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This is the first year linear algebra question.

Let $P$ be the plane passing through the point $(1,1,0)$ with normal vector $(2,-1,1)$. Which of the following points is in $P$?

a) $(-1,1,1)$
b) $(-1,-2,0)$
c) $(-1,1,3)$
d) $(3,4,-1)$
e) $(1,-2,1)$

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3 Answers

$$P:\;\;2x-y+z+d=0$$

and since we're given that point in the plane:

$$P:\;\;2-1+d=0$$

So you find $\,d\,$ and then check, substituting, what points are on the plane.

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Hint The vector in the plane (from the point (1,1,0) to the 'test point') must be orthogonal (perpendicular) to the normal vector.

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The vector $(x,y,z)$ is in $P$ iff the vector $(x,y,z)-(1,1,0)=(x-1,y-1,z)$ is normal to the vector $(2,-1,1)$ iff $$\langle(x-1,y-1,z),(2,-1,1)\rangle=0$$ with $\langle.,.\rangle$ the usual inner product in $\mathbb R^3$.

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