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## Homework Statement

Let W be the plane [itex]3x + 2y - z = 0[/itex] in R

^{3}. Find a basis for [itex]W^{\perp}[/itex]

## Homework Equations

N/A

## The Attempt at a Solution

Firstly, I take some arbitrary vector [itex] u = \begin{bmatrix}a\\b\\c\end{bmatrix}[/itex]

that is in [itex]W^{\perp}[/itex]. Then I note that W can be rewritten in terms of the natural basis for R

^{3}:

[itex]W = 3\begin{bmatrix}1\\0\\0\end{bmatrix} + 2 \begin{bmatrix}0\\1\\0\end{bmatrix} - 1 \begin{bmatrix}0\\0\\1\end{bmatrix}[/itex]

Then, by the inner product: [itex]\langle u,w\rangle = 3a + 2b - c = 0[/itex]. Thus,

[itex] a = -\frac{2}{3}b + \frac{1}{3}c[/itex] and hence the vector u in [itex]W^{\perp}[/itex] has the form:

[itex] u = \begin{bmatrix}-\frac{2}{3}b+\frac{1}{3}c\\b\\c\end{bmatrix} = b\begin{bmatrix}\frac{-2}{3}\\1\\0\end{bmatrix} + c u = \begin{bmatrix}\frac{1}{3}\\0\\1\end{bmatrix}[/itex]

Thus, the set [itex]\begin{Bmatrix}

\begin{bmatrix}-\frac{2}{3}\\1\\0\end{bmatrix}

\begin{bmatrix}\frac{1}{3}\\0\\1\end{bmatrix}

\end{Bmatrix}

[/itex] spans [itex]W^{\perp}[/itex].

Since both vectors are linearly independent, that is, one cannot be written as a linear combination of the other, I conclude that the set forms a basis for [itex]W^{\perp}[/itex]. After checking the solutions manual for the answer, the book came up with a completely different solution. What did I do wrong? Thanks in advance.