10.2 Hyperplanes and Nearest Points

Section 10.1 described a line by a parametrization, one free parameter. In the plane a line is also the solution set of one equation $a_1 x_1 + a_2 x_2 = c$ . This section takes one linear equation $\langle a, x \rangle = c$ in any $\mathbb{R}^n$ and names its solution set, a hyperplane. Distances to a hyperplane, and to a line, then have exact formulas. The main tool is the expansion of Chapter 9: squaring and expanding turns a condition on distances into an equation that can be solved, a linear one in the equidistant case.

10.2.1. Hyperplanes

Chapter 9 met equations of this kind twice. The vectors orthogonal to a fixed $a$ are the solutions of $\langle a, x \rangle = 0$ (Definition 9.3.2). The points equidistant from two given points satisfy an equation $\langle x, b - a \rangle = \text{constant}$ (Exercise (B3)). In $\mathbb{R}^2$ such a set is a line. In $\mathbb{R}^3$ it is not: Section 10.1 found three points of $\{ x_1 + x_2 + x_3 = 1 \}$ on no common line. The solution set of one equation deserves its own name.

Here is a small case. Take $a = (1, 1, 1)$ and $c = 1$ in $\mathbb{R}^3$ , so the equation is $x_1 + x_2 + x_3 = 1$ . The points $(1, 0, 0)$ , $(0, 1, 0)$ and $(0, 0, 1)$ satisfy it. The point $0$ gives $0 < 1$ , and the point $(1, 1, 1)$ gives $3 > 1$ . So these two points lie on opposite sides of the solution set.

A hyperplane is the set where one non-zero linear expression takes one value.

Definition 10.2.1 (Hyperplane)

Let $a \in \mathbb{R}^n$ with $a \ne 0$ , and let $c \in \mathbb{R}$ . The set $H_{a,c} \mathrel{:=} \{ x \in \mathbb{R}^n : \langle a, x \rangle = c \}$ is a hyperplane, and $a$ is a normal vector of it. The sets $\{ x : \langle a, x \rangle \le c \}$ and $\{ x : \langle a, x \rangle \ge c \}$ are the two closed half-spaces of $H_{a,c}$ ; with $<$ and $>$ strictly, they are the two open half-spaces, or the two sides of $H_{a,c}$ .

In words: one vector $a$ , which must be non-zero, and one number $c$ give one equation, and the hyperplane is everything that satisfies it. A subset of $\mathbb{R}^n$ is a hyperplane if it equals $H_{a,c}$ for some such pair.

A hyperplane is never empty. The point $p_0 \mathrel{:=} \frac{c}{\left\| a \right\|^2} \, a$ is defined, since $\left\| a \right\|^2 > 0$ by Proposition 9.4.1 (a), (c). It lies on the hyperplane: $\langle a, p_0 \rangle = (c/\left\| a \right\|^2)\left\| a \right\|^2 = c$ by linearity (Proposition 9.3.1 (b)). By linearity again, $\langle a, x \rangle - c = \langle a, x - p_0 \rangle$ , so $x \in H_{a,c} \iff x - p_0 \perp a .$ The hyperplane is the set of points reached from $p_0$ by a step orthogonal to $a$ .

Examples.

$n = 1$ . For $a \ne 0$ in $\mathbb{R}$ , $H_{a,c} = \{ c/a \}$ : a hyperplane of the line is one point. Its sides are the two open rays.
$n = 2$ . $H_{(1,1),1}$ is the line $x_1 + x_2 = 1$ . We will see shortly that every hyperplane of $\mathbb{R}^2$ is a line.
Coordinate hyperplanes. $H_{e_k, 0} = \{ x : x_k = 0 \}$ by Proposition 9.3.1 (d). In $\mathbb{R}^3$ , $H_{e_3, 0}$ is the plane of the first two axes.

What goes wrong when $a = 0$ . Change $a = (1, 1)$ to $a = (0, 0)$ . Then $\langle 0, x \rangle = 0$ for every $x$ (Proposition 9.3.1 (e)), so $\{ x : \langle 0, x \rangle = c \}$ is $\mathbb{R}^n$ when $c = 0$ and empty when $c \ne 0$ . Neither looks like the sets above, and the formula for $p_0$ would divide by $0$ . The clause $a \ne 0$ is what fails.

Why “a” normal and not “the” normal. The equations $x_1 + x_2 = 1$ and $2x_1 + 2x_2 = 2$ have the same solutions. So one hyperplane has many pairs $(a, c)$ , and the next result says exactly how many.

Proposition 10.2.1 (The Normals of a Hyperplane)

Let $a, a' \in \mathbb{R}^n$ be non-zero and $c, c' \in \mathbb{R}$ . Then $H_{a,c} = H_{a',c'}$ if and only if $(a', c') = (\lambda a, \lambda c)$ for some real $\lambda \ne 0$ .

Idea. The direction $(\Leftarrow)$ multiplies an equation by $\lambda$ . For $(\Rightarrow)$ , the equation $\langle a', x \rangle = c'$ must hold at every $p_0 + w$ with $w \perp a$ , so $a'$ is orthogonal to everything orthogonal to $a$ . Subtracting the projection (Example (Method: subtract the projection)) produces a vector orthogonal to $a$ from $a'$ itself, and that vector must be $0$ .

