Section 1 Consolidation: Vectors and Geometry

≈ 90 min

What this section established

Nothing new is introduced here. This lesson is the exit assessment for Section 1, and its purpose is to test whether the three preceding lessons have fused into a single working picture rather than three separate procedures.

The section put three pieces on the table.

$\mathbf{R}^n$ and its arithmetic. An element of $\mathbf{R}^n$ is an ordered $n$ -tuple, written by default as a column and abbreviated in running text by the transpose, so that $(3,-1,4)^{\mathsf{T}}$ is a column of $\mathbf{R}^3$ . Addition and scalar multiplication act one component at a time and obey the commutative, associative and distributive laws, with $\mathbf{0}$ as the additive identity. The norm $\|\mathbf{v}\| = \sqrt{v_1^2 + \cdots + v_n^2}$ is Pythagoras in $n$ slots; it satisfies $\|c\mathbf{v}\| = |c|\,\|\mathbf{v}\|$ and the triangle inequality $\|\mathbf{u}+\mathbf{v}\| \leq \|\mathbf{u}\| + \|\mathbf{v}\|$ , and for $\mathbf{v} \neq \mathbf{0}$ the vector $\mathbf{v}/\|\mathbf{v}\|$ has norm $1$ .
Linear combination and span. A linear combination of $\mathbf{v}_1, \ldots, \mathbf{v}_k$ is any $c_1\mathbf{v}_1 + \cdots + c_k\mathbf{v}_k$ with real coefficients, and $\operatorname{span}\{\mathbf{v}_1,\ldots,\mathbf{v}_k\}$ is the set of all of them. Taking every coefficient zero shows $\mathbf{0}$ lies in every span. The span of one non-zero vector is a line through the origin; of two non-parallel vectors in $\mathbf{R}^3$ , a plane through the origin; and of two parallel vectors, a line again. Asking whether a given vector lies in a given span is asking whether a particular linear system has a solution.
Lines and planes. A line is $\mathbf{r} = \mathbf{p} + t\mathbf{d}$ with $\mathbf{d} \neq \mathbf{0}$ : an offset added to a span. The same line has infinitely many such descriptions, since any point of the line and any non-zero multiple of $\mathbf{d}$ will serve. A plane is $\mathbf{r} = \mathbf{p} + s\mathbf{u} + t\mathbf{v}$ with $\mathbf{u}$ and $\mathbf{v}$ non-parallel, or equivalently $\mathbf{n} \cdot (\mathbf{r} - \mathbf{p}) = 0$ , which expands to $ax + by + cz = d$ with $\mathbf{n} = (a,b,c)^{\mathsf{T}}$ . The dot product entered for one purpose only: $\mathbf{a} \cdot \mathbf{b} = 0$ records perpendicularity.

What you should now be able to do unaided

Before starting the exercises, read this list and mark honestly which items you could carry out on a blank page with no prompt. The exercises below are ordered so that the early ones test the first three items and the last ones test the final three.

Add, scale and subtract vectors componentwise, and compute a norm, a unit vector and the norm of a scalar multiple without hesitating over signs.
Evaluate a linear combination, and go the other way: given a target vector and a spanning set, set up and solve the system that decides membership.
Say what the span of a given set looks like geometrically, and detect the collapse that occurs when the vectors are parallel.
Produce a parametrisation of a line from two points, recognise when two different parametrisations describe the same line, and convert between parametric, Cartesian and normal forms.
Use $\mathbf{n} \cdot \mathbf{d} = 0$ to decide when a line is parallel to a plane, and substitute a parametrisation into a Cartesian equation to find where a line meets a plane.
Distinguish a span from an affine translate of a span, and justify the distinction by the position of $\mathbf{0}$ .

How the three lessons are one idea

The section has a single spine: take a few vectors, scale them, add them, and see what set you get.

Lesson 1 supplied the two operations. Lesson 2 applied them without restriction and named the resulting set the span. Lesson 3 did the same thing but started somewhere other than the origin, which is why every object in it has the shape offset plus span: a line is $\mathbf{p} + \operatorname{span}\{\mathbf{d}\}$ and a plane is $\mathbf{p} + \operatorname{span}\{\mathbf{u},\mathbf{v}\}$ . When $\mathbf{p} = \mathbf{0}$ , or more generally when $\mathbf{p}$ already lies in the span, the offset contributes nothing and the set is a span after all. When it does not, the set cannot be a span, because it fails to contain $\mathbf{0}$ . That one observation accounts for most of the distinctions this section asks you to make.

