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Overview

This lesson covers the second part of lesson 3-3, "Vector multiplication". After it you should be able to:
define and calculate the cross product and apply the right-hand rule;
use it for parallelism, perpendicular vectors, and areas of parallelograms and triangles;
calculate the scalar triple product and use it for volumes and coplanarity.

🌳 Concept tree

The Vector (Cross) Product and the Scalar Triple Product

├── 3-3 (g) Definition of the vector product

│ ├── Definition

│ ├── Direction→the right-hand rule

│ ├── The unit vectors

│ └── Norm of the product

├── 3-3 (h) The cross product in components

│ └── Determinant form

├── 3-3 (i) Properties of the vector product

│ └── List

├── 3-3 (j) Parallel vectors

│ └── Condition

├── 3-3 (k) Geometric meaning→areas

│ ├── Parallelogram and triangle

│ └── Link with mechanics

├── 3-3 (l) The scalar triple product

│ ├── Definition

│ ├── Determinant form

│ └── Properties

└── 3-3 (m) Geometric meaning→volume

├── Parallelepiped and tetrahedron

└── Summary

The Vector (Cross) Product and the Scalar Triple Product · 1 / 14

3-3 (g) Definition of the vector product

Definition

Let $\vec{A}$ and $\vec{B}$ be two vectors in space with angle $θ$ between them ( $0^{\circ} \leq θ \leq 18 0^{\circ}$ ).
Their vector product (cross product) is
$\vec{A} \times \vec{B} = (∥ \vec{A} ∥ ∥ \vec{B} ∥ \sin θ) \hat{n}$
$\hat{n}$ is the unit vector perpendicular to the plane containing $\vec{A}$ and $\vec{B}$ .
The result is a vector, in contrast with the scalar product which is a number.

The Vector (Cross) Product: a box with three perpendicular edges shows the vector product and the scalar triple product: a x b = (0, 0, 6) and volume 24.

Direction: the right-hand rule

Curl the fingers of the right hand from the first vector $\vec{A}$ towards the second vector $\vec{B}$ through the smaller angle.
The thumb points along $\hat{n}$ , the direction of $\vec{A} \times \vec{B}$ .
Swapping the order reverses the direction B⃗×A⃗=−(A⃗×B⃗).

The unit vectors

Comparison aspect	Main idea	Keywords
For a right-handed system	i^×j^=k^ j^×k^=i^ k^×i^=j^
In the reverse order the sign changes	j^×i^=−k^ k^×j^=−i^ i^×k^=−j^
•	$\hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = \vec{0}$ .

Norm of the product

$∥ \vec{A} \times \vec{B} ∥ = ∥ \vec{A} ∥ ∥ \vec{B} ∥ \sin θ$ .
It is zero when the vectors are parallel ( $θ = 0^{\circ}$ or $18 0^{\circ}$ ) and largest when they are perpendicular.

✍️ Worked examples

Example 1 Two vectors in a plane have norms $15$ and $17.5$ and the angle between them is $7 0^{\circ}$ . Find $∥ \vec{A} \times \vec{B} ∥$ .
Solution:
$∥ \vec{A} \times \vec{B} ∥ = 15 \times 17.5 \times \sin 7 0^{\circ} = 262.5 \times 0.9397 \approx 246.7$ .
Example 2 If $\vec{A} \times \vec{B} = - 65 \hat{n}$ , $∥ \vec{A} ∥ = 5$ and $∥ \vec{B} ∥ = 26$ , find the angle between them.
Solution:
The norm of the product is $65$ , so $5 \times 26 \times \sin θ = 65$ .
$\sin θ = \frac{65}{130} = \frac{1}{2}$ , so $θ = 3 0^{\circ}$ or $15 0^{\circ}$ .
The minus sign only says that $\hat{n}$ points the other way; it does not affect the angle.