Proof. $(\Leftarrow)$ If $a' = \lambda a$ and $c' = \lambda c$ with $\lambda \ne 0$ , then $\langle a', x \rangle = \lambda\langle a, x \rangle$ by linearity, and $\lambda\langle a, x \rangle = \lambda c$ holds if and only if $\langle a, x \rangle = c$ , multiplying by $\lambda$ or by $1/\lambda$ .

$(\Rightarrow)$ Suppose $H_{a,c} = H_{a',c'}$ , and let $p_0 = (c/\left\| a \right\|^2) a$ . For every $w \perp a$ , the point $p_0 + w$ lies in $H_{a,c}$ , hence in $H_{a',c'}$ : $\langle a', p_0 + w \rangle = c'$ . Taking $w = 0$ gives $\langle a', p_0 \rangle = c'$ , and subtracting, $\langle a', w \rangle = 0$ for every $w \perp a$ .

Let $\mu \mathrel{:=} \langle a', a \rangle/\left\| a \right\|^2$ and $w \mathrel{:=} a' - \mu a = a' - \operatorname{proj}_a a'$ . By Theorem 9.6.1 (a), $w \perp a$ , so $\langle a', w \rangle = 0$ . Hence, by linearity, $\left\| w \right\|^2 = \langle a' - \mu a, w \rangle = \langle a', w \rangle - \mu \langle a, w \rangle = 0 - 0 = 0 ,$ and $w = 0$ , since only the zero vector has norm $0$ . So $a' = \mu a$ . Since $a' \ne 0$ , $\mu \ne 0$ . Finally $c' = \langle a', p_0 \rangle = \mu \langle a, p_0 \rangle = \mu c$ . This proves the claim with $\lambda = \mu$ .

Uses: Proposition 9.3.1 (b): linearity; Proposition 9.4.1 (a), (c): norm zero only at 0, $\left\| w \right\|^2 = \langle w, w \rangle$ .

Warning. A hyperplane has many normal vectors, and $a = 0$ gives none. $H_{(1,1),1} = H_{(2,2),2} = H_{(-1,-1),-1}$ . “The normal” of a hyperplane means “a normal”: it is fixed only up to a non-zero factor (Proposition 10.2.1). So a formula that uses a normal, such as the distance and nearest-point formulas below, is only meaningful if it gives the same answer for $(\lambda a, \lambda c)$ as for $(a, c)$ . We check that it does. And $a = 0$ is excluded on purpose: $\{ x : \langle 0, x \rangle = c \}$ is $\emptyset$ or $\mathbb{R}^n$ , and neither is a hyperplane. Indeed, every $H_{a,c}$ contains $p_0$ and misses $p_0 + a$ , where the value is $c + \left\| a \right\|^2 \ne c$ .

In the plane, hyperplanes and lines are the same sets, and each description converts into the other by swapping two coordinates and changing one sign.

Example (Lines of the Plane Are Its Hyperplanes)

Work in $\mathbb{R}^2$ .

(a) For $a \ne 0$ , $H_{a,c} = p_0 + \mathbb{R}(-a_2, a_1)$ , where $p_0 = (c/\left\| a \right\|^2) a$ .
(b) For $v \ne 0$ and any $p$ , $p + \mathbb{R} v = H_{a,c}$ with $a \mathrel{:=} (-v_2, v_1)$ and $c \mathrel{:=} \langle a, p \rangle$ .

Solution.

(a) Since $a \ne 0$ , also $(-a_2, a_1) \ne 0$ , so the right side is a line (Definition 10.1.1). $(\supseteq)$ $\langle a, (-a_2, a_1) \rangle = -a_1 a_2 + a_2 a_1 = 0$ , so $\langle a, p_0 + t(-a_2, a_1) \rangle = c + 0 = c$ . $(\subseteq)$ Let $x \in H_{a,c}$ and $w \mathrel{:=} x - p_0$ . Then $a_1 w_1 + a_2 w_2 = 0$ . Put $t \mathrel{:=} (a_1 w_2 - a_2 w_1)/\left\| a \right\|^2$ . Using $a_2 w_2 = -a_1 w_1$ and $a_1 w_1 = -a_2 w_2$ , $\begin{aligned} -t a_2 &= \frac{a_2^2 w_1 - a_1 (a_2 w_2)}{\left\| a \right\|^2} = \frac{(a_2^2 + a_1^2) w_1}{\left\| a \right\|^2} = w_1 , \\ t a_1 &= \frac{a_1^2 w_2 - a_2 (a_1 w_1)}{\left\| a \right\|^2} = \frac{(a_1^2 + a_2^2) w_2}{\left\| a \right\|^2} = w_2 . \end{aligned}$ So $x = p_0 + t(-a_2, a_1)$ .
(b) $a = (-v_2, v_1) \ne 0$ since $v \ne 0$ , and $\langle a, v \rangle = -v_2 v_1 + v_1 v_2 = 0$ . So $\langle a, p + tv \rangle = \langle a, p \rangle = c$ : the line lies in $H_{a,c}$ . By (a), $H_{a,c} = p_0 + \mathbb{R}(-a_2, a_1) = p_0 + \mathbb{R}(-v)$ . As $p \in H_{a,c}$ , $p = p_0 - sv$ for some $s$ , so $p_0 = p + sv \in p + \mathbb{R} v$ ; and $-v = (-1)v$ . By Proposition 10.1.1, $p_0 + \mathbb{R}(-v) = p + \mathbb{R} v$ .

