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Find all vectors $\vec{v}=\begin{bmatrix}x\\y\\z\end{bmatrix}$ orthogonal to both $\vec{u_1}=\begin{bmatrix}2\\0\\-1\end{bmatrix}$ and $\vec{u_2}=\begin{bmatrix}-4\\0\\2\end{bmatrix}$

$\vec{u_1}$ and $\vec{u_2}$ are parallel, so the cross product will be $\vec{0}$. This won't help. The other thing I tried was based on the dot product, so:

$\vec{u_1}\cdot\vec{v}=0\\ \vec{u_2}\cdot\vec{v}=0$

leading to the system of equations:

$2x-z=0\\-4x+2z=0$

and then

$\vec{v}=s\begin{bmatrix}1\\0\\2\end{bmatrix}+t\begin{bmatrix}0\\1\\0\end{bmatrix}$

which makes sense to me because the orthogonal vectors would "rotate" around $\vec{u_1}$ and $\vec{u_2}$ in infinitely many directions.

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  • $\begingroup$ The set of all vectors orthogonal to a given vector are orthogonal to all scalar multiples of that vector. Therefore giving you two vectors which are scalar multiples of each other is unnecessary. What geometric object defines that set of all vectors orthogonal to a given vector? $\endgroup$
    – Archaick
    Commented Apr 22, 2015 at 1:01
  • $\begingroup$ I just made a simple mistake when creating the system of equations. My question was based on that error and is now irrelevant. I'll keep the question up (with the error corrected) in case anyone else is interested in a solution to this kind of question $\endgroup$
    – yroc
    Commented Apr 22, 2015 at 2:11

2 Answers 2

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Note that actually your two equations are the same equation. (Divide the second one by $-4$.) So you have $$ 2x-z=0 $$ as your constraint, so any vector parallel to $$ \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} $$ will be orthogonal to both vectors (as will any with just a $y$ component, and by linearity, any linear combination thereof).

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Here, it's sufficient to use the triple product equal to 0. However, your method should be satisfying.

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