The Vector (Cross) Product and the Scalar Triple Product · 2 / 14

3-3 (h) The cross product in components

Determinant form

Comparison aspect	Main idea	Keywords
For A⃗=(Ax,Ay,Az) and B⃗=(Bx,By,Bz)	A⃗×B⃗=∣i^j^k^AxAyAzBxByBz∣ =(AyBz−AzBy) i^−(AxBz−AzBx) j^+(AxBy−AyBx) k^	AxAyAzBxByBzAyBzAzByAxBz
•	The middle term has a minus sign from expanding along the first row	middletermminussign
In the xy-plane (Az=Bz=0) only the last term survives	A⃗×B⃗=(AxBy−AyBx) k^	AxByAyBx

✍️ Worked examples

Example 3 Find $\vec{A} \times \vec{B}$ for $\vec{A} = (- 2, 3, 1)$ and $\vec{B} = (1, 2, 4)$ , and a unit vector perpendicular to both.
Solution:
i^ coefficient 3×4−1×2=10.
j^ coefficient −[(−2)(4)−(1)(1)]=−(−9)=9.
k^ coefficient (−2)(2)−(3)(1)=−7.
$\vec{A} \times \vec{B} = (10, 9, - 7)$ , with norm $\sqrt{100 + 81 + 49} = \sqrt{230}$ .
Unit vector perpendicular to the plane 1230(10,9,−7) (the opposite vector also works).
Example 4 Find a unit vector perpendicular to the plane of $\vec{A} = (2, - 3, 1)$ and $\vec{B} = (- 1, 2, - 3)$ .
Solution:
i^ (−3)(−3)−(1)(2)=7.
j^ −[(2)(−3)−(1)(−1)]=−(−5)=5.
k^ (2)(2)−(−3)(−1)=1.
$\vec{A} \times \vec{B} = (7, 5, 1)$ and its norm is $\sqrt{75} = 5 \sqrt{3}$ .
Unit vector 153(7,5,1).
Check (7,5,1)⋅A⃗=14−15+1=0 and (7,5,1)⋅B⃗=−7+10−3=0.
Example 5 For $\vec{A} = (3, 2)$ and $\vec{B} = (1, 4)$ in the plane, find $\vec{A} \times \vec{B}$ .
Solution:
(3)(4)−(2)(1)=10, so A⃗×B⃗=10 k^ the product points along the positive z-axis, since B⃗ is anticlockwise from A⃗.
Example 6:
$A B C D$ is a square of side 12 cm, labelled anticlockwise, and $\hat{e}$ is the unit vector perpendicular to its plane (out of the page).
Find $\vec{A B} \times \vec{A C}$ and $\vec{A B} \times \vec{B C}$ .
Solution:
The angle between $\vec{A B}$ and $\vec{A C}$ is $4 5^{\circ}$ and $A C = 12 \sqrt{2}$ .
$\vec{A B} \times \vec{A C} = 12 \times 12 \sqrt{2} \times \sin 4 5^{\circ} \hat{e} = 144 \hat{e}$ .
AB→⊥BC→ AB→×BC→=12×12×1 e^=144 e^.

The Vector (Cross) Product and the Scalar Triple Product · 3 / 14

3-3 (i) Properties of the vector product

List

Comparison aspect	Main idea	Keywords
Anti-commutative	$\vec{A} \times \vec{B} = - \vec{B} \times \vec{A}$ (so it is not commutative).
Distributive	$\vec{A} \times (\vec{B} + \vec{C}) = \vec{A} \times \vec{B} + \vec{A} \times \vec{C}$ .
Scalars	$(k \vec{A}) \times \vec{B} = \vec{A} \times (k \vec{B}) = k (\vec{A} \times \vec{B})$ .
Self product	$\vec{A} \times \vec{A} = \vec{0}$ .
Perpendicular to both	$\vec{A} \times \vec{B}$ is perpendicular to $\vec{A}$ and to $\vec{B}$ , so $(\vec{A} \times \vec{B}) \cdot \vec{A} = 0$ .
Zero product	$\vec{A} \times \vec{B} = \vec{0}$ means $\vec{A} ∥ \vec{B}$ , or one of them is the zero vector.	meansthemvector
The cross product is not associative in general	(A⃗×B⃗)×C⃗≠A⃗×(B⃗×C⃗)

The Vector (Cross) Product and the Scalar Triple Product · 4 / 14

3-3 (j) Parallel vectors

Condition

Two non-zero vectors are parallel if and only if $\vec{A} \times \vec{B} = \vec{0}$ , that is
$\frac{A_{x}}{B_{x}} = \frac{A_{y}}{B_{y}} = \frac{A_{z}}{B_{z}} = k$
Equivalent form A⃗=kB⃗. If k>0 they have the same direction; if k<0 they are opposite.
If a component of $\vec{B}$ is zero, the matching component of $\vec{A}$ must also be zero.

