Vectors: Introduction & Geometric Intuition

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# Vectors: Introduction & Geometric Intuition ## 🎯 Learning Objectives After this topic you will be able to: - Visualise a vector as a directed arrow in 2D/3D and interpret it as an $n$\-tuple of real numbers - Add vectors geometrically (tip-to-tail) and algebraically component-wise - Multiply a vector by a scalar...

Vectors: Introduction & Geometric Intuition

🎯 Learning Objectives

After this topic you will be able to:

Visualise a vector as a directed arrow in 2D/3D and interpret it as an $n$ -tuple of real numbers
Add vectors geometrically (tip-to-tail) and algebraically component-wise
Multiply a vector by a scalar and explain how it stretches or flips the arrow
Compute the dot product of two vectors in $\mathbb{R}^n$ and use it to find the angle between them
Compute the cross product of two vectors in $\mathbb{R}^3$ and interpret it geometrically
Decompose a vector into components parallel and perpendicular to another vector

📋 Prerequisites

Basic algebra: arithmetic with real numbers, solving simple equations
Coordinate geometry (from BSMA1001): plotting points $(x,y)$ in the plane
This topic builds your geometric intuition for everything that follows in the course

1. What is a Vector?

1.1 Intuition: The Displacement Arrow

Imagine you are standing at a street corner. You walk 3 steps east and then 4 steps north. Where do you end up? The "3 steps east, 4 steps north" is a vector — it carries magnitude (5 steps, as the crow flies) and direction (north-east). Every vector has two ingredients:

Magnitude (length) — how far
Direction — which way In mathematics we represent this idea in two equivalent ways:

Geometric arrow — a directed line segment from a starting point $A$ to an ending point $B$ , written $\overrightarrow{AB}$
Algebraic $n$ -tuple — an ordered list of numbers, e.g. $(3, 4)$ in the plane or $(3, 4, -2)$ in 3D space The same vector can appear anywhere in space — translating it does not change it. Only its length and orientation matter.

1.2 Formal Definition

Definition (Vector in $\mathbb{R}^n$ ). A vector is an ordered $n$ -tuple of real numbers

> \mathbf{v} = (v_1, v_2, \dots, v_n) \quad \text{or} \quad \mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix} >

where each $v_i \in \mathbb{R}$ . The set of all such $n$ -tuples is $\mathbb{R}^n$ . The first notation (row vector) and second (column vector) are both common. In this course we mostly use column vectors for matrix multiplication compatibility. Key symbols and terminology:

Symbol	Meaning	Example
$\mathbf{v}$	a vector (bold lowercase)	$\mathbf{v} = (1, -2, 5)$
$v_i$	the $i$ -th component (coordinate)	$v_2 = -2$
$\mathbb{R}^n$	$n$ -dimensional Euclidean space	$\mathbb{R}^2$ = the plane
$\mathbf{0}$	the zero vector $(0,0,\dots,0)$	$(0,0)$ in $\mathbb{R}^2$

1.3 Geometric Representation

(Diagram) A vector

\mathbf{v} = (v_1, v_2)

\mathbb{R}^2

is drawn as an arrow from the origin to the point

(v_1, v_2)

. Its length (magnitude) is:

\|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2}

For

\mathbf{v} = (3,4)

\|\mathbf{v}\| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5

\mathbb{R}^3

\mathbf{v} = (v_1, v_2, v_3)

has length

\sqrt{v_1^2 + v_2^2 + v_3^2}

2. Vector Operations

2.1 Scalar Multiplication

Multiplying a vector

\mathbf{v}

by a scalar (real number)

c

changes its length but not its direction (unless

c

is negative, which also flips the direction).

c\mathbf{v} = (c v_1, c v_2, \dots, c v_n)

Geometric effect:

$c > 1$ : stretches the arrow
$0 < c < 1$ : shrinks the arrow
$c = 0$ : collapses to the zero vector
$c < 0$ : flips the arrow to point opposite (and stretches/shrinks by $|c|$ )

Example 1: Scalar multiplication

Let $\mathbf{v} = (2, -1, 3)$ .

