Shortest Distance from Origin - 4D

Let the shortest distance of the given hyperplane from the origin be $d$ .

For a hyperplane $a x + by + cz + dw + e=0$ in $\mathbb{R}^{4}$ , the shortest distance from a point $(x_{0}, y_{0}, z_{0}, w_{0} )$ has the formula

$\text{Shortest Distance} = \frac{|ax_{0} + by_{0} + cz_{0} + dw_{0} + e | }{\sqrt{ a^{2} + b^{2}+ c^{2} +d^{2}}}$

The coordinates of the origin are $(0,0,0,0)$ . Our hyperplane is $4x - 3y + 12z - 84w -425= 0$ . We can plug these values in the formula to get the distance.

$\begin{aligned} d & = \frac{|4 \times 0 - 3 \times 0 + 12 \times 0 - 84 \times 0 -425 | }{\sqrt{ 4^{2} + (-3)^{2} + 12^{2} + (-84)^{2}}} \\ & = \frac{|0 + 0 +0 + 0- 425 | }{\sqrt{7225}} \\ & = \frac{425}{85} = \boxed{5} ~~~~ _\square \end{aligned}$

The general form of this formula is proved in this note .

An infinite plane (a 2D object itself) in 3D ( $\mathbb{R}^3$ ) is a normalized normal from the origin to the closest point and that signed distance normalized is the constant term. Therefore, normalize the coefficients' vector and the constant term by the same factor. The length of the coefficients' vector as stated is 85. Therefore, the closest distance is $\frac{425}{85}=5$ . the same logic works in any dimension of the equivalent . $\mathbb{R}^{n-1}$ object.

The normalized equation is $\frac{4}{85}x+\frac{-3}{85}y+\frac{12}{85}z+\frac{-84}{85}w+(-5)=0$ .