✍️ Worked examples

Example 7 Find $m$ and $k$ if $\vec{A} = (2, - 3, m)$ is parallel to $\vec{B} = (1, k, 8)$ .
Solution:
$\frac{2}{1} = \frac{- 3}{k} = \frac{m}{8}$ .
$\frac{- 3}{k} = 2$ gives $k = - \frac{3}{2}$ .
$\frac{m}{8} = 2$ gives $m = 16$ .
The vectors are in the same direction since the ratio $2$ is positive.
Example 8 A vector $\vec{B}$ is parallel to $\vec{A} = (2, - 3)$ and $∥ \vec{B} ∥ = 3 \sqrt{13}$ . Find $\vec{B}$ .
Solution:
$\vec{B} = k (2, - 3)$ and $∥ \vec{B} ∥ = ∣ k ∣ \sqrt{13} = 3 \sqrt{13}$ , so $k = \pm 3$ .
$\vec{B} = (6, - 9)$ or $(- 6, 9)$ .
Example 9 Decide whether $\vec{A} = (- 2, 1, 2)$ and $\vec{B} = (- 8, - 4, 8)$ are parallel, perpendicular, or neither.
Solution:
Ratios −2−8=14, 1−4=−14: not equal, so not parallel.
Dot product 16−4+16=28≠0, so not perpendicular.
They are neither.

The Vector (Cross) Product and the Scalar Triple Product · 5 / 14

3-3 (k) Geometric meaning: areas

Parallelogram and triangle

Let $\vec{A}$ and $\vec{B}$ be adjacent sides of a parallelogram. Its height is $∥ \vec{B} ∥ \sin θ$ above the base $∥ \vec{A} ∥$ .
Hence
$Area of parallelogram = ∥ \vec{A} \times \vec{B} ∥$
$Area of triangle with sides \vec{A}, \vec{B} = \frac{1}{2} ∥ \vec{A} \times \vec{B} ∥$
For a triangle ABC area =12∥AB→×AC→∥.

✍️ Worked examples

Comparison aspect	Main idea	Keywords
Example 10	Find the area of the parallelogram with adjacent sides $\vec{A} = (- 3, 1, 2)$ and $\vec{B} = (3, 4, - 1)$ .	Findareaparallelogramadjacent
•	Solution	Solution
i^	(1)(−1)−(2)(4)=−9
j^	−[(−3)(−1)−(2)(3)]=−(3−6)=3
k^	(−3)(4)−(1)(3)=−15
•	$\vec{A} \times \vec{B} = (- 9, 3, - 15)$ and the norm is $\sqrt{81 + 9 + 225} = \sqrt{315} = 3 \sqrt{35}$ .	norm
•	Area $= 3 \sqrt{35} \approx 17.75$ square units.	Areasquareunits
Example 11	Find the area of the triangle whose two sides are $\vec{A} = (1, 2, - 4)$ and $\vec{B} = (0, 5, - 1)$ .	Findareatrianglewhose
•	Solution	Solution
i^	(2)(−1)−(−4)(5)=18
j^	−[(1)(−1)−(−4)(0)]=1
k^	(1)(5)−(2)(0)=5
Norm	324+1+25=350=514
•	Area $= \frac{5 \sqrt{14}}{2} \approx 9.35$ square units.	Area5142squareunits
Example 12	Find the area of triangle $A B C$ with $A (1, 1, 0)$ , $B (3, 1, 2)$ , $C (1, 4, 1)$ .	Findareatriangle
•	Solution	Solution
•	$\vec{A B} = (2, 0, 2)$ and $\vec{A C} = (0, 3, 1)$ .
AB→×AC→	i^: 0×1−2×3=−6 j^: −(2×1−2×0)=−2 k^: 2×3−0=6
Norm	36+4+36=76=219
•	Area $= \sqrt{19} \approx 4.36$ square units.	Areasquareunits

Link with mechanics

The moment of a force $\vec{F}$ about a point is $\vec{M} = \vec{r} \times \vec{F}$ , where $\vec{r}$ points from the point to the line of action.
Its magnitude $∥ \vec{r} ∥ ∥ \vec{F} ∥ \sin θ$ is the force times the perpendicular distance.