So in $\mathbb{R}^2$ the two descriptions are interchangeable: a direction $v$ gives the normal $(-v_2, v_1)$ , and a normal $a$ gives the direction $(-a_2, a_1)$ . In $\mathbb{R}^3$ this fails. The hyperplane $H_{(1,1,1),1}$ contains $(1, 0, 0)$ , $(0, 1, 0)$ and $(0, 0, 1)$ , and Section 10.1 showed, after the definition of collinear points, that no line contains all three. A hyperplane of $\mathbb{R}^3$ is not a line.

Quick check. Write the line through $(1, 3)$ and $(4, -1)$ as an equation $a_1 x_1 + a_2 x_2 = c$ .

Solution. A direction is $v = (4, -1) - (1, 3) = (3, -4)$ . By Example (Lines of the Plane Are Its Hyperplanes) (b), a normal is $(4, 3)$ and $c = 4 \cdot 1 + 3 \cdot 3 = 13$ . So the line is $4x_1 + 3x_2 = 13$ . Check at the second point: $16 - 3 = 13$ .

Uses: Proposition 10.1.2: direction $q - p$ .

10.2.2. Equidistant points

At the end of Section 10.1 we found a point at equal distances from $p$ and $q$ that is not their midpoint, and asked for all such points. Chapter 9 found their equation in Exercise (B3). Here that set is named as a hyperplane. Its two sides are the points nearer to $p$ and the points nearer to $q$ .

A small case first. In $\mathbb{R}^2$ , take $p = (0, 0)$ and $q = (2, 0)$ . A point $x$ is equally far from both exactly when $x_1^2 + x_2^2 = (x_1 - 2)^2 + x_2^2$ , that is, when $4x_1 = 4$ , or $x_1 = 1$ . This is a line through the midpoint $(1, 0)$ . The points with $x_1 < 1$ are nearer to $p$ .

Proposition 10.2.2 (The Perpendicular Bisector)

Let $p, q \in \mathbb{R}^n$ with $p \ne q$ , and let $c \mathrel{:=} \frac12 \bigl( \left\| q \right\|^2 - \left\| p \right\|^2 \bigr)$ . Then $\{ x \in \mathbb{R}^n : \left\| x - p \right\| = \left\| x - q \right\| \} = H_{q - p, \, c} ,$ a hyperplane that contains the midpoint $\frac12 (p + q)$ . Moreover, $\left\| x - p \right\| < \left\| x - q \right\|$ if and only if $\langle q - p, x \rangle < c$ .

This hyperplane is the perpendicular bisector of $p$ and $q$ .

Idea. Expand the square (Example (Method: expand the square)): compare squares instead of norms. The two squared norms both contain $\left\| x \right\|^2$ , which cancels, and what is left of their difference is linear in $x$ .

Proof. For non-negative reals $s, u$ , we have $s < u$ if and only if $s^2 < u^2$ . Since the order of $\mathbb{R}$ is total, also $s = u$ if and only if $s^2 = u^2$ . Norms are non-negative, so both claims may be checked on squares. By Eq. 9.4.1 and linearity, $\begin{aligned} \left\| x - p \right\|^2 - \left\| x - q \right\|^2 &= \bigl( \left\| x \right\|^2 - 2\langle x, p \rangle + \left\| p \right\|^2 \bigr) - \bigl( \left\| x \right\|^2 - 2\langle x, q \rangle + \left\| q \right\|^2 \bigr) \\ &= 2\langle q - p, x \rangle - 2c . \end{aligned}$ So $\left\| x - p \right\|^2 = \left\| x - q \right\|^2$ if and only if $\langle q - p, x \rangle = c$ , and $\left\| x - p \right\|^2 < \left\| x - q \right\|^2$ if and only if $\langle q - p, x \rangle < c$ . Since $p \ne q$ , $q - p \ne 0$ , so $H_{q - p, c}$ is a hyperplane. For the midpoint, linearity and symmetry give $\langle q - p, p + q \rangle = \left\| q \right\|^2 - \left\| p \right\|^2$ , so $\langle q - p, \frac12 (p + q) \rangle = c$ .

Uses: Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (g), (i): squares keep order, order is total; Proposition 9.4.1 (a): norms are non-negative; Proposition 9.3.1 (a), (b): symmetry, linearity.

The open half-space $\langle q - p, x \rangle < c$ is the side of the bisector that contains $p$ : the points strictly nearer to $p$ than to $q$ . The midpoint is the only point of the segment $[p, q]$ on the bisector. Indeed, for a point of $[p, q]$ the two distances add up to $\left\| p - q \right\|$ , by Theorem 10.1.4. If they are also equal, the point is the midpoint (Corollary 10.1.5). For $n \ge 2$ , though, the bisector has many other points.

