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Find parametric equations of the plane that is parallel to the plane $$3x + 2y - z = 1$$ and passes through the point $$P(1, 1, 1)$$.

How would one approach this problem?

EDIT:

The normal vector is $$(3,2,-1)$$ So I was able to get two vectors $$(0,1,-1) $$ And $$(2,0,-2)$$

But what would be the translation of the two vectors?

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    What have you done? Can you write down the parametric form of a plane parallel to the given one? – 2017-02-01
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    Find two vectors spanning the plane $3(x-1)+2(y-1)-(z-1)=0$. Then $r(u,v)=u \vec V_1+v \vec V_2+(1,1,1)$ is enough to form the plane. – 2017-02-01
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    @Macavity See my edit – 2017-02-01
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    That is a normal vector to what? Anyway, a hint then: consider the family of planes $3x+2y-z=\lambda$. – 2017-02-01
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    @Macavity Unfortunately I do not understand your hint. – 2017-02-01
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    Can you see that the normal vector you have mentioned, is also a normal vector for any plane that I mentioned in the family above? i.e. That entire family is parallel to the given plane. So now which of those pass through the point given? – 2017-02-01
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    @Macavity Well you can subtract the point from any vector that satisfies the equation and that gives you the vectors that passes through the point. But since the plane may not lie in the origin, I don't know how you could find the vector that it should be translated by. – 2017-02-01
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    Are you convinced the family of planes I put down is parallel to the given plane? – 2017-02-01
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    @Macavity Yes, so any translation is therefore parallel? – 2017-02-01
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    Any translation along the normal vector generates parallel planes. All you need is to find which of those pass through the point by simple substitution, the translation vector is not needed for the stated problem. – 2017-02-01

2 Answers 2

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For two parallel planes, the normal vector is th same, so the new plane has normal vector $\vec{n}=(3,2,-1)$, and plane have to passes through the point $$P(1, 1, 1)$$ then the equation of the plane is $$3(x-1)+2(y-1)-(z-1)=0$$ and

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You can interpret $$ 1 = 3x + 2y-z = (3,2,-1)^T \cdot (x,y,z)^T \iff \\ d := \frac{1}{\lVert (3,2,-1)^T\rVert} = \frac{(3,2,-1)^T}{\lVert (3,2,-1)^T\rVert} \cdot (x,y,z)^T \iff \\ d = n \cdot v $$ as equation of a plane with a unit normal vector $n$ and signed distance $d$ to the origin.

A parallel plane differs by $d$.

Finally choose $d$ such that $v = (x,y,z)=(1,1,1)$ satisfies the above equation: \begin{align} d &= \frac{(3,2,-1)^T}{\lVert (3,2,-1)^T\rVert} \cdot (1,1,1)^T \\ &= \frac{4}{\sqrt{14}} \end{align} So $$ 3x + 2y - z = 4 $$ is the sought parallel plane.

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