If I have a vector say:
$$
\begin{bmatrix}
-1 \\
-\frac{1}{2} \\
1
\end{bmatrix}
$$
How can I identify a vector orthogonal to this by inspection?
For example if I have this vector:
$
\begin{bmatrix}
1 \\
1
\end{bmatrix}
$ I know that $\begin{bmatrix} 1 \\ -1 \end{bmatrix}$ is orthogonal.
So if I change the the sign of the third entry of the initial vector of interest such that it now looks like this:
$$ \begin{bmatrix}
-1 \\
-\frac{1}{2} \\
-1
\end{bmatrix}
$$
Would that suffice to be orthogonal? Is there a general rule to use here?
$\endgroup$ 52 Answers
$\begingroup$Your example : $(-1,-1/2,1)$ , choose, for example, $(-1/2,1,0).$
The scalar product: $(-1)(-1/2) + (-1/2)1 + 0 =0.$
Scalar product:
$\vec A:= (a,b,c) \ne 0$, $ \vec B =(d,e,f) \ne 0$, then:
$\vec A \cdot \vec B =$
$(a,b,c) \cdot (d,e,f) =$
$ad + be +cf = 0$,
$\rightarrow$ the 2 vectors are orthogonal.
See also:
$\endgroup$ 2 $\begingroup$You've found a way to make two of the terms of the dot product sum to zero — what you need to do is also make the rest of the terms of the dot product sum to zero, so that the overall sum is zero.
$(-\frac{1}{2} ) \cdot (-\frac{1}{2}) \neq 0$, so that doesn't work; your dot product is nonzero.
Fortunately, this has an easy solution: those components to zero in the vector you're constructing.
$\endgroup$ 1
