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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
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2 Answers

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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