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The line $\vec{r}=\vec{a} \times \lambda \vec{b}$ will not meet the plane $\vec{r} \cdot \vec{n}=q$, provided: |
$\vec{b} . \vec{n}=0, \vec{a} . \vec{n}=q$ $\vec{b} . \vec{n} \neq 0, \vec{a} . \vec{n} \neq q$ $\vec{b} . \vec{n}=0, \vec{a} . \vec{n} \neq q$ $\vec{b} . \vec{n} \neq 0, \vec{a} . \vec{n}=q$ |
$\vec{b} . \vec{n}=0, \vec{a} . \vec{n} \neq q$ |
We must have $\vec{b} . \vec{n}=0$ and $\vec{a} . \vec{n} \neq q$ Hence (3) is correct answer. |
