IB Mathematics Applications & Interpretation Topic…

Why Multiply Two Vectors: Two Products For Two Different Jobs

You can already add vectors and scale them, but multiplying two vectors turns out to mean two genuinely different things.
The scalar (dot) product combines two vectors into a single number that measures how much they point the same way.
The vector (cross) product combines two 3D vectors into a new vector at right angles to both.
1. In modelling, the dot product measures alignment, angles, and the work done by a force, while the cross product gives areas and the normal direction to a surface.
Both appear on every AI HL paper, and since a GDC is required on every paper, the calculator does the arithmetic while you supply the correct set-up.

Hint

The dot product returns a scalar (a plain number) and is written $\mathbf{v} \cdot \mathbf{w}$.
The cross product returns a vector and is written $\mathbf{v} \times \mathbf{w}$.

The Scalar (Dot) Product: Two Formulas, Both In The Booklet

The component form is the one you compute with: multiply matching components and add. $$\mathbf{v} \cdot \mathbf{w} = v_1 w_1 + v_2 w_2 + v_3 w_3$$
The geometric form is the one you reason with, linking the product to the angle $\theta$ between the vectors. $$\mathbf{v} \cdot \mathbf{w} = |\mathbf{v}|\,|\mathbf{w}|\cos\theta$$
A large positive value means the vectors are well aligned, a value near zero means they are close to perpendicular, and a negative value means they point in broadly opposite directions.

Note

Both forms are printed in the formula booklet under AHL 3.13, so the exam work is choosing the right one and reading the angle correctly.

Computing The Dot Product By Hand And On Your GDC

By hand, pair up matching components, multiply each pair, and add the results to one number.
On the GDC, store the two vectors and use the built-in dot-product command; stating that command is the method, and it is the fast, error-free route on Paper 2 and Paper 3.

Example question

Given $\mathbf{v} = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}$ and $\mathbf{w} = \begin{pmatrix} 4 \\ 5 \\ -2 \end{pmatrix}$, find $\mathbf{v} \cdot \mathbf{w}$.

Solution

Multiply matching components and add: $$\mathbf{v} \cdot \mathbf{w} = 3(4) + (-1)(5) + 2(-2)$$
Evaluate: $$12 - 5 - 4 = 3$$
On the GDC, entering both vectors and applying the dot-product command returns the same value, $3$.

Finding The Angle Between Two Vectors

Rearranging the geometric form makes the angle the subject. $$\cos\theta = \frac{\mathbf{v} \cdot \mathbf{w}}{|\mathbf{v}|\,|\mathbf{w}|} \qquad\Rightarrow\qquad \theta = \arccos\!\left(\frac{\mathbf{v} \cdot \mathbf{w}}{|\mathbf{v}|\,|\mathbf{w}|}\right)$$
1. This rearrangement is not printed separately in the booklet; it follows directly from the geometric form given under AHL 3.13.
The recipe is always the same: dot product on top, the two magnitudes on the bottom, then $\arccos$.
1. Set your GDC to the angle mode the question wants; angles between vectors are usually reported in degrees, so check the mode before you evaluate the $\arccos$.

Example question

Find the angle between $\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$, giving your answer to the nearest degree.

Solution

Dot product: $$\mathbf{a} \cdot \mathbf{b} = 1(2) + 2(-1) + 2(2) = 4$$
Magnitudes: $$|\mathbf{a}| = \sqrt{1+4+4} = 3$ and $|\mathbf{b}| = \sqrt{4+1+4} = 3$$
Substitute into the set-up: $$\cos\theta = \dfrac{4}{3 \times 3} = \dfrac{4}{9}$$
So $\theta = \arccos\tfrac{4}{9} \approx 64^\circ$

Perpendicular And Parallel Vectors: The Two Extreme Angles

Perpendicular: The Dot Product Is Zero

If $\theta = 90^\circ$ then $\cos\theta = 0$, so the whole product collapses to zero. $$\mathbf{v} \cdot \mathbf{w} = 0 \iff \mathbf{v} \perp \mathbf{w} \quad(\text{for non-zero vectors})$$
This is the fastest perpendicularity check in the course: no magnitudes and no $\arccos$, just one dot product.

Example question

Find the value of $t$ for which $\begin{pmatrix} 1 \\ t \\ 2 \end{pmatrix}$ and $\begin{pmatrix} 3 \\ -1 \\ t \end{pmatrix}$ are perpendicular.