> 2\mathbf{v} = (4, -2, 6), \qquad \frac{1}{2}\mathbf{v} = \left(1, -\frac{1}{2}, \frac{3}{2}\right), \qquad -3\mathbf{v} = (-6, 3, -9) >

2.2 Vector Addition (and Subtraction)

Add two vectors component-wise:

\mathbf{v} + \mathbf{w} = (v_1 + w_1, v_2 + w_2, \dots, v_n + w_n)

Geometric intuition: place the tail of

\mathbf{w}

at the head of

\mathbf{v}

; the sum is the arrow from the tail of

\mathbf{v}

to the head of

\mathbf{w}

(the tip-to-tail rule). (Diagram)

Example 2: Vector addition

$\mathbf{v} = (1, 2, -1)$ , $\mathbf{w} = (3, -1, 4)$

> \mathbf{v} + \mathbf{w} = (1+3,\; 2+(-1),\; -1+4) = (4, 1, 3) >

Subtraction: $\mathbf{v} - \mathbf{w} = \mathbf{v} + (-\mathbf{w}) = (1-3,\; 2-(-1),\; -1-4) = (-2, 3, -5)$

2.3 Properties of Vector Operations

These properties hold for all vectors

\mathbf{u}, \mathbf{v}, \mathbf{w} \in \mathbb{R}^n

and scalars

a, b \in \mathbb{R}

Property	Addition	Scalar Multiplication
Commutativity	$\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}$	—
Associativity	$(\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w})$	$(ab)\mathbf{v} = a(b\mathbf{v})$
Identity	$\mathbf{v} + \mathbf{0} = \mathbf{v}$	$1\mathbf{v} = \mathbf{v}$
Inverse	$\mathbf{v} + (-\mathbf{v}) = \mathbf{0}$	—
Distributive	$(a+b)\mathbf{v} = a\mathbf{v} + b\mathbf{v}$	$a(\mathbf{u}+\mathbf{v}) = a\mathbf{u} + a\mathbf{v}$

These eight rules form the axioms of a vector space — we will meet them again in Week 3.

3. Dot Product (Inner Product)

3.1 Intuition: How Much Do Two Vectors Align?

The dot product measures alignment. Two vectors pointing in roughly the same direction have a positive dot product; opposite directions give a negative dot product; perpendicular vectors give zero.

3.2 Definition

Definition (Dot Product). For $\mathbf{u} = (u_1, \dots, u_n)$ and $\mathbf{v} = (v_1, \dots, v_n)$ in $\mathbb{R}^n$ ,

> \mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + \cdots + u_n v_n = \sum_{i=1}^n u_i v_i >

The result is a scalar (a real number), not a vector.

3.3 Relationship to Magnitude

The dot product of a vector with itself gives the squared length:

\mathbf{v} \cdot \mathbf{v} = \|\mathbf{v}\|^2

Example 3: Dot product

$\mathbf{u} = (2, -1, 3)$ , $\mathbf{v} = (1, 4, -2)$

> \begin{aligned} \mathbf{u} \cdot \mathbf{v} &= 2(1) + (-1)(4) + 3(-2) \\ &= 2 - 4 - 6 = -8 \end{aligned} >

$\|\mathbf{u}\| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{4+1+9} = \sqrt{14}$

Check: $\mathbf{u} \cdot \mathbf{u} = 4 + 1 + 9 = 14 = (\sqrt{14})^2$ ✓

3.4 Geometric Formula and the Angle Between Vectors

The dot product relates to the angle

\theta

between two vectors:

\mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\| \|\mathbf{v}\| \cos\theta

Therefore:

\cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|}, \quad 0 \leq \theta \leq \pi

Example 4: Angle between two vectors

Find the angle between $\mathbf{u} = (1, 0, 1)$ and $\mathbf{v} = (0, 1, 1)$ .

