Prove that the set $W^{\perp}$ is a subspace of $V$ and use this to find $W^{\perp}$ when $W$ is the span of $(1,2,3)$ in $V=\mathbb{R}^{3}$

Question

My Linear Algebra book (Larson, Eight Edition) has a two-part exercise that I'm trying to answer. I was able to do the first [proving] part on my own but need help tackling the second part of the problem.

Question: (A) Let $W$ be a subspace of the inner product space $V$. Prove that the set
$$W^{\perp} = \left \{ v \in V:\left \langle v,w \right \rangle=0\text{ for all }w\in W \right \}$$ is a subspace of $V$.
(B) Use the result of this to find $W^{\perp}$ when $W$ is the span of (1,2,3) in $V=\mathbb{R}^{3}$.

I was able to finish Part A (I proved that $W^{\perp}$ is a non-empty set and satisfies the two closure properties), but haven't encountered anything similar to the second part of the problem, so I don't really know where to start. Any input on this will be greatly appreciated!

Answer 1

We want to find vectors which are perpendicular to the span of W. Let $$W^{⊥}= \begin{pmatrix} a\\ b\\ c \end{pmatrix} $$ By perpendicular, the inner product of $W$ and $W^{⊥}$ is $0$, i.e $<W^{⊥},W>=0$ $$ \begin{pmatrix} a\\ b\\ c \end{pmatrix}. \begin{pmatrix} 1\\ 2\\ 3 \end{pmatrix}=a+2b+3c=0 $$ Let $b=k$,$c=l$, then $a=-2k-3l$, so $$W^{⊥}= \begin{pmatrix} a\\ b\\ c \end{pmatrix}=\begin{pmatrix} -2k-3l\\ k\\ l \end{pmatrix}=k\begin{pmatrix} -2\\ 1\\ 0 \end{pmatrix}+l\begin{pmatrix} -3\\ 0\\ 1 \end{pmatrix} $$ $W^{⊥}$ is spanned by the 2 vector above

Answer 2

We know that $(x,y,z) \in W^\perp$ if and only if $ \langle(x,y,z), a(1,2,4)\rangle=0$ for all $a$ in $\mathbb{R}$. The inner product is linear so we get $w \in W^\perp \iff \langle(x,y,z), (1,2,4)\rangle=0 $. Solve this equation: $$x+2y+3z=0$$ and you will get that $(-2y-3z,y,z) \in W^\perp$. In other words $span\left((-2,1,0),(-3,0,1)\right) \subset W^\perp$. Now use what you proved in A to conclude that $span\left((-2,1,0),(-3,0,1)\right) = W^\perp$