8. Affine and orthogonal geometry in higher dimensions

Learning goals and definitions

Separate points from directions, describe affine sets in $\mathbb R^n$, solve line–hyperplane intersections, and prove distance formulas. For $u,v\in\mathbb R^n$, use $u\cdot v=\sum u_iv_i$ and $\|u\|=\sqrt{u\cdot u}$. A vector subspace contains zero and is closed under sums and scalar multiples. An affine subspace is a translate $p+V$ of a vector subspace $V$; it need not contain zero.

A line is $p+tv$ with $v\ne0$. A hyperplane is $H=\{x:n\cdot x=c\}$ with normal $n\ne0$. In dimension three it is a plane; in dimension two it is a line. Two descriptions may name the same set with different base points or normal multiples. A hyperplane contains every direction perpendicular to $n$.

An orthonormal family $e_1,\ldots,e_k$ satisfies $e_i\cdot e_j=0$ for $i\ne j$ and $\|e_i\|=1$. Its vectors are linearly independent: dotting $\sum a_ie_i=0$ with $e_j$ gives $a_j=0$.

Theorem: closest point on a hyperplane

For $q\in\mathbb R^n$, the unique closest point of $H$ is

$$p=q-\frac{n\cdot q-c}{\|n\|^2}n, \qquad \operatorname{dist}(q,H)=\frac{|n\cdot q-c|}{\|n\|}.$$

Proof. Direct substitution gives $n\cdot p=c$, so $p\in H$. For any $x\in H$, the vector $x-p$ is perpendicular to $n$, while $q-p$ is a multiple of $n$. Therefore the cross term vanishes in

$$\|q-x\|^2=\|(q-p)-(x-p)\|^2=\|q-p\|^2+\|x-p\|^2.$$

The distance is minimized exactly at $x=p$. Taking the length of the displayed formula for $q-p$ gives the distance. Nonzero $n$ ensures both valid division and uniqueness.

More generally, if $V$ has orthonormal basis $e_1,\ldots,e_k$, then the projection onto $V$ is $\sum_j(q\cdot e_j)e_j$. Subtracting it from $q$ gives a vector perpendicular to every $e_j$, hence to all of $V$. The same squared-distance argument proves it is the unique nearest vector in $V$. For $p+V$, first project the displacement $q-p$, then add $p$.

Worked examples

Higher-dimensional distance. In $\mathbb R^4$, let $H$ be $x_1+x_2+x_3+x_4=2$ and $q=(2,0,0,2)$. Here $n=(1,1,1,1)$, $n\cdot q-c=2$ and $\|n\|^2=4$. The foot is $(3/2,-1/2,-1/2,3/2)$ and the distance is $2/2=1$.

Intersection classification. A line $p+tv$ meets a hyperplane when $n\cdot p+t(n\cdot v)=c$. If $n\cdot v\ne0$, there is one intersection with $t=(c-n\cdot p)/(n\cdot v)$. If $n\cdot v=0$, the whole line lies in the hyperplane when $n\cdot p=c$ and otherwise no intersection exists. “Parallel” alone does not distinguish containment from disjointness.

Graded practice

1. Foundation. Find a line through $(1,2,3)$ parallel to $(2,-1,1)$, and decide whether it passes through $(5,0,5)$.
2. Core. Find the distance from $(1,-1,2)$ to $2x-y+2z=3$.
3. Core. For orthonormal vectors $e_1=(1,1,0)/\sqrt2$ and $e_2=(0,0,1)$, project $q=(2,0,3)$ onto their span.
4. Proof. Prove that an affine subspace $p+V$ contains the line segment between any two of its points. Does the converse assertion characterize affine subspaces?

Full solutions

1. The line is $(1,2,3)+t(2,-1,1)$. At $t=2$ it reaches $(5,0,5)$, so the point lies on it.
2. The normal is $(2,-1,2)$ with length $3$. Its dot product with the point is $7$, so the distance is $|7-3|/3=4/3$.
3. The dot products are $\sqrt2$ and $3$. The projection is $\sqrt2e_1+3e_2=(1,1,3)$. The residual $(1,-1,0)$ is perpendicular to both basis vectors.
4. Write the points as $p+u,p+v$ with $u,v\in V$. For $0\le t\le1$, their convex combination is $p+((1-t)u+tv)$, still in $p+V$. The converse fails: a closed ball contains the segment between any two points but is not an affine subspace. Segment closure is convexity, a weaker property.

Summary and term definitions

An affine set is a translated linear space. A hyperplane is one nontrivial linear equation in $n$ variables. An orthonormal basis combines perpendicularity with unit lengths. A convex set contains every segment joining its points. Orthogonal decomposition proves distance statements in every finite dimension; it does not require a three-dimensional cross product or a diagram.