Quick check. Find the points of $\mathbb{R}^2$ equidistant from $(0, 0)$ and $(2, 4)$ .

Solution. Here $q - p = (2, 4)$ and $c = \frac12 (20 - 0) = 10$ , so the set is $2x_1 + 4x_2 = 10$ , that is, $x_1 + 2x_2 = 5$ (the same hyperplane, by Proposition 10.2.1 with $\lambda = \frac12$ ). The midpoint $(1, 2)$ satisfies it: $1 + 4 = 5$ .

10.2.3. Nearest points

Chapter 9 found the point of a line through the origin nearest to a given point (Theorem 9.6.1). A general line $p + \mathbb{R} v$ is the line $\mathbb{R} v$ slid by $p$ , and distances do not change under sliding (Proposition 9.4.2 (c)). So the answer carries over: slide $x$ by $-p$ , project, slide back.

For example, take the horizontal line $(0, 1) + \mathbb{R}(1, 0)$ in $\mathbb{R}^2$ and the point $x = (3, 5)$ . The point of the line nearest to $x$ is $(3, 1)$ , straight below $x$ , at distance $4$ . The step from it to $x$ is $(0, 4)$ , orthogonal to the direction $(1, 0)$ .

Proposition 10.2.3 (The Nearest Point on a Line)

Let $L = p + \mathbb{R} v$ with $v \ne 0$ , let $x \in \mathbb{R}^n$ , and put $t_0 \mathrel{:=} \frac{\langle x - p, v \rangle}{\left\| v \right\|^2} , \qquad m \mathrel{:=} p + t_0 v .$ Then $x - m \perp v$ , and for every $t \in \mathbb{R}$ , $\left\| x - (p + tv) \right\|^2 = \left\| x - m \right\|^2 + (t - t_0)^2 \left\| v \right\|^2 .$ Hence $m$ is the unique point of $L$ nearest to $x$ .

Proof. Apply Theorem 9.6.1 to the vectors $x - p$ and $v \ne 0$ . Its number is $t_0$ , and $x - m = (x - p) - t_0 v$ . Part (a) gives $x - m \perp v$ . Part (b), since $x - (p + tv) = (x - p) - tv$ , gives the identity, and with it $\left\| x - (p + tv) \right\| \ge \left\| x - m \right\|$ , with equality if and only if $t = t_0$ . So a nearest point $p + tv$ has $t = t_0$ , and is $m$ .

Since $m$ is described as the nearest point, it depends only on $L$ and $x$ , not on the parametrization used to compute it. The number $\left\| x - m \right\|$ is the distance from $x$ to $L$ . For $p = 0$ this is Theorem 9.6.1. Two cases of it were computed in Example (Distance to a line through the origin).

For a hyperplane the roles turn around. On a line we search along the direction $v$ and the error is orthogonal to $v$ . On a hyperplane the error should be parallel to the normal $a$ : from $x$ , move along $a$ until the equation $\langle a, x \rangle = c$ holds.

For example, take $x_1 + x_2 = 2$ in $\mathbb{R}^2$ , so $a = (1, 1)$ , and the point $x = (3, 3)$ . Moving to $x - s(1, 1)$ changes the value $x_1 + x_2 = 6$ by $-2s$ . The value is $2$ when $s = 2$ , at the point $(1, 1)$ . Its distance from $x$ is $\left\| (2, 2) \right\| = 2\sqrt 2$ .

Proposition 10.2.4 (Distance to a Hyperplane)

Let $a \ne 0$ , $c \in \mathbb{R}$ and $x \in \mathbb{R}^n$ , and put $x^* \mathrel{:=} x - \frac{\langle a, x \rangle - c}{\left\| a \right\|^2} \, a .$ Then $x^* \in H_{a,c}$ , and for every $y \in H_{a,c}$ , $\left\| x - y \right\|^2 = \left\| x - x^* \right\|^2 + \left\| x^* - y \right\|^2 .$ Hence $x^*$ is the unique point of $H_{a,c}$ nearest to $x$ , and the distance from $x$ to $H_{a,c}$ is $\left\| x - x^* \right\| = \frac{\left| \langle a, x \rangle - c \right|}{\left\| a \right\|} .$

Proof. Write $s \mathrel{:=} (\langle a, x \rangle - c)/\left\| a \right\|^2$ , so $x^* = x - sa$ . By linearity, $\langle a, x^* \rangle = \langle a, x \rangle - s\left\| a \right\|^2 = c$ , so $x^* \in H_{a,c}$ . Let $y \in H_{a,c}$ . Then $x - y = (x - x^*) + (x^* - y)$ , where $x - x^* = sa$ and $\langle a, x^* - y \rangle = c - c = 0$ . Hence $\langle x - x^*, x^* - y \rangle = s \cdot 0 = 0$ , and Pythagoras (Proposition 9.4.4) gives the identity. The identity shows $\left\| x - y \right\|^2 \ge \left\| x - x^* \right\|^2$ . Equality holds if and only if $\left\| x^* - y \right\| = 0$ , that is, $y = x^*$ . Comparing squares of non-negative numbers, as in the proof of Proposition 10.2.2, gives the same for the norms. Finally, since $\left\| sa \right\| = \left| s \right| \, \left\| a \right\|$ , $\left\| x - x^* \right\| = \left| s \right| \, \left\| a \right\| = \frac{\left| \langle a, x \rangle - c \right|}{\left\| a \right\|^2} \, \left\| a \right\| = \frac{\left| \langle a, x \rangle - c \right|}{\left\| a \right\|} ,$ using $\left| u/w \right| = \left| u \right|/w$ for $w > 0$ .