What will be rebuilt later, and why

Three of these ideas are deliberately provisional. It is worth knowing now which parts of your understanding are permanent and which are scaffolding.

Span. Here it is described by its pictures: a line, a plane, or a collapsed case. In Section 5 the same object is characterised by its closure properties — a set containing $\mathbf{0}$ and closed under addition and scalar multiplication — and given the name subspace. Nothing you have learned will be withdrawn, but the emphasis moves from what it looks like to what it is closed under, and that shift is what makes the theory work in $\mathbf{R}^n$ where no picture exists.
Offset plus span. The shape $\mathbf{p} + \operatorname{span}\{\ldots\}$ is not a curiosity of geometry. In Section 2 it reappears as the description of the solution set of every consistent linear system: one particular solution, plus all solutions of the associated homogeneous system. The lines and planes of Lesson 3 are the first instances of a structure that governs the whole subject.
The dot product. It has been used here only as a test for perpendicularity, which is all that the normal form of a plane requires. Section 9 develops it as an inner product, with length, angle, projection and orthogonality derived from it systematically. Until then, resist reading anything more into $\mathbf{a} \cdot \mathbf{b}$ than the statement that it vanishes exactly when the two vectors are perpendicular.

Let $\mathbf{u} = (2,-5,1)^{\mathsf{T}}$ and $\mathbf{v} = (-3,4,6)^{\mathsf{T}}$ , and set $\mathbf{w} = 3\mathbf{u} - \mathbf{v}$ . Enter the third component $w_3$ .

Compute $\|\mathbf{v}\|$ for $\mathbf{v} = (2,-3,6)^{\mathsf{T}}$ .

Describe $\operatorname{span}\{(2,-1,3)^{\mathsf{T}}\}$ geometrically, in a few words.

Let $\mathbf{v} = (0,-5,12)^{\mathsf{T}}$ and let $\hat{\mathbf{v}} = \mathbf{v}/\|\mathbf{v}\|$ be the unit vector in its direction. Enter the third component of $\hat{\mathbf{v}}$ as a decimal, correct to at least three decimal places.

Let $\mathbf{a} = (1,0,-2)^{\mathsf{T}}$ , $\mathbf{b} = (4,-1,3)^{\mathsf{T}}$ and $\mathbf{c} = (0,5,1)^{\mathsf{T}}$ . Enter the second component of the linear combination $2\mathbf{a} - 3\mathbf{b} + \mathbf{c}$ .

A line is given by $\mathbf{r} = (1,-2,4)^{\mathsf{T}} + t(3,0,-1)^{\mathsf{T}}$ . Exactly one point of this line has third coordinate equal to $1$ . Enter the first coordinate of that point.

Let $A = (2,-1,5)$ and $B = (-3,4,5)$ . Which of the following is a valid direction vector for the line through $A$ and $B$ ?

$(-5,-5,0)^{\mathsf{T}}$

$(2,-1,5)^{\mathsf{T}}$

$(1,-1,0)^{\mathsf{T}}$

$(-3,4,5)^{\mathsf{T}}$

A vector $\mathbf{v}$ satisfies $\|\mathbf{v}\| = 3$ . Compute $\|-4\mathbf{v}\|$ .

Which of the following vectors belongs to $\operatorname{span}\{(2,-1)^{\mathsf{T}}\}$ ?

$(2,1)^{\mathsf{T}}$

$(-6,3)^{\mathsf{T}}$

$(0,1)^{\mathsf{T}}$

$(4,-1)^{\mathsf{T}}$

Find the distance between the points $P = (1,-2,2)$ and $Q = (4,2,-10)$ in $\mathbf{R}^3$ .

Let $L$ be the line through $P = (1,1,3)$ and $Q = (4,-2,6)$ . Exactly one of the following points lies on $L$ . Which one?