> \begin{aligned} \mathbf{u} \cdot \mathbf{v} &= 1(0) + 0(1) + 1(1) = 1 \\ \|\mathbf{u}\| &= \sqrt{1+0+1} = \sqrt{2} \\ \|\mathbf{v}\| &= \sqrt{0+1+1} = \sqrt{2} \\ \cos\theta &= \frac{1}{\sqrt{2}\sqrt{2}} = \frac{1}{2} \\ \theta &= \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} = 60^\circ \end{aligned} >

3.5 Orthogonal (Perpendicular) Vectors

Two vectors are orthogonal (perpendicular) when

\mathbf{u} \cdot \mathbf{v} = 0

. This is because

\cos(90^\circ) = 0

, so

\mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\| \|\mathbf{v}\| \cdot 0 = 0

Example 5: Orthogonal vectors

$\mathbf{u} = (1, 2, -3)$ , $\mathbf{v} = (4, 1, 2)$

$\mathbf{u} \cdot \mathbf{v} = 1(4) + 2(1) + (-3)(2) = 4 + 2 - 6 = 0$ ✓

Therefore $\mathbf{u} \perp \mathbf{v}$ .

3.6 Properties of the Dot Product

Property	Formula
Commutativity	$\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}$
Linearity	$\mathbf{u} \cdot (a\mathbf{v} + b\mathbf{w}) = a(\mathbf{u} \cdot \mathbf{v}) + b(\mathbf{u} \cdot \mathbf{w})$
Positive definiteness	$\mathbf{v} \cdot \mathbf{v} \geq 0$ , and $=0$ only when $\mathbf{v} = \mathbf{0}$
Cauchy-Schwarz	$

4. Cross Product (in ℝ³)

4.1 Intuition: The Perpendicular Area Vector

While the dot product tells us how much two vectors align, the cross product produces a third vector perpendicular to both. Its magnitude equals the area of the parallelogram spanned by the two input vectors.

4.2 Definition

Definition (Cross Product). For $\mathbf{u} = (u_1, u_2, u_3)$ and $\mathbf{v} = (v_1, v_2, v_3)$ in $\mathbb{R}^3$ ,

> \mathbf{u} \times \mathbf{v} = (u_2 v_3 - u_3 v_2,\; u_3 v_1 - u_1 v_3,\; u_1 v_2 - u_2 v_1) >

The result is a vector in $\mathbb{R}^3$ (unlike the dot product). A handy mnemonic using a determinant:

\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix}

where

\mathbf{i}, \mathbf{j}, \mathbf{k}

are the standard basis vectors

(1,0,0)

(0,1,0)

(0,0,1)

4.3 Geometric Interpretation

Direction: $\mathbf{u} \times \mathbf{v}$ is perpendicular to both $\mathbf{u}$ and $\mathbf{v}$ (right-hand rule)
Magnitude: $\|\mathbf{u} \times \mathbf{v}\| = \|\mathbf{u}\| \|\mathbf{v}\| \sin\theta$ = area of parallelogram (Diagram)

Example 6: Cross product

$\mathbf{u} = (1, 2, 3)$ , $\mathbf{v} = (4, 5, 6)$

> \begin{aligned} \mathbf{u} \times \mathbf{v} &= (2\cdot6 - 3\cdot5,\; 3\cdot4 - 1\cdot6,\; 1\cdot5 - 2\cdot4) \\ &= (12 - 15,\; 12 - 6,\; 5 - 8) \\ &= (-3, 6, -3) \end{aligned} >

Check orthogonality: $\mathbf{u} \cdot (\mathbf{u} \times \mathbf{v}) = 1(-3) + 2(6) + 3(-3) = -3 + 12 - 9 = 0$ ✓

$\mathbf{v} \cdot (\mathbf{u} \times \mathbf{v}) = 4(-3) + 5(6) + 6(-3) = -12 + 30 - 18 = 0$ ✓

4.4 Properties of Cross Product

Property	Formula
Anti-commutativity	$\mathbf{u} \times \mathbf{v} = -(\mathbf{v} \times \mathbf{u})$
Linearity	$\mathbf{u} \times (a\mathbf{v} + b\mathbf{w}) = a(\mathbf{u} \times \mathbf{v}) + b(\mathbf{u} \times \mathbf{w})$
Scalar triple product	$\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = \det[\mathbf{u}\; \mathbf{v}\; \mathbf{w}]$ (volume)
Self cross	$\mathbf{v} \times \mathbf{v} = \mathbf{0}$