Uses: Proposition 9.3.1 (b): linearity; Proposition 9.4.1 (a), (b): norm zero only at 0, $\left\| sa \right\| = \left| s \right|\left\| a \right\|$ ; Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.3 (b): $\left| u/w \right| = \left| u \right|/\left| w \right|$ .

The formula does not depend on which normal is used. Replacing $(a, c)$ by $(\lambda a, \lambda c)$ multiplies the numerator $\left| \langle a, x \rangle - c \right|$ by $\left| \lambda \right|$ and the denominator $\left\| a \right\|$ by $\left| \lambda \right|$ ; in $x^*$ the coefficient is multiplied by $\lambda/\lambda^2$ and the vector by $\lambda$ . Nothing changes. The sign of $\langle a, x \rangle - c$ , which the absolute value discards, says on which side of the hyperplane $x$ lies.

Quick check. Find the distance from $(1, 2, 3)$ to the hyperplane $x_1 + x_2 + x_3 = 0$ , and the nearest point.

Solution. Here $a = (1, 1, 1)$ , $c = 0$ , $\langle a, x \rangle - c = 6$ and $\left\| a \right\|^2 = 3$ . So $x^* = (1, 2, 3) - 2(1, 1, 1) = (-1, 0, 1)$ , which satisfies $-1 + 0 + 1 = 0$ , and the distance is $6/\sqrt 3 = 2\sqrt 3$ (as $(2\sqrt 3)^2 = 12 = 36/3$ ).

10.2.4. Spheres meet lines; segments cross hyperplanes

A sphere is the set of points at one fixed distance from a center. In the plane, the points at distance $1$ from the origin form the unit circle $x_1^2 + x_2^2 = 1$ . On the number line, the points at distance $r$ from $c$ are the two points $c \pm r$ .

Definition 10.2.2 (Sphere)

For $c \in \mathbb{R}^n$ and a real $r > 0$ , the sphere with center $c$ and radius $r$ is $S(c, r) \mathrel{:=} \{ x \in \mathbb{R}^n : \left\| x - c \right\| = r \}$ . In $\mathbb{R}^2$ a sphere is called a circle.

So the unit circle is $S(0, 1)$ in $\mathbb{R}^2$ .

How many points can a line share with a sphere? Try the unit circle and the horizontal line $x_2 = h$ . A common point has $x_1^2 = 1 - h^2$ . So there are two common points if $\left| h \right| < 1$ , one if $\left| h \right| = 1$ , and none if $\left| h \right| > 1$ . Here $\left| h \right|$ is the distance from the center to the line. In general, too, the answer depends only on that distance.

Proposition 10.2.5 (A Line Meets a Sphere)

Let $L = p + \mathbb{R} v$ with $v \ne 0$ , let $c \in \mathbb{R}^n$ and $r > 0$ , and let $m = p + t_0 v$ be the point of $L$ nearest to $c$ and $d \mathrel{:=} \left\| c - m \right\|$ its distance (Proposition 10.2.3). Then $L \cap S(c, r)$ is

(a) empty if $d > r$ ;
(b) $\{ m \}$ if $d = r$ ;
(c) the two points $p + \bigl( t_0 \pm \sqrt{r^2 - d^2}/\left\| v \right\| \bigr) v$ if $d < r$ .

Idea. Parametrize (Example (Method: parametrize and solve)): the point $p + tv$ lies on the sphere exactly when one equation in $t$ holds. The identity of Proposition 10.2.3 makes that equation a square equal to a number, $(t - t_0)^2 = k$ , and the sign of $k$ decides everything.

Proof. Apply Proposition 10.2.3 to the point $c$ . Since $\left\| y - c \right\| = \left\| c - y \right\|$ for every $y$ , its identity reads $\left\| (p + tv) - c \right\|^2 = d^2 + (t - t_0)^2 \left\| v \right\|^2 \quad \text{for every } t \in \mathbb{R} .$ Both $\left\| (p + tv) - c \right\|$ and $r$ are non-negative, so we may compare squares, as in the proof of Proposition 10.2.2. Thus $p + tv \in S(c, r)$ if and only if $(t - t_0)^2 = k , \qquad k \mathrel{:=} \frac{r^2 - d^2}{\left\| v \right\|^2} .$ Since $d, r \ge 0$ , Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (g) and totality give: $k < 0$ if $d > r$ , $k = 0$ if $d = r$ , $k > 0$ if $d < r$ .

