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At student demonstrations yesterday we were solving this task:

"Find the equation of the plane parallel to the plane $x - 2y + 2z - 5 = 0$ and distanced 2 units from the plane."

We read the normal from the equation, which is $\vec n = (1, -2 ,2)$. The length of $\vec n$ is $9$, so we conclude that $\vec n$ is not of unit length. We get that $\vec n_0 = (\frac 13, - \frac 23, \frac 23)$.

Now here's the part that I don't understand:

The demonstrator wrote this equation:

$\frac 13 |\delta| = 2$

and from that we get that $\delta = 6$.

I'm not seeing how he reached this conslusion that led him to write $\frac 13 |\delta| = 2$. Isn't $\delta$ supposed to be the distance from the origin? Can someone clarify?

Thanks in advance!

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  • $\begingroup$ What is $\delta$? $\endgroup$ – amd Mar 30 '17 at 20:21
  • $\begingroup$ The problem as posed is ambiguous, by the way, and has two solutions. Two units in which direction? $\endgroup$ – amd Mar 30 '17 at 20:23
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The constant $d$ in an equation of the form $ax+by+cz+d=0$ is the (negative of the signed) distance of the plane from the origin scaled by the length of the normal vector $\langle a,b,c\rangle$, which in this case is $3$. So, to shift the plane by $2$ units, $d$ must be changed by $3\cdot2=6$.

Or, if you prefer, we’re looking for the value of $\delta$ such that the difference between the distances of the planes from the origin is two, i.e., $$\left|{-d+\delta\over\sqrt{a^2+b^2+c^2}}-{-d\over\sqrt{a^2+b^2+c^2}}\right|={|\delta|\over\sqrt{a^2+b^2+c^2}}=\frac13|\delta|=2.$$

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  • $\begingroup$ Could you explain the formula? $\endgroup$ – NumberSymphony Apr 1 '17 at 5:14
  • $\begingroup$ @NumberSymphony The explanation is pretty much in the first sentence and ultimately comes from the formula for the projection of a vector $\mathbf u$ onto a vector $\mathbf v$: ${\mathbf u\cdot\mathbf v\over\mathbf v\cdot\mathbf v}\mathbf v$. The normal form of equation for a plane just says that $\mathbf n\cdot\mathbf x$ is constant, specifically $\|\mathbf n\|$ times the orthogonal distance of the plane from the origin. $\endgroup$ – amd Apr 2 '17 at 21:52
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The normal vector is equal to $[1,-2,2],$ thus the vector $\left[\frac{2}{3} , \frac{-4}{3},\frac{4}{3}\right]$ is normal and of lenght $2$. Thus the equation is of the form $$\left(x\pm \frac{2}{3} \right) -2\left(y\pm \frac{-4}{3}\right) +2\left(y\pm \frac{4}{3}\right)-5=0$$

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I would write $$x-2y+2z+n=0$$ for the searched plane.From the other plane we take this point:$$P(1,-2,2)$$ thus we get the distance by the Hession normal form: $$\frac{|1+4+n|}{\sqrt{1+4+4}}=\frac{|9+n|}{3}$$ and this must be $2$ I hope you will be able to solve the problem now

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