5. Vector Projection

5.1 Intuition: How Much of v Points in the Direction of u?

Projection decomposes a vector

\mathbf{v}

into two parts: one parallel to

\mathbf{u}

and one perpendicular to

\mathbf{u}

5.2 Formula

The scalar projection (component) of

\mathbf{v}

onto

\mathbf{u}

is:

\text{comp}_{\mathbf{u}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{u}}{\|\mathbf{u}\|}

The vector projection of

\mathbf{v}

onto

\mathbf{u}

is:

\text{proj}_{\mathbf{u}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{u}}{\|\mathbf{u}\|^2} \mathbf{u}

The perpendicular component is:

\mathbf{v}_{\perp} = \mathbf{v} - \text{proj}_{\mathbf{u}} \mathbf{v}

Example 7: Projection

Find the projection of $\mathbf{v} = (3, 1, -2)$ onto $\mathbf{u} = (1, 2, 2)$ .

> \begin{aligned} \mathbf{v} \cdot \mathbf{u} &= 3(1) + 1(2) + (-2)(2) = 3 + 2 - 4 = 1 \\ \|\mathbf{u}\|^2 &= 1^2 + 2^2 + 2^2 = 1 + 4 + 4 = 9 \\ \text{proj}_{\mathbf{u}} \mathbf{v} &= \frac{1}{9}(1, 2, 2) = \left(\frac{1}{9}, \frac{2}{9}, \frac{2}{9}\right) \\ \mathbf{v}_{\perp} &= (3, 1, -2) - \left(\frac{1}{9}, \frac{2}{9}, \frac{2}{9}\right) = \left(\frac{26}{9}, \frac{7}{9}, -\frac{20}{9}\right) \end{aligned} >

6. Edge Cases & Gotchas

Situation	What Happens	Why It Matters
Zero vector $\mathbf{0}$	Has no direction, length 0	Dot product with anything is 0; cross product with anything is $\mathbf{0}$
Division by zero in angle formula	If $\	\mathbf{u}\
Cross product only in $\mathbb{R}^3$	The 3D cross product does not exist in $\mathbb{R}^2$ or $\mathbb{R}^n$ for $n \neq 3$	In $\mathbb{R}^2$ , use the determinant or the "perp dot product" $u_1 v_2 - u_2 v_1$
Vectors vs. points	A vector is a displacement, not a location	The same vector can be drawn anchored anywhere

7. Common Pitfalls

❌ Pitfall 1: Adding vectors of different dimensions

You cannot add

(1,2,3) + (4,5)

— dimensions must match.

❌ Pitfall 2: Dot product produces a scalar

\mathbf{u} \cdot \mathbf{v}

is a number, not a vector. A common mistake is writing

\mathbf{u} \cdot \mathbf{v} = (u_1 v_1, u_2 v_2, \dots)

— that is the Hadamard (element-wise) product, not the dot product.

❌ Pitfall 3: Confusing projection formulas

Scalar projection:

\frac{\mathbf{v} \cdot \mathbf{u}}{\|\mathbf{u}\|}

(a scalar). Vector projection:

\frac{\mathbf{v} \cdot \mathbf{u}}{\|\mathbf{u}\|^2} \mathbf{u}

(a vector). The difference is dividing by

\|\mathbf{u}\|

vs.

\|\mathbf{u}\|^2

8. Formula Summary Table

Concept	Formula	Notes
Magnitude	$\	\mathbf{v}\
Dot product	$\mathbf{u} \cdot \mathbf{v} = \sum u_i v_i$	Result is a scalar
Angle	$\cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\	\mathbf{u}\
Orthogonality	$\mathbf{u} \cdot \mathbf{v} = 0$	Perpendicular vectors
Cross product (ℝ³)	$\mathbf{u} \times \mathbf{v} = (u_2 v_3 - u_3 v_2, \dots)$	Result is a vector
Projection	$\text{proj}_{\mathbf{u}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{u}}{\	\mathbf{u}\
Cauchy-Schwarz	$	\mathbf{u} \cdot \mathbf{v}

9. 📝 Practice Questions

Q1: Basic operations

Let $\mathbf{u} = (2, -1, 3)$ and $\mathbf{v} = (0, 4, -2)$ . Compute: (a) $2\mathbf{u} - 3\mathbf{v}$ (b) $\mathbf{u} \cdot \mathbf{v}$ (c) $\mathbf{u} \times \mathbf{v}$

Strategy: Work component-wise for (a), use sum of products for (b), use determinant formula for (c).