(a) A square is never negative (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (e)), so for $k < 0$ no $t$ works.
(b) For $k = 0$ , $(t - t_0)^2 = 0$ forces $t = t_0$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (f)), and the only common point is $m$ .
(c) For $k > 0$ , let $\rho \mathrel{:=} \sqrt k$ , the non-negative real whose square is $k$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 6.1.4). Then $\rho > 0$ , since $\rho \ge 0$ and $\rho^2 = k \ne 0$ . The equation $(t - t_0)^2 = \rho^2$ says $(t - t_0 - \rho)(t - t_0 + \rho) = 0$ . So $t = t_0 + \rho$ or $t = t_0 - \rho$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (f)), and both work. They differ, as $\rho > 0$ . Distinct parameters give distinct points (Section 10.1). Finally $\rho = \sqrt{r^2 - d^2}/\left\| v \right\|$ , since the right side is non-negative and its square is $k$ . Here we use that the non-negative square root is unique.

Uses: Proposition 9.4.2 (b): distance is symmetric. Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 6.1.4: a non-negative square root exists and is unique. Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 5.5.2: $\mathbb{R}$ is complete.

When $d = r$ the line is tangent to the sphere at $m$ . Exercise B3 shows that the tangent lines at a point $m$ of the sphere are exactly the lines through $m$ whose direction is orthogonal to $m - c$ .

The last result of the section is about segments. If $p$ and $q$ are on opposite sides of a hyperplane, the segment between them must cross it, and it crosses exactly once. For example, $p = (0, 0)$ and $q = (4, 0)$ lie on opposite sides of the line $x_1 = 1$ . The point $(1 - t)p + tq = (4t, 0)$ is on the line only for $t = 1/4$ , at $(1, 0)$ .

Proposition 10.2.6 (A Segment Crosses a Hyperplane Once)

Let $a \ne 0$ , $c \in \mathbb{R}$ , and let $p, q \in \mathbb{R}^n$ with $\langle a, p \rangle < c < \langle a, q \rangle$ . Then $[p, q]$ meets $H_{a,c}$ in exactly one point, namely $(1 - t_1)p + t_1 q$ with $t_1 \mathrel{:=} \frac{c - \langle a, p \rangle}{\langle a, q \rangle - \langle a, p \rangle} \in (0, 1) .$

Proof. Write $\alpha \mathrel{:=} \langle a, p \rangle$ and $\beta \mathrel{:=} \langle a, q \rangle$ , so $\alpha < c < \beta$ . Every point of $[p, q]$ is $(1 - t)p + tq$ with $t \in [0, 1]$ (Definition 10.1.3). By linearity (Proposition 9.3.1 (b)), $\langle a, (1 - t)p + tq \rangle = \alpha + t(\beta - \alpha) .$ This equals $c$ if and only if $t(\beta - \alpha) = c - \alpha$ , and since $\beta - \alpha > 0$ , if and only if $t = t_1$ . Moreover $0 < c - \alpha < \beta - \alpha$ , so $0 < t_1 < 1$ . Hence the point with parameter $t_1$ lies on $[p, q]$ and on $H_{a,c}$ , and every common point has parameter $t_1$ , so it is that point.

Along the segment, the value of $\langle a, x \rangle$ grows from $\alpha$ to $\beta$ at a constant rate in $t$ , so it passes the level $c$ once. When $p$ and $q$ are on the same side, the segment does not reach $H_{a,c}$ at all. Exercise B4 (d) treats the side $\langle a, x \rangle < c$ ; the other side follows by replacing $(a, c)$ with $(-a, -c)$ . Section 10.3 calls a set convex if it contains the segment between any two of its points.

10.2.5. Exercises

A. Check your understanding

Exercise (A1)

Consider the hyperplane $2x_1 - x_2 + 2x_3 = 6$ in $\mathbb{R}^3$ .

(a) Give a normal vector, and a second normal vector different from the first.
(b) Find the distance from $0$ to the hyperplane.
(c) Find the point of the hyperplane nearest to $0$ .

Solution.

(a) $a = (2, -1, 2)$ ; for example $-a = (-2, 1, -2)$ , with the equation $-2x_1 + x_2 - 2x_3 = -6$ (Proposition 10.2.1).
(b) $\left\| a \right\| = \sqrt{4 + 1 + 4} = 3$ , so by Proposition 10.2.4 the distance is $\left| 0 - 6 \right|/3 = 2$ .
(c) $x^* = 0 - \frac{0 - 6}{9} a = \frac23 (2, -1, 2) = (4/3, -2/3, 4/3)$ . Check: $8/3 + 2/3 + 8/3 = 6$ , and $\left\| x^* \right\| = \frac23 \cdot 3 = 2$ .

Exercise (A2)

Are $0$ and $(3, 0, 1)$ on the same side of the hyperplane $2x_1 - x_2 + 2x_3 = 6$ ? If not, where does the segment between them cross it?

