# What does it mean to find the distance from the origin to a plane in $\mathbb R^3$?

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?

• I think so, yes. Consider the intersection of the plane with a line through the origin parallel to the vector $(a,b,c)^T$, which is normal to the plane. Commented Jan 20, 2019 at 20:14

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

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 $$\sqrt{2}$$.

For the plane $$ax+by+cz + d = 0$$, the normal vector is: $$\hat n = $$ 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: $$\frac{|d|}{\sqrt {a^2+b^2+c^2}}$$

• How does projection of $v$ onto $n$ give the minimum distance? Commented Jan 21, 2019 at 5:01
• Think of the plane as a wall, the shortest way to walk to a wall is always perpendicular to the wall. The projection onto the normal vector gives this perpendicular distance which is the closest the point (in your case the origin) is to the plane. Hope this helps! Commented Jan 26, 2019 at 21:20

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.