For a line, there is only one way to be parallel, but infinitely many ways to be orthogonal (think: every vector parallel to the xy-plane is orthogonal to the z-axis). Therefore the vector we want, the direction vector, is parallel to the line.

For a plane, conversely, there are infinitely many ways to be parallel, but only one way to be orthogonal (any vector orthogonal to the xy-plane is parallel to the z-axis). Therefore the vector we want, the normal vector, is orthogonal to the plane.

You may obtain these two pieces in many ways. In addition to being given the point and direction vector immediately, the following are enough to determine a line:

• two points on the line
• a point and a parallel line
• a point and two nonparallel vectors orthogonal to the line
• a point and two nonparallel lines orthogonal to the desired line
• a point and an orthogonal plane
• two intersecting (nonidentical) planes (the line of intersection)
• two intersecting (nonidentical) lines (the line through their point of intersection and orthogonal to both)

The following are enough to define a plane, in addition to being given a point and normal vector directly:

• three points in the plane
• a point and an orthogonal line
• a point and a line in the plane not containing that point
• two lines in the plane
• a point and a parallel plane
• a point and two planes orthogonal to the desired plane

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2. Whenever we want to isolate the direction of something, we use a unit vector.

3. While it is useful to think of linearly independent sets as “small” and spanning sets as “large”, be careful not to try to apply the converse. {0} is very small, but linearly dependent. {v, 2v, 3v, 4v, 5v, …} is very large, but spans only a dimension 1 space.

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Coordinate vectors and sweets
A restaurant serves 2 desserts, cannoli and cookies. Bill and Bob walk in. Bill says, “I’ll take a cannoli and two cookies.” Bob says, “I’ll take 3 cannoli, no cookies for me.” Finally, Xavier walks into the restaurant and says, “Hmm, I’ll take 3 of what Bill’s having and 2 of what Bob’s having.” To figure how much to give Xavier, the server thinks “He needs 3*1 + 2*3 cannoli and 3*2 + 2*0 cookies.”

[of course the server really thinks “what is wrong with this guy?” but we’re in math fantasy land, so suspend your disbelief.]

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• all linear transformations take 0 to 0
• images of subspaces are subspaces (special example: image)
• preimages of subspaces are subspaces (special example: kernel)
• closing a set of vectors before or after applying a transformation gives the same vector space
• we may unambiguously determine the entirety of a linear transformation from its action on a basis (linear extension) — which is what makes matrix representation possible!
• having an inverse is equivalent to being bijective
• the rank of a matrix is definable by its rows as well as its columns
• inverses of isometries are also isometries (also uses bilinearity of inner product)

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A) U = (R x R, +_U, *_U)
where (a₁, a₂) +_U (b₁, b₂) = (a₁ + 2b₁, a₂ + 3b₂)
and *_U is the usual scalar multiplication

B) V = (R x R, +_V, *_V)
where (a₁, a₂) +_V (b₁, b₂) = (a₁ + b₁ + 1, a₂ + b₂)
and *_V is the usual scalar multiplication

C) W = (R x R, +_W, *_W)
where (a₁, a₂) +_W (b₁, b₂) = (a₁ + b₁ + 1, a₂ + b₂ + 1)
and r*_W(a₁, a₂) = (ra₁ + r – 1, ra₂ + r – 1)

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Throughout this entry I am going to be conflating points and their position vectors, or alternatively vectors and the location of their heads in standard position.

Curves in 2-space that are graphs of some y = f(x) are immediately parametrizable by x = t, y = f(t).

The easiest curve to parametrize in 3-space is a line. If P is a point on the line and v is a nonzero vector parallel to the line, then every point on the line is some multiple of v added to P. We represent the variable multiples using the parameter t: r(t) = P + tv.

If the line is given by two points v₁ and v₂, r(t) = tv₂ + (1-t)v₁ = v₁ + t(v₂ – v₁). The first formulation comes from the point of view of parametrizing just the line segment from v₁ to v₂, hitting the former point as t = 0 and the latter at t = 1. If you remove the bounds on t you get the entire line containing that line segment. The second formulation comes from computing a direction vector by v₂ – v₁ and using v₁ as the point anchoring the line in space.

Circles and ellipses in 2-space are all based on the trigonometric identity sin²t + cos²t = 1. Generally x is some modification of cos t and y of sin t. For a circle of radius r, for instance, we need x² + y² = r². By multiplying the identity by r² and rewriting a bit, we get (r sin t)² + (r cos t)² = r², so letting x = r cos t and y = r sin t will work. These are not the only possible parametrizations, but they are typically the easiest to get at and work with.

If your circle is moved, with equation, say, (x – 4)² + (y + 2)² = 25, you need to parametrize so the left hand side becomes 25(cos²t + sin²t). This is easily taken care of by making the multiplicative adjustment as usual and then “undoing” the additive adjustment: x = 5 cos t + 4; y = 5 sin t – 2.

For ellipses, the multiplicative adjustment is different for each variable. If the axes of the ellipse are not parallel to the coordinate axes, life gets difficult, but if they are, it’s essentially the same process as above: (5x/2 – 1)² + y² = 1 (always solve for 1 on the right) may be parametrized as x = 2/5 cos t + 1; y = sin t.

A note on arc length versus the integral of a vector-valued function: you integrate velocity to get displacement, but you integrate speed to get distance. That is why you integrate the magnitude of the derivative to get the length of the curve.

Curvature may be thought of in these terms as well: the curvature κ at time t is |T'(t)|/|r'(t)| = speed of change in tangent direction divided by speed of change in position. You use the unit tangent T to isolate the tangent direction.

Note that every curve has infinitely many possible parametrizations. As t ranges over the real numbers, the same points in space will be traced out. Different parametrizations, however, may hit different points at a given value of t, trace out more or less distance in a unit of t change, and proceed in different directions as t changes positively or negatively.

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Exercise: show that if a vector space (over the real numbers) has at least two vectors, it has infinitely many.

Simple proof: If V has two vectors, one of them is nonzero. The span of such a vector is infinite because there are infinitely many real numbers, and its span is contained in V, so V is infinite.

Sharp-eyed student question: how do we know that aX and bX can’t be the same, for a, b distinct scalars and X a nonzero vector?

After a certain amount of trying to produce a counterexample, a quest I knew was doomed to fail but thought might be illuminating, I produced a two step proof of the infinitude of the span.

Lemma: If for some vector X there is a nonzero real number a such that aX = 0, then for any real number b, bX = 0.

Proof: Suppose X, a are as in the lemma. Since a is nonzero we may divide by it. Then for any scalar b, bX = ((b/a)a)X = (b/a)(aX) = (b/a)0 = 0.

Claim: If X is a nonzero vector and a, b are distinct real numbers, then aX, bX are distinct.

Proof: By the lemma, since 1X is not 0 (this is an axiom of vector spaces), no nonzero real number can give 0 when multiplied by X. Suppose a, b are real numbers that give the same product with X. Then by adding (-b)X to each side of the equality and factoring, we see (a-b)X = (b-b)X = 0, which implies a = b.

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