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If I have three vectors $u_1 = (1,3,-1)$, $u_2 = (0,2,4)$, $u_3 = (2,1,-1)$. How can I decide all vectors perpendicular to $u_3$ in a plane spanned by $u_1$ and $u_2$?

Any help is greatly appreciated, I've been working on it for a while without success

J.st
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2 Answers2

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Any vector in the plane spanned by $u_1$ and $u_2$ must be orthogonal to the cross product ${\bf n} = u_1 \times u_2$, and so any vector in the plane that is also orthogonal to $u_3$ must be parallel to the cross product $u_3 \times {\bf n}$.

B. Mehta
  • 12,774
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Condition 1: $\vec v \perp \vec u_3 \iff \vec v\cdot \vec u_3 = 0$.

So let $\vec v=(v_1,v_2,v_3)$. Then $\vec v$ satisfies $$2v_1 + v_2 - v_3 = 0$$

Condition 2: $\vec v\in\operatorname{span}(\vec u_1, \vec u_2) \iff \vec v\wedge \vec u_1\wedge \vec u_2 = 0$.

So additionally, $\vec v$ satisfies: $$(v_1\hat e_1 + v_2\hat e_2 + v_3\hat e_3)\wedge(2\hat e_1\wedge \hat e_2 + 4\hat e_1\wedge \hat e_3 + 14\hat e_2\wedge \hat e_3) = 0 \\ (14v_1-4v_2+2v_3)e_1\wedge e_2\wedge e_3=0$$

Then the solution to the system of equations $$\begin{cases} 2v_1 + v_2 - v_3 = 0 \\ 14v_1-4v_2+2v_3 = 0\end{cases}$$ is $\operatorname{span}(1,9,11)$. So all vectors parallel to $(1,9,11)$ will satisfy your conditions.