Solution (a):

> 2\mathbf{u} = (4, -2, 6), \quad 3\mathbf{v} = (0, 12, -6) >

> 2\mathbf{u} - 3\mathbf{v} = (4 - 0, -2 - 12, 6 - (-6)) = (4, -14, 12) >

Solution (b):

> \mathbf{u} \cdot \mathbf{v} = 2(0) + (-1)(4) + 3(-2) = 0 - 4 - 6 = -10 >

Solution (c):

> \begin{aligned} \mathbf{u} \times \mathbf{v} &= ((-1)(-2) - 3(4),\; 3(0) - 2(-2),\; 2(4) - (-1)(0)) \\ &= (2 - 12,\; 0 + 4,\; 8 - 0) \\ &= (-10, 4, 8) \end{aligned} >

Q2: Angle between vectors

Find the angle between $\mathbf{u} = (1, 1, 1)$ and $\mathbf{v} = (1, 0, 2)$ .

Strategy: Use $\cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|}$ .

Solution:

> \mathbf{u} \cdot \mathbf{v} = 1(1) + 1(0) + 1(2) = 3 >

> \|\mathbf{u}\| = \sqrt{1+1+1} = \sqrt{3}, \quad \|\mathbf{v}\| = \sqrt{1+0+4} = \sqrt{5} >

> \cos\theta = \frac{3}{\sqrt{3}\sqrt{5}} = \frac{3}{\sqrt{15}} \approx 0.7746 >

> \theta = \cos^{-1}\left(\frac{3}{\sqrt{15}}\right) \approx 39.2^\circ >

Q3: Orthogonality test

Are $\mathbf{u} = (2, -1, 4)$ and $\mathbf{v} = (3, 2, -1)$ orthogonal?

Strategy: Compute dot product. If zero, they are orthogonal.

Solution:

> \mathbf{u} \cdot \mathbf{v} = 2(3) + (-1)(2) + 4(-1) = 6 - 2 - 4 = 0 >

Yes, they are orthogonal. Q4: Vector projection

Find the vector projection of $\mathbf{v} = (4, 1, -2)$ onto $\mathbf{u} = (2, -1, 3)$ .

Strategy: Use $\text{proj}_{\mathbf{u}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{u}}{\|\mathbf{u}\|^2} \mathbf{u}$ .

Solution:

> \mathbf{v} \cdot \mathbf{u} = 4(2) + 1(-1) + (-2)(3) = 8 - 1 - 6 = 1 >

> \|\mathbf{u}\|^2 = 4 + 1 + 9 = 14 >

> \text{proj}_{\mathbf{u}} \mathbf{v} = \frac{1}{14}(2, -1, 3) = \left(\frac{1}{7}, -\frac{1}{14}, \frac{3}{14}\right) >

Q5: Cross product area

Find the area of the parallelogram spanned by $\mathbf{u} = (1, 0, 2)$ and $\mathbf{v} = (0, 3, 1)$ .

Strategy: Area = $\|\mathbf{u} \times \mathbf{v}\|$ .

Solution:

> \mathbf{u} \times \mathbf{v} = (0(1) - 2(3),\; 2(0) - 1(1),\; 1(3) - 0(0)) = (-6, -1, 3) >

> \text{Area} = \sqrt{(-6)^2 + (-1)^2 + 3^2} = \sqrt{36 + 1 + 9} = \sqrt{46} >

Q6: Unit vector

Find a unit vector in the direction of $\mathbf{v} = (2, -2, 1)$ .

Strategy: Divide the vector by its magnitude.

