that line is in the third plane, the intersection of all three planes is the line $L$.

  • If $L$ is parallel to the third plane, the intersection of the three planes is empty.

  • And if $L$ cuts through the third plane, the intersection of the three planes is a point, the point of intersection of $L$ and the third plane.

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