Solution

Perpendicular means the dot product is zero.
Compute it: $1(3) + t(-1) + 2(t) = 3 + t$
Set equal to zero and solve: $3 + t = 0 \Rightarrow t = -3$

Parallel: One Vector Is A Multiple Of The Other

Two non-zero vectors are parallel when one is a scalar multiple of the other, $\mathbf{v} = k\mathbf{w}$.
Then $\theta = 0^\circ$ or $180^\circ$, so $\cos\theta = \pm 1$ and the size of the dot product is the full product of the magnitudes. $$|\mathbf{v} \cdot \mathbf{w}| = |\mathbf{v}|\,|\mathbf{w}| \iff \mathbf{v} \parallel \mathbf{w}$$

The Angle Between Two Lines: Use Their Direction Vectors

A straight line's tilt in space is carried entirely by its direction vector, not by the point it passes through.
So the angle between two lines is found by applying the dot-product angle formula to their two direction vectors.
If the formula returns an obtuse angle, subtract from $180^\circ$ to report the acute angle between the lines.

Example question

Two zip-lines run along the lines $\mathbf{r}_1 = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + s\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$ and $\mathbf{r}_2 = \begin{pmatrix} 0 \\ 3 \\ 1 \end{pmatrix} + u\begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix}$. Find the acute angle between them, to the nearest degree.

Solution

Use the direction vectors $\mathbf{d}_1 = (1,2,2)$ and $\mathbf{d}_2 = (2,2,1)$.
Dot product: $$\mathbf{d}_1 \cdot \mathbf{d}_2 = 1(2) + 2(2) + 2(1) = 8$$
Magnitudes: $$|\mathbf{d}_1| = 3$ and $|\mathbf{d}_2| = 3$$
So $\cos\theta = \dfrac{8}{9}$, giving $\theta \approx 27^\circ$, which is already acute.

The Component (Scalar Projection) Of One Vector On Another

The component of $\mathbf{v}$ in the direction of $\mathbf{w}$ is the length of the shadow $\mathbf{v}$ casts along $\mathbf{w}$, which is $|\mathbf{v}|\cos\theta$.
Replacing $\cos\theta$ using the geometric form turns this into a formula you can compute directly. $$|\mathbf{v}|\cos\theta = \frac{\mathbf{v} \cdot \mathbf{w}}{|\mathbf{w}|}$$
This projection formula is not printed in the booklet, so you derive it from the geometric dot-product form on the day.

Example question

A force $\mathbf{F} = \begin{pmatrix} 6 \\ 8 \end{pmatrix}\,\mathrm{N}$ acts on a crate. Find the component of $\mathbf{F}$ along the direction $\mathbf{d} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}$.

Solution

Apply the scalar projection $\dfrac{\mathbf{F} \cdot \mathbf{d}}{|\mathbf{d}|}$.
Dot product: $$\mathbf{F} \cdot \mathbf{d} = 6(4) + 8(3) = 48$$
Magnitude: $$|\mathbf{d}| = \sqrt{16 + 9} = 5$$
Component $= \dfrac{48}{5} = 9.6\,\mathrm{N}$

The Vector (Cross) Product: A Product That Returns A Vector

The component form is the one you compute with. $$\mathbf{v} \times \mathbf{w} = \begin{pmatrix} v_2 w_3 - v_3 w_2 \\ v_3 w_1 - v_1 w_3 \\ v_1 w_2 - v_2 w_1 \end{pmatrix}$$
The geometric form gives the length and direction of the output vector. $$|\mathbf{v} \times \mathbf{w}| = |\mathbf{v}|\,|\mathbf{w}|\sin\theta$$
The direction is perpendicular to both $\mathbf{v}$ and $\mathbf{w}$, fixed by the right-hand rule: curl the fingers of your right hand from $\mathbf{v}$ to $\mathbf{w}$ and your thumb points along $\mathbf{v} \times \mathbf{w}$.

Note

Both the component form and $|\mathbf{v} \times \mathbf{w}| = |\mathbf{v}|\,|\mathbf{w}|\sin\theta$ are given in the formula booklet under AHL 3.13.
Each output component skips its own row and cross-multiplies the other two, and the middle component carries a sign flip, so write it carefully.