$(7,-5,10)$

$(0,0,0)$

$(7,-5,9)$

$(-2,4,1)$

Let $\mathbf{p} = (1,2,2)^{\mathsf{T}}$ and $\mathbf{q} = (0,1,2)^{\mathsf{T}}$ , and let $\mathbf{w} = 2\mathbf{p} - 3\mathbf{q}$ . Compute $\|\mathbf{w}\|$ .

The vector $\mathbf{w} = (-1,8)^{\mathsf{T}}$ can be written as $\mathbf{w} = a\mathbf{u} + b\mathbf{v}$ where $\mathbf{u} = (1,2)^{\mathsf{T}}$ and $\mathbf{v} = (2,-1)^{\mathsf{T}}$ . Enter the value of $b$ .

A plane is given in normal form by $\mathbf{n} \cdot (\mathbf{r} - \mathbf{p}) = 0$ with $\mathbf{n} = (2,-1,3)^{\mathsf{T}}$ and $\mathbf{p} = (1,0,2)^{\mathsf{T}}$ . A line is given by $\mathbf{r} = (0,1,-1)^{\mathsf{T}} + t(1,1,1)^{\mathsf{T}}$ . The line meets the plane in exactly one point. Enter the value of $t$ at that point.

Let $S = \{(1,2,0)^{\mathsf{T}} + t(0,1,1)^{\mathsf{T}} : t \in \mathbf{R}\}$ . Which statement about $S$ is correct?

$S$ is a plane through the origin, since it is described by two vectors.

$S$ is a line that does not pass through the origin, so $S$ is not the span of any set of vectors.

$S = \operatorname{span}\{(1,2,0)^{\mathsf{T}}, (0,1,1)^{\mathsf{T}}\}$ , since both vectors appear in its description.

$S$ is a line through the origin, namely $\operatorname{span}\{(0,1,1)^{\mathsf{T}}\}$ .

A plane is parametrised through $A = (2,1,0)$ , $B = (3,3,1)$ and $C = (1,2,2)$ as $\mathbf{r} = \mathbf{a} + s(\mathbf{b}-\mathbf{a}) + t(\mathbf{c}-\mathbf{a})$ . The point $D = (5,4,0)$ lies on this plane. Enter the value of $s$ for which the parametrisation produces $D$ .

The plane through $\mathbf{p} = (1,4,-1)^{\mathsf{T}}$ with normal $\mathbf{n} = (3,-2,5)^{\mathsf{T}}$ has Cartesian equation $3x - 2y + 5z = d$ . Enter $d$ .

Let $L$ be the line $\mathbf{r} = (1,1,1)^{\mathsf{T}} + t(2,3,-1)^{\mathsf{T}}$ and let $\Pi$ be the plane $4x - 2y + 2z = 7$ . What is the relationship between $L$ and $\Pi$ ?

They meet in exactly one point.

$L$ lies entirely inside $\Pi$ .

$L$ is parallel to $\Pi$ and has no point in common with it.

$L$ is perpendicular to $\Pi$ .

There is exactly one vector $\mathbf{w}$ in $\operatorname{span}\{(2,-1,2)^{\mathsf{T}}\}$ with $\|\mathbf{w}\| = 6$ and negative first component. Enter the second component of $\mathbf{w}$ .

Let $L$ be the line $\mathbf{r} = (1,2)^{\mathsf{T}} + t(2,-1)^{\mathsf{T}}$ in $\mathbf{R}^2$ . Which of the following describes the same line $L$ ?

$\mathbf{r} = (1,2)^{\mathsf{T}} + s(4,-1)^{\mathsf{T}}$

$\mathbf{r} = (3,1)^{\mathsf{T}} + s(2,1)^{\mathsf{T}}$

$\mathbf{r} = s(2,-1)^{\mathsf{T}}$

$\mathbf{r} = (3,1)^{\mathsf{T}} + s(-4,2)^{\mathsf{T}}$

The line $\mathbf{r} = (1,4)^{\mathsf{T}} + t(3,-2)^{\mathsf{T}}$ in $\mathbf{R}^2$ has Cartesian equation $2x + 3y = k$ . Enter $k$ .

Describe $\operatorname{span}\{(1,-2,3)^{\mathsf{T}}, (-2,4,-6)^{\mathsf{T}}\}$ geometrically.

A plane through the origin, because two vectors are given.

A line through the origin.

All of $\mathbf{R}^3$ .

