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Say $x \in P$, where $P = \text{span } \{v_1, v_2 \}$ is a plane. Then let $n$ be the normal vector of $P$ and let $y = cn$ where $c$ is a scalar.

Is it true that $x \cdot y = 0$?

This must be true I think because

$$x \cdot y = (c_1v_1 + c_2v_2) \cdot cn = c_1c (v_1 \cdot n) + c_2c (v_2 \cdot n) = 0$$

??

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Your reasoning is perfectly correct.

However, the normal vector to a plane does not exist, it is nonunique! Even if you ask it to be unitary. In fact, there are exactly two unitary normal vectors to a plane.

Perhaps, if you want to define the unitary normal vector to a plane spanned by two vectors $v_1,v_2$, you can equip your vector space with an orientation and chose the normal vector $n$ such that $(v_1,v_2,n)$ is direct.

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    normal vector is the cross product of $v_1, v_2$, so it is unique isnt it? – 2017-02-06
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    If you defined it like this, this is okay then! In fact, the cross product is exactly defined by the construction I described in my last paragraph. Note that you also have to ask it to have length the area of the parallelogram spanned by $v_1,v_2$. – 2017-02-06
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    I would note that while $P$ was produced by taking the span of $v_1$ and $v_2,$ there are many other pairs of vectors that can produce the exact same plane $P.$ Taking the cross product of any other such pair would give you some scalar multiple of $v_1\times v_2,$ but not necessarily an equal vector and not necessarily even a positive multiple. – 2017-02-06
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    Indeed, this is worth mentionning! Thank you for this precision! – 2017-02-06