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Let

$W = \{\begin{bmatrix}x\\y\\z\end{bmatrix} \in R^3\mid 3x+2y-z=0\}$ be a plane in $R^3$.

Find a basis for the subspace $W^\perp$.

I'm not sure how to do it, but I'm guessing you're supposed to find all the solutions to $3x+2y-z=0$. Then, the set of solutions is the basis of $U^\perp$?

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2 Answers 2

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you need to find a vector normal to the plane. Any such vector will be a basis for the $1d$ subspace $W^\perp$. An example of such a vector is $$(3,2,-1)$$

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$W$ is a plane which passes through the origin, so $W^\perp$ is the set of vectors orthogonal to that plane. In particular $W$ is the image of the orthogonal projection $(x,y,z)\mapsto (x,y,3x+2y)$, so $\dim W$ and $\mathbb R^3=W\oplus W^\perp$, hence $\dim W^\perp=1$. Therefore a basis for $W^\perp$ is any nonzero element in $W^\perp$, for example $(3,2,-1)$.

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