The single point $\mathbf{0}$ .

Vectors $\mathbf{u}, \mathbf{v} \in \mathbf{R}^n$ satisfy $\|\mathbf{u}\| = 5$ and $\|\mathbf{v}\| = 12$ . What is the largest value $\|\mathbf{u}+\mathbf{v}\|$ can take?

The points $A = (1,2,-1)$ , $B = (3,-1,2)$ and $C = (7,k,8)$ are collinear. Find $k$ .

For which value of $k$ does $(1,k,5)^{\mathsf{T}}$ lie in $\operatorname{span}\{(1,1,2)^{\mathsf{T}}, (0,1,1)^{\mathsf{T}}\}$ ?

Consider the claim: if $\mathbf{w} \in \operatorname{span}\{\mathbf{u},\mathbf{v}\}$ then $\|\mathbf{w}\| \leq \|\mathbf{u}\| + \|\mathbf{v}\|$ . Must this be true?

Yes — it is exactly the triangle inequality applied to $\mathbf{u}$ and $\mathbf{v}$ .

No — with $\mathbf{u} = (1,0)^{\mathsf{T}}$ , $\mathbf{v} = (0,1)^{\mathsf{T}}$ and $\mathbf{w} = (3,0)^{\mathsf{T}}$ , we have $\mathbf{w} \in \operatorname{span}\{\mathbf{u},\mathbf{v}\}$ but $\|\mathbf{w}\| = 3 > 2 = \|\mathbf{u}\|+\|\mathbf{v}\|$ .

No — with $\mathbf{u} = (1,0)^{\mathsf{T}}$ , $\mathbf{v} = (0,1)^{\mathsf{T}}$ and $\mathbf{w} = (1,1)^{\mathsf{T}}$ , we have $\|\mathbf{w}\| = \sqrt{2} > 2 = \|\mathbf{u}\|+\|\mathbf{v}\|$ .

Yes — every element of a span is of the form $\mathbf{u}+\mathbf{v}$ , so the bound always applies.

The line $L$ is given by $\mathbf{r} = (2,-1)^{\mathsf{T}} + t(4,6)^{\mathsf{T}}$ . Construct a second parametrisation of the same line, of the form $\mathbf{r} = (6,5)^{\mathsf{T}} + s(2,c)^{\mathsf{T}}$ . Enter the value of $c$ .

Every span contains $\mathbf{0}$ . Which of the following correctly states the reason and one consequence?

The reason is that $\mathbf{0}$ is perpendicular to every vector; the consequence is that no span can contain a non-zero vector.

The reason is that spans are closed under addition; the consequence is that only spans of two or more vectors contain $\mathbf{0}$ .

The reason is that taking every coefficient equal to $0$ gives the combination $0\mathbf{v}_1 + \cdots + 0\mathbf{v}_k = \mathbf{0}$ ; the consequence is that the line $\{(x,y)^{\mathsf{T}} : y = 2x+1\}$ is not the span of any set of vectors in $\mathbf{R}^2$ , since $(0,0)^{\mathsf{T}}$ does not satisfy $y = 2x+1$ .

The reason is that $\mathbf{v} - \mathbf{v} = \mathbf{0}$ for any $\mathbf{v}$ ; the consequence is that every line in $\mathbf{R}^2$ is a span, since $\mathbf{0}$ can always be adjoined to it.

The following argument is offered.

Claim. For $\mathbf{u} = (3,-4)^{\mathsf{T}}$ and $\mathbf{v} = (-6,8)^{\mathsf{T}}$ , $\|\mathbf{u}+\mathbf{v}\| = 15$ . Step 1. $\|\mathbf{u}\| = \sqrt{3^2 + (-4)^2} = \sqrt{25} = 5$ . Step 2. $\|\mathbf{v}\| = \sqrt{(-6)^2 + 8^2} = \sqrt{100} = 10$ . Step 3. By the triangle inequality, $\|\mathbf{u}+\mathbf{v}\| = \|\mathbf{u}\| + \|\mathbf{v}\| = 5 + 10 = 15$ . Step 4. Hence the unit vector in the direction of $\mathbf{u}+\mathbf{v}$ is $\tfrac{1}{15}(\mathbf{u}+\mathbf{v}) = \tfrac{1}{15}(-3,4)^{\mathsf{T}}$ .