The Vector (Cross) Product and the Scalar Triple Product · 6 / 14

3-3 (l) The scalar triple product

Definition

For three vectors $\vec{A}$ , $\vec{B}$ , $\vec{C}$ the scalar triple product is
$\vec{A} \cdot (\vec{B} \times \vec{C})$
The brackets are essential in the sense that $\vec{B} \times \vec{C}$ is calculated first (the other order, $(\vec{A} \cdot \vec{B}) \times \vec{C}$ , is meaningless since $\vec{A} \cdot \vec{B}$ is a number).
The result is a number.

If $\vec{A} = (A_{x}, A_{y}, A_{z})$ , $\vec{B} = (B_{x}, B_{y}, B_{z})$ , $\vec{C} = (C_{x}, C_{y}, C_{z})$ , then
$\vec{A} \cdot (\vec{B} \times \vec{C}) = ∣ \begin{array}{ccc} A_{x} & A_{y} & A_{z} \\ B_{x} & B_{y} & B_{z} \\ C_{x} & C_{y} & C_{z} \end{array} ∣$
$= A_{x} (B_{y} C_{z} - B_{z} C_{y}) - A_{y} (B_{x} C_{z} - B_{z} C_{x}) + A_{z} (B_{x} C_{y} - B_{y} C_{x})$

Cyclic permutation keeps the value A⃗⋅(B⃗×C⃗)=B⃗⋅(C⃗×A⃗)=C⃗⋅(A⃗×B⃗).
Interchanging two vectors changes the sign (two rows of the determinant swap).
$\vec{A} \cdot (\vec{B} \times \vec{C}) = (\vec{A} \times \vec{B}) \cdot \vec{C}$ , so the dot and the cross may be exchanged.
If two of the vectors are equal or parallel, the product is zero.
If the three vectors lie in one plane (coplanar), the product is zero, and conversely.

✍️ Worked examples

Example 13 Calculate $\vec{A} \cdot (\vec{B} \times \vec{C})$ for $\vec{A} = (2, 1, 3)$ , $\vec{B} = (- 1, 3, 2)$ , $\vec{C} = (1, 1, - 2)$ .
Solution:
Determinant 2[3(−2)−2(1)]−1[(−1)(−2)−2(1)]+3[(−1)(1)−3(1)].
$= 2 (- 8) - 1 (0) + 3 (- 4) = - 16 - 0 - 12 = - 28$ .
Example 14 Show that $\vec{A} = (1, 2, 3)$ , $\vec{B} = (0, 1, 4)$ , $\vec{C} = (1, 3, 7)$ are coplanar.
Solution:
Note that $\vec{C} = \vec{A} + \vec{B}$ .
Determinant 1[1×7−4×3]−2[0×7−4×1]+3[0×3−1×1]=−5+8−3=0.
The triple product is zero, so the vectors are coplanar.

The Vector (Cross) Product and the Scalar Triple Product · 7 / 14

Parallelepiped and tetrahedron

If $\vec{A}$ , $\vec{B}$ , $\vec{C}$ are three non-coplanar edges meeting at a vertex of a parallelepiped, then
$V = ∣ \vec{A} \cdot (\vec{B} \times \vec{C}) ∣$
Reason ∥B⃗×C⃗∥ is the area of the base and the component of A⃗ perpendicular to the base is the height.
A tetrahedron on the same three edges has volume $\frac{1}{6} ∣ \vec{A} \cdot (\vec{B} \times \vec{C}) ∣$ .
The absolute value matters the sign only shows whether the three vectors form a right-handed or left-handed set.