Solution. $\langle a, 0 \rangle = 0 < 6$ and $\langle a, (3, 0, 1) \rangle = 6 + 0 + 2 = 8 > 6$ : opposite sides. By Proposition 10.2.6, $t_1 = (6 - 0)/(8 - 0) = 3/4$ , and the crossing point is $\frac34 (3, 0, 1) = (9/4, 0, 3/4)$ . Check: $9/2 + 3/2 = 6$ .

B. Practice

Exercise (B1)

Find the point of the line $(1, 0, 0) + \mathbb{R}(1, 1, 1)$ nearest to $(0, 3, 0)$ , and the distance between them.

Solution. By Proposition 10.2.3 with $p = (1, 0, 0)$ , $v = (1, 1, 1)$ , $x = (0, 3, 0)$ : $x - p = (-1, 3, 0)$ , $\langle x - p, v \rangle = 2$ , $\left\| v \right\|^2 = 3$ , so $t_0 = 2/3$ and $m = (5/3, 2/3, 2/3)$ . Then $x - m = (-5/3, 7/3, -2/3)$ , orthogonal to $v$ as it should be ( $-5 + 7 - 2 = 0$ ), and $\left\| x - m \right\|^2 = \frac{25 + 49 + 4}{9} = \frac{26}{3} .$ Hence the distance is $\sqrt{26/3}$ .

Exercise (B2)

Find the points where the line $\mathbb{R}(1, 2, 2)$ meets the sphere $S((3, 0, 0), 3)$ .

Solution. Parametrize (Example (Method: parametrize and solve)): $t(1, 2, 2) - (3, 0, 0) = (t - 3, 2t, 2t)$ , whose squared norm is $(t - 3)^2 + 8t^2 = 9t^2 - 6t + 9$ . Both norms being non-negative, the point is on the sphere if and only if $9t^2 - 6t + 9 = 9$ , that is, $3t(3t - 2) = 0$ , so $t = 0$ or $t = 2/3$ . The points are $0$ and $(2/3, 4/3, 4/3)$ .

The same answer from Proposition 10.2.5: $t_0 = \langle (3, 0, 0), (1, 2, 2) \rangle/9 = 1/3$ , $m = (1/3, 2/3, 2/3)$ , $d^2 = \left\| (8/3, -2/3, -2/3) \right\|^2 = 72/9 = 8 < 9$ , so there are two points, at $t = 1/3 \pm \sqrt{9 - 8}/3$ .

Exercise (B3)

Let $m \in S(c, r)$ and $v \ne 0$ . Prove that the line $m + \mathbb{R} v$ meets $S(c, r)$ only at $m$ if and only if $\langle m - c, v \rangle = 0$ .

Solution. Expand the square (Eq. 9.4.1), and use $\left\| tv \right\| = \left| t \right|\,\left\| v \right\|$ and $\left\| m - c \right\| = r$ : $\left\| (m + tv) - c \right\|^2 = r^2 + 2t\langle m - c, v \rangle + t^2 \left\| v \right\|^2 .$ Comparing squares of non-negative numbers, $m + tv \in S(c, r)$ if and only if this equals $r^2$ , that is, $t \bigl( 2\langle m - c, v \rangle + t\left\| v \right\|^2 \bigr) = 0$ , that is, $t = 0$ or $t = t' \mathrel{:=} -2\langle m - c, v \rangle/\left\| v \right\|^2$ . Distinct parameters give distinct points (Section 10.1). The point $m$ has parameter $t = 0$ , so it is the only common point if and only if $t' = 0$ , that is, if and only if $\langle m - c, v \rangle = 0$ .

A line through $m$ meets $S(c, r)$ at $m$ , so by Proposition 10.2.5 case (a) is impossible and case (c) gives two points. So the line meets $S(c, r)$ only at $m$ exactly when $d = r$ , that is, when it is tangent there, and then $m$ is its point nearest to $c$ . Hence the tangent lines at $m$ are those whose direction is orthogonal to the radius $m - c$ .

Exercise (B4: Prove or disprove)

Throughout, $a \ne 0$ .

(a) If $c \ne c'$ , then $H_{a,c} \cap H_{a,c'} = \emptyset$ .
(b) Any two distinct hyperplanes of $\mathbb{R}^3$ have a common point.
(c) For $n \ge 2$ and $p \ne q$ in $\mathbb{R}^n$ , the only point equidistant from $p$ and $q$ is their midpoint.
(d) If $\langle a, p \rangle < c$ and $\langle a, q \rangle < c$ , then $[p, q]$ does not meet $H_{a,c}$ .
(e) The distance from $0$ to $H_{a,c}$ is $\left| c \right|$ .

Solution.

