In do Carmo, one exercise gives a plane in $\mathbb R^3$, $ax +by +cz+d = 0$, and tells us to show that $|d|/\sqrt{a^2 + b^2 + c^2}$ measures the distance from the plane to the origin.
However, this seems a bit ambiguous since we don't know what the plane actually is.
By distance, does he mean minimal distance?
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$\begingroup$By distance, the author presumably implies minimal distance to the plane, which is achieved by finding a vector that is perpendicular to the plane, and then using that vector to find the distance between the origin and the plane.
More precisely, given a general plane of equation ax+by+cz+d=0, we can find the perpendicular vector (a,b,c).
The reason as to why the minimal distance is the formula provided can be seen in a much better exposition here
$\endgroup$ 0 $\begingroup$Yes he does mean minimal distance. I'm working on a proof for the $\mathbb R ^n$ case, and will update once this is done.
Consider the analog in $\mathbb R^2$, distance $= |c| / (\sqrt{a^2+b^2})$, for a line $ax+by+c=0$.
Example: $x+y+1=0 \Rightarrow y=-1-x$. Clearly the minimum (Euclidean) distance from the origin is $\sqrt{2}$.
$\endgroup$ $\begingroup$For the plane $ax+by+cz + d = 0$, the normal vector is: $$\hat n = <a,b,c>$$The vector from the plane to any arbitrary point is: $$\hat v = <(x-x_0),(y-y_0),(z-z_0)>$$If we consider the origin in particular, $(x_0,y_0,z_0) = (0,0,0)$ so, $$\hat v = <(x),(y),(z)>$$The MINIMUM distance from the origin to the plane is the projection of v onto n: $$\frac{|\hat n\cdot \hat v|}{|\hat n|}$$ Plugging in the vectors and computing the dot product yields:$$\frac{ax+by+cz}{\sqrt {a^2+b^2+c^2}}$$From the plane's equation, $-d = ax+by+cz$
Thus the minimum distance from the origin to a plane is given by: $$\frac{|d|}{\sqrt {a^2+b^2+c^2}}$$
$\endgroup$ 2 $\begingroup$A different way using optimization:\begin{align} \min\ & x^2+y^2+z^2 & \text{(squared distance to the origin)}\\ & ax+by+cz+d=0 & \text{(the point belongs to the plane)} \end{align}The Lagrangian is $$ \mathcal{L}(x,y,z)= x^2+y^2+z^2 + \lambda(ax+by+cz+d) $$Solving $\nabla\mathcal{L}=0$ (where the gradient is taken w.r.t. to $x,y,z$) you get:$$ \nabla\mathcal{L} =2\left(\begin{array}{c}x\\y\\z\end{array}\right)+\lambda\left(\begin{array}{c}a\\b\\c\end{array}\right)=0 $$Now observe that like the point $(x,y,z)$ belongs to the plane you have:$$\langle \left(\begin{array}{c}x\\y\\z\end{array}\right),\left(\begin{array}{c}a\\b\\c\end{array}\right) \rangle =-d $$Thus$$ \langle \nabla\mathcal{L}, \left(\begin{array}{c}a\\b\\c\end{array}\right) \rangle = -2d+\lambda(a^2+b^2+c^2)=0 $$which gives you the multiplier value: $$\lambda = \frac{2d}{a^2+b^2+c^2}$$Going back to the initial equation $\nabla\mathcal{L}=0$ you get the point $(x_\star,y_\star,z_\star)$ that realizes the minimum:$$ \left(\begin{array}{c}x_\star\\y_\star\\z_\star\end{array}\right) =\frac{d}{a^2+b^2+c^2}\left(\begin{array}{c}a\\b\\c\end{array}\right) $$Taking the norm you get its distance to the origin:$$ \sqrt{x_\star^2+y_\star^2+z_\star^2}=\frac{|d|}{\sqrt{a^2+b^2+c^2}} $$To answer to your question, yes this formula is the minimal distance to a point $(x,y,z)$ in your plane to the origin.
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