I am trying to find the basis and dimensions of the the space orthogonal $S$ which is in $\mathbb R^3$.

$$S = \begin{bmatrix}1\\2\\3\end{bmatrix}$$

So the dimension would be two because it is $3 - 1$.

The problem I am having is finding the basis. Would I do

$$ 1x+2y+3z = 0$$

Which would give me different basis. Am I on the correct track?


2 Answers 2


You just need find two independent vectors $[x~y~z]^T$ that satisfies $1x+2y+3z=0$.

For example, let $x_1=-3, y_1=0,z_1=1$, $\Rightarrow [x_1~y_1~z_1]^TS=-3\times 1+0\times 2+3\times 1=0$, also let $x_2=2, y_1=-1,z_1=0$, then $[x_2~y_2~z_2]^TS=2\times 1+(-1)\times 2+3\times 0=0$.

Therefore $\begin{bmatrix}x_1\\y_1\\z_1\end{bmatrix}=\begin{bmatrix}-3\\0\\1\end{bmatrix}$, and $\begin{bmatrix}x_2\\y_2\\z_2\end{bmatrix}=\begin{bmatrix}2\\-1\\0\end{bmatrix}$ are such two vectors, because they are also independent:

if $\alpha\begin{bmatrix}-3\\0\\1\end{bmatrix}+\beta\begin{bmatrix}2\\-1\\0\end{bmatrix}=0$, then $\begin{bmatrix}-3\alpha\\0\\\alpha\end{bmatrix}+\begin{bmatrix}2\beta\\-\beta\\0\end{bmatrix}=0$ , $$\Rightarrow \begin{bmatrix}-3\alpha+2\beta\\-\beta\\\alpha\end{bmatrix}=0$$, which implies $\alpha=\beta=0$


Well, the equation $1x+2y+3z=0$ indeed says that the inner product of vectors $S$ and $(x,y,z)^T$ is $0$, that is, they are orthogonal.

So we are looking for a basis among its solutions: e.g. you can take $y=1,\ z=0$ then $y=0,\ z=1$ and calculate the $x$'s for them.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.