Solution:

> \|\mathbf{v}\| = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 >

> \hat{\mathbf{v}} = \frac{\mathbf{v}}{\|\mathbf{v}\|} = \left(\frac{2}{3}, -\frac{2}{3}, \frac{1}{3}\right) >

Q7: Dot product properties

If $\mathbf{u}$ and $\mathbf{v}$ are unit vectors and $\mathbf{u} \cdot \mathbf{v} = \frac{1}{2}$ , find $\|\mathbf{u} - \mathbf{v}\|$ .

Strategy: Use $\|\mathbf{u} - \mathbf{v}\|^2 = (\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} - \mathbf{v}) = \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2 - 2\mathbf{u} \cdot \mathbf{v}$ .

Solution:

> \|\mathbf{u} - \mathbf{v}\|^2 = 1 + 1 - 2\left(\frac{1}{2}\right) = 2 - 1 = 1 >

> \|\mathbf{u} - \mathbf{v}\| = 1 >

Q8: Finding a perpendicular vector

Find a vector $\mathbf{w}$ in $\mathbb{R}^3$ that is perpendicular to both $\mathbf{u} = (1, 2, -1)$ and $\mathbf{v} = (3, 0, 4)$ .

Strategy: Use the cross product $\mathbf{w} = \mathbf{u} \times \mathbf{v}$ .

Solution:

> \begin{aligned} \mathbf{w} &= (2(4) - (-1)(0),\; (-1)(3) - 1(4),\; 1(0) - 2(3)) \\ &= (8 - 0,\; -3 - 4,\; 0 - 6) \\ &= (8, -7, -6) \end{aligned} >

Check: $\mathbf{u} \cdot \mathbf{w} = 1(8) + 2(-7) + (-1)(-6) = 8 - 14 + 6 = 0$ ✓ Q9: Scalar triple product

Find the volume of the parallelepiped spanned by $\mathbf{a} = (1, 0, 0)$ , $\mathbf{b} = (0, 2, 0)$ , $\mathbf{c} = (1, 1, 3)$ .

Strategy: Volume = $|\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|$ .

Solution:

> \mathbf{b} \times \mathbf{c} = (2(3) - 0(1),\; 0(1) - 0(3),\; 0(1) - 2(1)) = (6, 0, -2) >

> \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 1(6) + 0(0) + 0(-2) = 6 >

> \text{Volume} = |6| = 6 >

Q10: Decomposition

Decompose $\mathbf{v} = (5, 2, 1)$ into components parallel and perpendicular to $\mathbf{u} = (1, 1, 1)$ .

Strategy: Compute $\mathbf{v}_\parallel = \text{proj}_{\mathbf{u}} \mathbf{v}$ , then $\mathbf{v}_\perp = \mathbf{v} - \mathbf{v}_\parallel$ .

Solution:

> \mathbf{v} \cdot \mathbf{u} = 5 + 2 + 1 = 8, \quad \|\mathbf{u}\|^2 = 3 >

> \mathbf{v}_\parallel = \frac{8}{3}(1, 1, 1) = \left(\frac{8}{3}, \frac{8}{3}, \frac{8}{3}\right) >

> \mathbf{v}_\perp = \left(5 - \frac{8}{3},\; 2 - \frac{8}{3},\; 1 - \frac{8}{3}\right) = \left(\frac{7}{3}, -\frac{2}{3}, -\frac{5}{3}\right) >

Check: $\mathbf{v}_\parallel \cdot \mathbf{v}_\perp = \frac{8}{3}\cdot\frac{7}{3} + \frac{8}{3}\cdot\left(-\frac{2}{3}\right) + \frac{8}{3}\cdot\left(-\frac{5}{3}\right) = \frac{56-16-40}{9} = 0$ ✓

🔗 Cross-References

Next topic: Matrices Introduction — matrices as arrays of vectors
BSMA1001 (Maths 1): Review coordinate geometry and basic algebra
Week 7 (Inner Products): Generalises the dot product to abstract vector spaces
Week 3 (Vector Spaces): The eight axioms formalise what we've seen here
BSCS2004 (ML Foundations): Vectors are the data representation in ML Join Discord Next1.2 Matrices Introduction