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Let this be the question:

Suppose $\vec{v}$ and $\vec{w}$ are two vectors parallel to the plane $$x + 2y + 3z = 7.$$ Suppose furthermore that $\vec{v}$ is perpendicular to $\vec{w}$, $$‖v‖= 3, \ ‖w‖= 4.$$

How would you go about answering this question? I always reach a dead end when I try to solve $\vec{v}$ and $\vec{w}$ that satisfy the given information, I just don't know how to go about it. I tried to draw the plane and two parallel vectors, then I know the length of the cross product would be $12$. Then what do I do now? How could you solve this to find vectors $\vec{v}$ and $\vec{w}$?

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    $\begingroup$ The cross product is perpendicular to the plane. Therefore it is parallel to the vector (1,2,3) which is perpendicular to the plane. $\endgroup$
    – kmitov
    Commented Feb 26, 2022 at 4:18
  • $\begingroup$ @kmitov I see, so the cross product of ⟨v⟩ and ⟨w⟩ would be n(1,2,3), where n is some arbitrary value. we know the length is 12. What do you think I should do next? equate the equation with something? $\endgroup$
    – mb3
    Commented Feb 26, 2022 at 4:28

1 Answer 1

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You have: $\vec{u}\times \vec{v}= n(1,2,3)\implies |\vec{u}\times\vec{v}|=|n|\sqrt{1^2+2^2+3^2}=|n|\sqrt{14}=12\implies n = \pm\dfrac{12}{\sqrt{14}}\implies \vec{u}\times\vec{v}=\pm\dfrac{12}{\sqrt{14}}\left(1,2,3\right)$.

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  • $\begingroup$ Such a clear and concise answer. You made it look so easy haha! Thank you Wang!! $\endgroup$
    – mb3
    Commented Feb 26, 2022 at 18:22
  • $\begingroup$ Thank you as well , mb3. $\endgroup$
    – Wang YeFei
    Commented Feb 26, 2022 at 18:23

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