✍️ Worked examples

Example 15 Find the volume of the parallelepiped with edges $\vec{A} = (2, 1, 3)$ , $\vec{B} = (- 1, 3, 2)$ , $\vec{C} = (1, 1, - 2)$ .
Solution:
The triple product is $- 28$ (Example 13).
$V = ∣ - 28 ∣ = 28$ cubic units.
Example 16 Find the volume of the parallelepiped with edges $(3, 4, 1)$ , $(0, 2, 3)$ , $(3, 2, 2)$ .
Solution:
Determinant 3[2×2−3×2]−4[0×2−3×3]+1[0×2−2×3].
$= 3 (- 2) - 4 (- 9) + 1 (- 6) = - 6 + 36 - 6 = 24$ .
Volume $= 24$ cubic units.
Example 17 Find the volume of the tetrahedron with vertices $O (0, 0, 0)$ , $P (2, 0, 0)$ , $Q (0, 3, 0)$ , $R (0, 0, 4)$ .
Solution:
Edges from O (2,0,0), (0,3,0), (0,0,4); the determinant is 2×3×4=24.
Volume $= \frac{24}{6} = 4$ cubic units.
Example 18 Find $k$ such that the points $A (1, 0, 0)$ , $B (0, 1, 0)$ , $C (0, 0, 1)$ , $D (k, k, k)$ are in one plane.
Solution:
Vectors from A AB→=(−1,1,0), AC→=(−1,0,1), AD→=(k−1,k,k).
Triple product −1[0⋅k−1⋅k]−1[(−1)k−1(k−1)]+0=k−(−2k+1)⋅1=3k−1. Setting it to zero gives k=13.
The four points are coplanar for $k = \frac{1}{3}$ (the point $D$ then lies on the plane $x + y + z = 1$ ).

The Vector (Cross) Product and the Scalar Triple Product · 8 / 14

Summary

Point of comparison	Scalar product	Vector product	Scalar triple product
Result	number	vector	number
Formula	$∥ \vec{A} ∥ ∥ \vec{B} ∥ \cos θ$	$∥ \vec{A} ∥ ∥ \vec{B} ∥ \sin θ \hat{n}$	$\vec{A} \cdot (\vec{B} \times \vec{C})$
Order	commutative	anti-commutative	sign changes if two swap
Zero when	vectors perpendicular	vectors parallel	vectors coplanar
Gives	angle, component, work	area, perpendicular vector, moment	volume of a parallelepiped

The Vector (Cross) Product and the Scalar Triple Product · 9 / 14

📌 Comprehensive Scientific Terminology & Definitions:

Comparison aspect	Main idea	Keywords
Vector / Cross Product (الضرب الاتجاهي)	An operation taking two vectors A and B in 3D space and returning a vector C perpendicular to both: A × B = \|A\| \|B\| sin θ n where n is a unit vector perpendicular to the plane of A and B given by the Right-Hand Rule	operationtakingvectorsspace
Determinant Expansion Form	A × B = \| i j k; A_x A_y A_z ; B_x B_y B_z \| = (A_y B_z – A_z B_y) i – (A_x B_z – A_z B_x) j + (A_x B_y – A_y B_x) k.
Condition for Parallelism (التوازي)	Two non-zero vectors A and B are parallel (A ∥ B) if and only if A × B = 0 (since sin 0° = sin 180° = 0), or equivalently their components are proportional: A_x/B_x = A_y/B_y = A_z/B_z = k.	zerovectorssinceequivalently
Geometric Area Interpretations	Magnitude \|A × B\| equals the area of the parallelogram formed by adjacent sides A and B Area of triangle = (1/2) \|A × B\|	Magnitudeequalsparallelogramformed

Scalar Triple Product (الضرب الثلاثي القياسي)→A · (B × C) represented by 3×3 determinant: | A_x A_y A_z ; B_x B_y B_z ; C_x C_y C_z |, whose absolute value equals the volume of the parallelepiped formed by coterminous edges A, B, and C.