(a) True. A common point $x$ would give $c = \langle a, x \rangle = c'$ .
(b) False. $H_{e_1, 0} = \{ x_1 = 0 \}$ and $H_{e_1, 1} = \{ x_1 = 1 \}$ are disjoint by (a), so no point is common to both; they are distinct since both are non-empty.
(c) False. For $p = (-1, 0)$ and $q = (1, 0)$ , Proposition 10.2.2 gives the set $H_{(2, 0), 0} = \{ x_1 = 0 \}$ , which contains $(0, y)$ for every $y$ , not only the midpoint $(0, 0)$ .
(d) True. For $t \in [0, 1]$ , $\langle a, (1 - t)p + tq \rangle = (1 - t)\langle a, p \rangle + t\langle a, q \rangle$ . Each of $(1 - t)\langle a, p \rangle \le (1 - t)c$ and $t\langle a, q \rangle \le tc$ holds. One of $t$ , $1 - t$ is positive, and the inequality with that weight is strict. Adding the two inequalities gives a value $< c$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (b)). So no point of $[p, q]$ satisfies $\langle a, x \rangle = c$ .
(e) False. For $a = (2, 0)$ and $c = 2$ , the hyperplane is $\{ x_1 = 1 \}$ , at distance $\left| 0 - 2 \right|/\left\| a \right\| = 1 \ne 2$ from $0$ (Proposition 10.2.4). The correct value is $\left| c \right|/\left\| a \right\|$ .

C. Going deeper

Exercise (C1: Chapter check: the circle through three points)

Prove that three non-collinear points $p, q, s$ of $\mathbb{R}^2$ lie on exactly one circle. Find it for $(0, 0)$ , $(4, 0)$ , $(0, 2)$ .

Solution (Hint 1). Where must the center of such a circle lie, relative to $p$ and $q$ ? Relative to $q$ and $s$ ?

Solution (Hint 2). A center $z$ satisfies $\left\| z - p \right\| = \left\| z - q \right\|$ and $\left\| z - q \right\| = \left\| z - s \right\|$ ; by Proposition 10.2.2, $z$ lies on two lines of $\mathbb{R}^2$ . When do two lines of the plane meet in exactly one point?

Solution. The three points are distinct: if two coincided, the line through the two distinct ones among them (or any line through the point, if all three coincide) would contain all three.

Reduction. $S(z, \rho)$ contains $p, q, s$ if and only if $\left\| z - p \right\| = \left\| z - q \right\| = \left\| z - s \right\| = \rho$ . By Proposition 10.2.2 this puts $z$ on $B_1 \mathrel{:=} H_{q - p, c_1}$ and on $B_2 \mathrel{:=} H_{s - q, c_2}$ , with $c_1, c_2$ as there.

The normals are not multiples. If $s - q = \lambda(q - p)$ , then $s = p + (1 + \lambda)(q - p)$ lies on the line $p + \mathbb{R}(q - p)$ , which contains $p$ and $q$ too, against non-collinearity. So, by Exercise (B2, used later) (a), with $a \mathrel{:=} q - p \ne 0$ and $a' \mathrel{:=} s - q \ne 0$ , $D \mathrel{:=} a_1 a'_2 - a_2 a'_1 \ne 0$ .

The bisectors meet once. By Example (Lines of the Plane Are Its Hyperplanes) (a), $B_1$ and $B_2$ are lines with directions $(-a_2, a_1)$ and $(-a'_2, a'_1)$ . For these two directions, the expression of Exercise (B2, used later) (a) is $(-a_2) a'_1 - a_1 (-a'_2) = D \ne 0$ . So neither direction is a multiple of the other. By part (b) of the same exercise, the lines meet in exactly one point $z$ .

Uniqueness. If $p, q, s \in S(z', \rho')$ , then $z' \in B_1 \cap B_2 = \{ z \}$ by the reduction, and $\rho' = \left\| z - p \right\| = \rho$ .

The example. The point $(0, 2)$ is not on the line $x_2 = 0$ through the other two, so the three points are not collinear. $B_1$ : $q - p = (4, 0)$ , $c_1 = \frac12 (16 - 0) = 8$ , so $4x_1 = 8$ , $x_1 = 2$ . $B_2$ : $s - q = (-4, 2)$ , $c_2 = \frac12 (4 - 16) = -6$ , so $-4x_1 + 2x_2 = -6$ , and with $x_1 = 2$ , $x_2 = 1$ . The center is $(2, 1)$ , and $\rho = \left\| (2, 1) \right\| = \sqrt 5$ ; indeed $\left\| (2, 1) - (4, 0) \right\|^2 = 4 + 1$ and $\left\| (2, 1) - (0, 2) \right\|^2 = 4 + 1$ .

What a complete answer needs. (10 points)

2: reduces “ $p, q, s$ on $S(z, \rho)$ ” to “ $z$ on both perpendicular bisectors”, citing the bisector result or expanding the squares directly.
2: shows the normals $q - p$ , $s - q$ are not multiples, from non-collinearity, and carries this to the directions of the two lines.
1: concludes the two lines meet in exactly one point, by the two-line criterion of Section 10.1 or by solving the two equations directly with $D \ne 0$ .
2: existence: the common point is equidistant from all three, with $\rho > 0$ .
1: uniqueness of the center and hence of the radius.
2: the example: non-collinearity checked, center $(2, 1)$ , radius $\sqrt 5$ , with the three distances checked.

A route that solves the two bisector equations as a linear system in $z$ , with $D \ne 0$ shown, earns the second and third items at once (3 points); the existence item still needs $\rho > 0$ .

Slips that lose the step. Each of these earns 0 for its step: assuming the bisectors meet without using non-collinearity; checking only the example; forgetting that the radius must be positive.