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Suppose a plane has a unit normal vector of $\left(\frac {1}{\sqrt6} , \frac {1}{\sqrt6} , \frac {2}{\sqrt6}\right)$ and is at a signed distance of $\left(\frac {13}{\sqrt6}\right)$ from the origin. What is the equation of the plane?

So given $d$ = - ($n$ $\cdot$ $p$) and my $n$ = $\left(\frac {1}{\sqrt6} , \frac {1}{\sqrt6} , \frac {2}{\sqrt6}\right)$

Can I say that my $p$, which is $(x , y , z)$ , is $(1 , 1 , 2)$?

so my d = - $\left(\frac {1}{\sqrt6} , \frac {1}{\sqrt6} , \frac {2}{\sqrt6}\right)$ . (1 , 1 , 2) = -${\sqrt6}$

so my equation of plane is $\frac {1}{\sqrt6} , \frac {1}{\sqrt6} , \frac {2}{\sqrt6}-\sqrt6=0$.

Is this correct?

How can I work out my distance to be $\frac {13}{\sqrt6}$?

confusedbeginner
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3 Answers3

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The general plane with your unit normal $n$ has equation $$x+y+2z=a$$ for some real $a$. To find $a$ you must have the point $(13/\sqrt6)n$ on the plane. That point is $(13/6,13/6,13/3)$.

Angina Seng
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Since the distance between point $\;(a,b,c)\;$ to plane $\;Ax+By+Cz+D=0\;$ is given by

$$\frac{|Aa+Bb+Cc+D|}{\sqrt{A^2+B^2+C^2}}\;,\;\;\text{your plane is given by the conditions:}$$

$$\begin{cases}&\frac1{\sqrt6}x+\frac1{\sqrt6}y+\frac2{\sqrt6}z+d=0\\{}\\ &\cfrac{|d|}{\sqrt{\frac16+\frac16+\frac46}}=\cfrac{13}{\sqrt6}\end{cases}$$

So you can write your plane's equation as

$$x+y+2z+13=0$$

DonAntonio
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The equation of the plane whose unit normal is $\hat{\underline{n}}$ and which is at a signed distance $d$ from the origin is given by $$\underline{r}\cdot\hat{\underline{n}}=d$$

So the equation of this plane is $$x+y+2z=13$$

David Quinn
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