The Vector (Cross) Product and the Scalar Triple Product · 10 / 14

⚖️ Governing Scientific Laws, Principles & Mathematical Formulas:

Comparison aspect	Main idea	Keywords
(‘Anti-Commutative Property	A × B = – (B × A).’
(‘Parallelogram Area	Area = \|A × B\| Triangle Area = (1/2) \|A × B\|.’	Triangle
(‘Parallelepiped Volume	Volume = \|A · (B × C)\| Tetrahedron Volume = (1/6) \|A · (B × C)\|.’	Tetrahedron
(‘Coplanar Vectors Condition	and C lie in the same geometric plane if and only if A · (B × C) = 0.’	samegeometricplane
(“Lagrange’s Identity	\|A × B\|² + (A · B)² = \|A\|² \|B\|².”

The Vector (Cross) Product and the Scalar Triple Product · 11 / 14

🔍 In-Depth Explanations, Deductions & Core Concepts:

Comparison aspect	Main idea	Keywords
Why the cross product is anti-commutative	Swapping two vectors reverses the direction of the Right-Hand Rule thumb rotating the normal unit vector by 180° producing -n	Swappingvectorsreversesdirection
Right-hand cyclical rules for standard unit vectors	i × j = k j × k = i k × i = j reversing cycle yields negative signs: j × i = -k k × j = -i i × k = -j and self-cross products are zero: i × i = j × j = k × k = 0	reversingcycleyieldsnegative
Physical application to Moment of a Force (عزم القوة)	Moment vector MO about point O = r × F where r is position vector from O to any point on the line of action of force F	vectorpointpositionline

The Vector (Cross) Product and the Scalar Triple Product · 12 / 14

📊 Comparison Between Dot Product and Cross Product in 3D Space:

Property / Metric	Scalar (Dot) Product: A · B	Vector (Cross) Product: A × B
Result Type	Scalar (real number)	Vector (perpendicular to A and B)
Trigonometric Kernel	\|A\| \|B\| cos θ	\|A\| \|B\| sin θ n (Right-Hand Rule)
Commutativity	Commutative: A · B = B · A	Anti-commutative: A × B = -(B × A)
Zero Test Condition	A ⊥ B ⟺ A · B = 0	A ∥ B ⟺ A × B = 0
Physical Application	Mechanical Work: W = F · s	Moment / Torque: M = r × F

The Vector (Cross) Product and the Scalar Triple Product · 13 / 14

💡 High-Yield Exam Traps & Key Takeaways:

Comparison aspect	Main idea	Keywords
Lagrange Identity Trick	If given \|A\| = 5 \|B\| = 6 and A · B = 24 find \|A × B\| immediately using: \|A × B\|² = \|A\|²\|B\|² – (A · B)² = (25)(36) – 576 = 900 – 576 = 324 → \|A × B\| = 18	givenfindimmediatelyusing
Coplanar vectors trap	If four points A D are given test if vectors AB and AD have a scalar triple product equal to zero	fourpointsgiventest

The Vector (Cross) Product and the Scalar Triple Product · 14 / 14

Curriculum

Mathematics

Unit 1

Unit 2

Unit 3

Unit 4

Unit 5

Unit 6

Unit 7

The Vector (Cross) Product and the Scalar Triple Product

🌳 Concept tree

├── 3-3 (g) Definition of the vector product

│ ├── Definition

│ ├── The unit vectors

│ └── Norm of the product

├── 3-3 (h) The cross product in components

│ └── Determinant form

├── 3-3 (i) Properties of the vector product

│ └── List

├── 3-3 (j) Parallel vectors

│ └── Condition

│ ├── Parallelogram and triangle

│ └── Link with mechanics

├── 3-3 (l) The scalar triple product

│ ├── Definition

│ ├── Determinant form

│ └── Properties

├── Parallelepiped and tetrahedron

└── Summary

Definition

Direction: the right-hand rule

The unit vectors

Norm of the product

✍️ Worked examples

Determinant form

✍️ Worked examples

List

Condition

✍️ Worked examples

Parallelogram and triangle

✍️ Worked examples

Link with mechanics

Definition

✍️ Worked examples

✍️ Worked examples

Well done! You finished this lesson