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Which one of the following vectors in $\mathbb{R}^3$ is a unit vector that is parallel to the plane with general equation $2x+2y+z=1$?

$$a) (2/3, -1/3, -2/3) $$ $$b) (-3, 2, 2)$$ $$c) (2/3, 2/3, -1/3)$$ $$d) (1/2, 1/2, 1/\sqrt{2})$$

Would I be correct in choosing a) as the magnitude of the vector is $1$ and the dot product of $(2/3, -1/3, -2/3)$ and the normal vector to the plane is $0$?

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Technically you are completely right. However your approach seems a bit artificial to me, since you first find a vector which is perpendicular to the plane and then check whether a vector is perpendicular to this vector.

More directly you can see that every vector parallel to the plane must actually be in the plane after translation. Hence it must satisfy $$2x+2y+z=0.$$

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On the other hand, "first find a vector which is perpendicular to the plane" is quite trivial, because that vector is exactly what the coefficients in the plane's equation are. – Henning Makholm Nov 1 '12 at 12:45
Sure, saying that a vector satisfies $2x+2y+z=0$ is virtually the same as saying that it is orthogonal to $(2,2,1)^T$, so the effort is the same. All I am saying is that that the reasoning appears to be a bit intricate. – Simon Markett Nov 1 '12 at 13:27
x @Simon: Not significantly more intricate than imagining the equation for a translated plane, in my immediate opinion. – Henning Makholm Nov 1 '12 at 14:26
Actually, if you try to prove that the normal vector is given as $(2,2,1)^T$ you do exactly that. But maybe this is just the way I think about it and most certainly it is a matter of taste. – Simon Markett Nov 1 '12 at 14:37

Short answer: yes. Since the vector is perpendicular to the normal (which is itself perpendicular to the plane) and we are in $\mathbb{R^3}$ it must be parallel to the plane. The only other vector this works for is b) which isn't a unit vector. So yes you are correct.

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