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The question is: Find a basis for the plane x-2y+3z=0. Then find a basis for the intersection of that plane with the xy plane.

The basis for the plane is the nullspace of the matrix:

A=\begin{bmatrix} 1 & -2 & 3\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}*

which is null(A)=\begin{bmatrix} 2 &-3\\1 & 0\\0 & 1 \end{bmatrix}

My confusion is with the second part of the question. My approach was to find the intersection of this basis with a basis for the xy plane. E.G. I want to solve for a, b, c,d in the relation a[2, 1, 0] + b[ 3, 0, 1] = c[1, 0, 0] + d[0, 1, 0] which should be the nullspace of \begin{bmatrix} 2 &-3 &-1 & -1\\ 1 & 0 & 0 & -1\\ 0 & 1 & 0 & 0 \end{bmatrix}

which is [1, 0, 2, 1]. I interpreted this to mean that the basis for the intersection is [2, 1, 0]: or the linear combination of null(A) with a=1,b=0.

But the solution in my book says that the basis for the intersection is the first column of A, [1, 0, 0] since it lies in the xy plane. Where have I gone wrong?

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If you use basic analytic geometry it'll be much simpler: two planes in space are either disjoint or their intersection is a straight line. In your case, the intersection is when $\;z=0\;$ , i.e. simply the line

$$x-2y=0\iff x=2y\;,\;\;\text{with basis}\;\;\left\{\;\binom21\;\right\}\;,\;\;\text{say}.$$

or, if you want to see it embedded in space, a basis is

$$\left\{\;\;\begin{pmatrix}2\\1\\0\end{pmatrix}\right\}$$

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