Lecture 22
Geometry of Vectors

We have treated vectors in Rⁿ primarily from the algebraic point of view. Now two geometric notions, those of length and perpendicularity, will lead us to additional important algebraic operations on vectors, as well as being important in their own right. In both cases, we will extend to any number of dimensions the usual geometrical interpretations in two or three dimensions.

Def. For v in Rⁿ, the norm (or length) of v is ||v|| = (v₁² + · · · + v_n²)^1/2.

The usual notion of distance between two vectors can be expressed using the norm.

Def. For u, v in Rⁿ, the distance between u and v is ||u - v||.

Examples. These notions extend the usual one of Euclidean length and distance.

Now, the Pythagorean Theorem in R² (and, similarly, in R³) states that u is perpendicular to v if and only if || u - v||² = ||u||² +||v||². (Draw the triangle with legs u and v and hypotenuse u - v.) But this statement is equivalent to the following:

The analogous result holds in R³ . This result leads us to consider the following important number associated to a pair of vectors u and v.

Definition For u, v in Rⁿ, the dot product (or scalar product) of u and v is

Important Note: The dot product of two vectors is a scalar.

The condition on the dot product being zero (equivalent to the perpendicularity of the vectors) in R² or R³
leads us to define perpendicularity in any number of dimensions via the same condition.

Definition The vectors u and v in Rⁿ are called orthogonal (or perpendicular) if u^· v = 0.

Examples

The norm and dot product are new algebraic operations, and it is easy to verify that they satisy the following algebraic rules.

Theorem (Algebraic Rules for Norms and Dot Products) For u, v, w in Rⁿ, and c in R,
(a) u^·u = ||u||².
(b) u^·u = 0 if and only if u = 0.
(c) u^·v = v^·u. (Commutativity)
(d) u^· (v + w) = u^·v + u^· w (Distributive Law)
(e) (v + w)^·u = v^·u + w^·u. (Distributive Law)
(f) (c u)^·v = c(u ^·v) = u^·(cv). (Associative Law)
(g) ||c u|| =|c| ||u||

Given our definitions of norm and of orthogonality, the same proof of the equivalence of orthogonality and the Pythagorean formula holds in any number of dimensions, not just in dimensions two and three.

Pythagorean Theorem
Let u and v bevectors in R ⁿ . Then u and v are orthogonal if and only if

We can consider a geometric example with important physical applications: given a line and a vector, what portion of that vector lies in the direction parallel to the given line? An equivalent problem is to find the distance between the vector and the line. The natural idea is to "drop a perpendicular" from the vector to the line, that is, to write the vector v as the sum of one vector parallel to the line, and one vector perpendicular to the line. If we consider the line as being spanned by the vector u, then the problem is to write v = cu + (v - cu), since the first vector on the right is the generic vector parallel to u, and then to require in addition that (v - cu) be orthogonal to v. Thus we need to solve for the unknown c in the equation u^·(v - cu)= 0. This yields the value c = (v^· u)/(u^· u). Thus we have the following result.

Orthogonal Projection of a Vector along a Line
If v is a vector, and L is the line through the nonzero vector u, we have v = (v^·u)u/||u||² + (v - (v^·u)u/||u||²). The first vector on the right is parallel to L, and the second is orthogonal to L.

Finally, we consider two important inequalities.

Cauchy-Schwarz Inequality For u, v in Rⁿ, we have |u^·v| < ||u||||v||.

Triangle Inequality For u, v in Rⁿ, we have ||u + v|| < ||u|| + ||v||.

In order to prove the first result, note that if u = 0, then it is clear. Otherwise, we can write v = (v^·u)u/||u||² + z, with the two terms on the right hand side orthogonal to each other. The Pythagorean Theorem yields ||v||² = ||(v^·u)u/||u||² ||² + ||z||² > ||(v^·u)u/||u||² ||² = |v^·u|² /||u||². Hence ||v||² ||u||² > |v^·u|² and the result follows.

The triangle inequality can be deduced from the Cauchy-Schwarz inequality, because

250 Lecture Index