At which step does the argument first fail, and why?

Step 1 — the norm of $\mathbf{u}$ is $|3| + |-4| = 7$ .

Step 2 — the components of $\mathbf{v}$ have been squared incorrectly; $\|\mathbf{v}\| = 14$ .

Step 3 — the triangle inequality states $\|\mathbf{u}+\mathbf{v}\| \leq \|\mathbf{u}\|+\|\mathbf{v}\|$ , and it is an inequality; here $\mathbf{v} = -2\mathbf{u}$ , so $\mathbf{u}+\mathbf{v} = (-3,4)^{\mathsf{T}}$ and $\|\mathbf{u}+\mathbf{v}\| = 5$ .

Step 4 — dividing a vector by its norm does not produce a unit vector.

The following argument is offered.

Claim. The points $A = (1,1,1)$ , $B = (3,4,5)$ and $C = (5,7,9)$ determine a plane. Step 1. $\mathbf{b}-\mathbf{a} = (2,3,4)^{\mathsf{T}}$ . Step 2. $\mathbf{c}-\mathbf{a} = (4,6,8)^{\mathsf{T}}$ . Step 3. A plane through three points is $\mathbf{r} = \mathbf{a} + s(\mathbf{b}-\mathbf{a}) + t(\mathbf{c}-\mathbf{a})$ , so the plane is $\mathbf{r} = (1,1,1)^{\mathsf{T}} + s(2,3,4)^{\mathsf{T}} + t(4,6,8)^{\mathsf{T}}$ . Step 4. Taking $s = t = 1$ gives $(7,10,13)$ , so this point lies on the plane.

At which step does the argument first fail, and why?

Step 1 — the displacement has been computed as $\mathbf{a}-\mathbf{b}$ rather than $\mathbf{b}-\mathbf{a}$ .

Step 2 — the second direction should be $\mathbf{c}-\mathbf{b}$ , not $\mathbf{c}-\mathbf{a}$ .

Step 3 — the construction requires the three points to be non-collinear, but $\mathbf{c}-\mathbf{a} = 2(\mathbf{b}-\mathbf{a})$ ; the set described is the line through $A$ with direction $(2,3,4)^{\mathsf{T}}$ , not a plane.

Step 4 — the arithmetic is wrong: $s = t = 1$ gives $(7,10,12)$ .

Let $\mathbf{u}, \mathbf{v}, \mathbf{w} \in \mathbf{R}^n$ with $\mathbf{w} \in \operatorname{span}\{\mathbf{u},\mathbf{v}\}$ . Which statement about $\operatorname{span}\{\mathbf{u},\mathbf{v},\mathbf{w}\}$ and $\operatorname{span}\{\mathbf{u},\mathbf{v}\}$ is correct?

They are equal. Writing $\mathbf{w} = p\mathbf{u}+q\mathbf{v}$ , any $a\mathbf{u}+b\mathbf{v}+c\mathbf{w}$ equals $(a+cp)\mathbf{u}+(b+cq)\mathbf{v}$ , so the larger span sits inside the smaller; the reverse containment follows by taking $c = 0$ .

The first is strictly larger whenever $\mathbf{w} \neq \mathbf{0}$ , because it is built from three vectors rather than two.

They are equal only when $\mathbf{u}$ and $\mathbf{v}$ are parallel.

They are different in general: with $\mathbf{u} = (1,0)^{\mathsf{T}}$ , $\mathbf{v} = (0,1)^{\mathsf{T}}$ and $\mathbf{w} = (1,1)^{\mathsf{T}}$ , the first span is $\mathbf{R}^3$ and the second is only $\mathbf{R}^2$ .

Exactly one of the following subsets of $\mathbf{R}^3$ is the span of some set of vectors. Which one?

$\{(1,2,3)^{\mathsf{T}} + t(2,4,6)^{\mathsf{T}} : t \in \mathbf{R}\}$

$\{(1,0,0)^{\mathsf{T}} + t(0,1,0)^{\mathsf{T}} : t \in \mathbf{R}\}$

$\{(x,y,z)^{\mathsf{T}} : x+y+z = 3\}$

$\{(x,y,z)^{\mathsf{T}} : x \geq 0\}$