1. Vectors, projections and geometry

Learning goals and prerequisites

Use an orthonormal Cartesian coordinate system in $\mathbb R^n$, with $n\ge1$; in $\mathbb R^3$ choose its right-handed orientation. Compute vector operations, project onto a direction, and describe lines and planes. Prerequisites are arithmetic, square roots and elementary equations. Coordinates represent a vector after a choice of axes; a point and a displacement play different geometric roles.

Definitions and structure

A vector $u=(u_1,\ldots,u_n)$ in $\mathbb R^n$ is added componentwise and multiplied by a real scalar componentwise. The dot product and length are

$$u\cdot v=\sum_{j=1}^n u_jv_j,\qquad \|u\|=\sqrt{u\cdot u}.$$

Two nonzero vectors are perpendicular when their dot product is zero. A unit vector has length one. For nonzero $v$, the projection of $u$ onto the line spanned by $v$ is $(u\cdot v)v/\|v\|^2$. A line through a point $p$ with nonzero direction $v$ is $p+tv$, $t\in\mathbb R$. In $\mathbb R^3$, a plane with nonzero normal $n$ through $p$ is the set of points $x$ satisfying $n\cdot(x-p)=0$. The same equation in $\mathbb R^n$ defines a hyperplane, which has dimension $n-1$. To verify the dimension assertion, choose $j$ with $n_j\ne0$. The equation uniquely determines $x_j=p_j-\sum_{k\ne j}n_k(x_k-p_k)/n_j$ after the other $n-1$ coordinates have been chosen arbitrarily. These are exactly $n-1$ independent affine parameters.

Theorem: Cauchy–Schwarz and the triangle inequality

For $u,v\in\mathbb R^n$, $|u\cdot v|\le\|u\|\|v\|$ and $\|u+v\|\le\|u\|+\|v\|$. Equality in Cauchy–Schwarz holds exactly when the vectors are linearly dependent, including a zero vector.

Proof. If $v=0$, the result is immediate. Otherwise let $a=(u\cdot v)/\|v\|^2$. Expanding the nonnegative square gives

$$0\le\|u-av\|^2=\|u\|^2-\frac{(u\cdot v)^2}{\|v\|^2}.$$

Rearranging proves the first inequality; equality holds exactly when $u=av$. Now expand $\|u+v\|^2$ and bound $u\cdot v$ above by $\|u\|\|v\|$. Both sides are nonnegative, so taking square roots proves the second. This also ensures that defining an angle by $\cos\theta=(u\cdot v)/(\|u\|\|v\|)$ makes sense for nonzero vectors.

In $\mathbb R^3$, define

$$u\times v=(u_2v_3-u_3v_2,\ u_3v_1-u_1v_3,\ u_1v_2-u_2v_1).$$

Direct substitution gives $(u\times v)\cdot u=(u\times v)\cdot v=0$. Expanding squares gives $\|u\times v\|^2=\|u\|^2\|v\|^2-(u\cdot v)^2$. Thus the cross-product magnitude is the parallelogram area; reversing the order reverses orientation.

Worked examples

Projection and distance. Let $u=(3,1)$ and $v=(1,2)$. Then $u\cdot v=5$, $\|v\|^2=5$, so the projection is $(1,2)$. The residual $(2,-1)$ is perpendicular to $v$, and the distance from $u$ to its line is $\sqrt5$. The projection is a vector, not the scalar coefficient $1$.

A plane from three points. Through $A=(1,0,0)$, $B=(0,1,0)$ and $C=(0,0,2)$, take $B-A=(-1,1,0)$ and $C-A=(-1,0,2)$. Their cross product is $(2,2,1)$. The plane is $2x+2y+z=2$. The nonzero cross product verifies that the three points are not collinear, so the plane is unique.

Graded practice

1. Foundation. For $a=(2,-1,2)$ and $b=(1,2,0)$, compute $a\cdot b$, both lengths, and their angle.
2. Core. Find the closest point on $L=\{(1,1)+t(1,-1):t\in\mathbb R\}$ to $q=(4,2)$.
3. Core. Find the area of the triangle with vertices $(0,0,0),(1,2,0),(0,1,3)$.
4. Proof. Prove that the projection residual $u-av$ minimizes $\|u-tv\|$ over all real $t$, and that the minimizer is unique when $v\ne0$.

Full solutions

1. The dot product is $2-2=0$. The lengths are $3$ and $\sqrt5$. Both are nonzero, so the angle is $\pi/2$.
2. Project $q-(1,1)=(3,1)$ onto $v=(1,-1)$. The coefficient is $(3-1)/2=1$. The closest point is $(2,0)$; the residual $(2,2)$ is perpendicular to $v$ and has length $2\sqrt2$.
3. The cross product is $(6,-3,1)$, with length $\sqrt{46}$. A triangle has half the parallelogram area, so the answer is $\sqrt{46}/2$.
4. Put $a=(u\cdot v)/\|v\|^2$. Since $(u-av)\cdot v=0$, expansion gives $\|u-tv\|^2=\|u-av\|^2+(a-t)^2\|v\|^2$. The last term is nonnegative and vanishes exactly when $t=a$. This proves both claims without relying on a diagram.

Summary and term definitions

The dot product detects perpendicularity and supports metric estimates. Projection separates parallel and perpendicular parts; cross products encode oriented area in three dimensions. A normal is perpendicular to every direction in a plane; a projection is the closest component in a prescribed subspace; linear dependence means one vector is a scalar multiple of the other, or one is zero. Do not normalize a zero vector or divide by its squared length.