$\mathbf{Question:}$ Find set of vectors orthogonal to $\begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix}$
$\mathbf{My\ attempt:}$
The vector is in $R^3$ so we can let vector $\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \end{bmatrix} $ represent orthogonal vectors.
$$\begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix} . \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} $$
$$ x_1 + x_2 + x_3 = 0 $$
$$ x_1 = -x_2 - x_3 $$
$$ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \end{bmatrix} = \begin{bmatrix} -x_2 - x_3 \\ x_2 \\ x_3 \\ \end{bmatrix} = \begin{bmatrix} -1 \\ 1 \\ 0 \\ \end{bmatrix}x_2 + \begin{bmatrix} -1 \\ 0 \\ 1 \\ \end{bmatrix}x_3 $$
Therefore, the vectors in subspace $W = span(\begin{bmatrix} -1 \\ 1 \\ 0 \\ \end{bmatrix} , \begin{bmatrix} -1 \\ 0 \\ 1 \\ \end{bmatrix} )$ are orthogonal to vector $\begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix}$.
$\endgroup$ 82 Answers
$\begingroup$The space of vectors orthogonal to $\begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix}$ can be defined by $span(\begin{bmatrix} -1 \\ 1 \\ 0 \\ \end{bmatrix} , \begin{bmatrix} -1 \\ 0 \\ 1 \\ \end{bmatrix} )$ where the dimension of the space is 2.
$\endgroup$ $\begingroup$If you're working in with a non degenerate scalar product as this case the standard one you know that called $W$ a subspace of $V$ of dimension n, (in this case $W = $ span$(e_1+e_2+e_3)$ and $V = \mathbb{R}^3$) you have that
$$dim W + dim W^\perp = dim V$$
Hence if you know that you are looking for a two dimensional subspace orthogonal to $e:= e_1+e_2+e_3$ all you have to do is to find $2$ linearly independant vectors orthogonal to $e$, which by dimension will span $W^\perp$. In this case you can tell at glance.
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