Condition for vector to be in a plane

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Suppose $\ell$ and $m$ are non-parallel lines with direction vectors $v$ and $w$, respectively. Let $u$ be a vector of the line $\ell$. Then a vector $x$ is on the plane $V$ spanned by $ℓ$ and $m$ if and only if:

  • a) $x \bullet(v \times w) = u \bullet (v \times w) $
  • b) $x \bullet(v \times w) = u $
  • c) $v \bullet(x \times w) = u \bullet (v \times w) $
  • d) $x \bullet(v \times w) = 0 $

Are there multiple correct choices? I believe a) and d) are the same, since $u$ should be perpendicular to the normal vector of the plane. Is there something I am missing?

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By the geometric interpretation of triple product the two volumes

  • $V_1=x \bullet(v \times w)$
  • $V_2=u \bullet (v \times w)$

are the same since the components of $x$ and $u$ onto $v \times w$ are the same therefore the correct statement is "a)".

As noticed by Jaap Scherphuis in the comments, another way to see that if $u$ and $x$ both (when placed at the origin) point to points within the plane, it means $u−x$ is parallel to the plane. From this we can deduce that "a)" is the only correct anwer. Moreover, if the plane happens to go through the origin then (c) and (d) hold as well, but not in general.