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  "body": "\n## Statement\n\nLet $a$, $b$, $c$ and $d$ be real numbers with $(a, b, c) \\neq (0, 0, 0)$, and let $P = \\{(x, y, z) \\in \\mathbb{R}^3 : a x + b y + c z = d\\}$ be a [[plane|07SDJ5DNE4]].\nLet $X = (x_0, y_0, z_0)$ be a point of $\\mathbb{R}^3$. The distance from $X$ to $P$ is the smallest [[Euclidean distance|0NPPGHQNXQ]] $d(X, Q)$ for $Q \\in P$. Then\n$$\\operatorname{dist}(X, P) = \\frac{|a x_0 + b y_0 + c z_0 - d|}{\\sqrt{a^2 + b^2 + c^2}}.$$\n\n## Remarks\n\nThe smallest distance is reached at exactly one point of $P$, the orthogonal projection of $X$ on $P$. With $n = (a, b, c)$ this point is $X - \\frac{a x_0 + b y_0 + c z_0 - d}{a^2 + b^2 + c^2} \\, n$.\nThe number $(a x_0 + b y_0 + c z_0 - d) / \\|n\\|$ is the signed distance. Its sign tells on which side of the plane the point is.\nThe same formula holds for a hyperplane $\\{x : a \\cdot x = d\\}$ with $a \\neq 0$ in $\\mathbb{R}^n$: the distance from $x_0$ is $|a \\cdot x_0 - d| / \\|a\\|$.\n\n## Explanations\n\n### L1 human\n\nThe nearest point of a plane is straight below or above you, along the normal vector. For the floor plane $z = 0$, the formula gives $|z_0|$, your height above the floor. Tilted planes need the factor $\\sqrt{a^2 + b^2 + c^2}$, because $(a, b, c)$ need not be a unit vector.\n",
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