Computing A Cross Product By Hand And On Your GDC

For each output component, cover its own row and cross-multiply the remaining pairs, remembering the minus sign on the middle entry.
On the GDC, storing the two vectors and calling the cross-product command is the reliable route, and quoting that command counts as the method.

Example question

Given $\mathbf{v} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ and $\mathbf{w} = \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix}$, find $\mathbf{v} \times \mathbf{w}$ and verify it is perpendicular to $\mathbf{v}$.

Solution

Apply the component form: $$\mathbf{v} \times \mathbf{w} = \begin{pmatrix} 2(6)-3(5) \\ 3(4)-1(6) \\ 1(5)-2(4) \end{pmatrix}$$
Evaluate: $$\mathbf{v} \times \mathbf{w} = \begin{pmatrix} -3 \\ 6 \\ -3 \end{pmatrix}$$
Check with the dot product: $1(-3) + 2(6) + 3(-3) = 0$, so it is perpendicular to $\mathbf{v}$.

The Magnitude Is An Area: Parallelograms And Triangles

Take $\mathbf{v}$ as the base of a parallelogram; the height is the part of $\mathbf{w}$ at right angles to it, which is $|\mathbf{w}|\sin\theta$.
Base times height is exactly the magnitude of the cross product, and this area result is given in the formula booklet. $$\text{Area of parallelogram} = |\mathbf{v}|\,(|\mathbf{w}|\sin\theta) = |\mathbf{v} \times \mathbf{w}|$$
A triangle is half a parallelogram, so its area is half the magnitude. $$\text{Area of triangle} = \tfrac{1}{2}|\mathbf{v} \times \mathbf{w}|$$

Example question

Find the area of the triangle with vertices $A(1, 0, 0)$, $B(0, 2, 0)$ and $C(0, 0, 3)$.

Solution

Form two edge vectors from $A$: $$\vec{AB} = (-1, 2, 0)$ and $\vec{AC} = (-1, 0, 3)$$
Cross product: $$\vec{AB} \times \vec{AC} = (6, 3, 2)$$
Magnitude: $$|\vec{AB} \times \vec{AC}| = \sqrt{6^2 + 3^2 + 2^2} = \sqrt{49} = 7$$
Triangle area is half: $\tfrac{1}{2}(7) = 3.5$ square units

The Direction Of The Cross Product: Perpendicular To Both, Zero When Parallel

The single most useful fact for the exam is that $\mathbf{v} \times \mathbf{w}$ is perpendicular to both $\mathbf{v}$ and $\mathbf{w}$.
1. Given two vectors lying in a plane, their cross product is therefore a normal to that plane, which is the bridge to equations of planes.
If the vectors are parallel they span no area, so the magnitude is zero. $$\mathbf{v} \times \mathbf{w} = \mathbf{0} \iff \mathbf{v} \parallel \mathbf{w} \quad(\text{for non-zero vectors})$$

Common Mistake

Do not mix the two products: the dot product returns a scalar, the cross product returns a vector.
The dot product is zero for perpendicular vectors, but the cross product is zero for parallel vectors, so the zero conditions are opposite.
The dot product is commutative, but the cross product is anticommutative, $\mathbf{v} \times \mathbf{w} = -(\mathbf{w} \times \mathbf{v})$.

Dot Versus Cross: A Side-By-Side Summary

Feature	Dot product	Cross product
Output	Scalar (a number)	Vector (3D)
Component form	v1w1 + v2w2 + v3w3	(v2w3-v3w2, v3w1-v1w3, v1w2-v2w1)
Size formula	\|v\|\|w\|cos(theta)	\|v\|\|w\|sin(theta)
Geometric meaning	Alignment and projection	Area of parallelogram
Zero when	Vectors perpendicular	Vectors parallel
Order	Commutative	Anticommutative

Active recall

Write both booklet forms of $\mathbf{v} \cdot \mathbf{w}$ and state whether the output is a scalar or a vector.
State the quick test that tells you two non-zero vectors are perpendicular.
How do you find the angle between two lines from their equations?
What does $|\mathbf{v} \times \mathbf{w}|$ represent geometrically, and how do you get a triangle's area from it?
For non-zero vectors, when is $\mathbf{v} \cdot \mathbf{w} = 0$ and when is $\mathbf{v} \times \mathbf{w} = \mathbf{0}$?

Why Multiply Two Vectors: Two Products For Two Different Jobs

You can already add vectors and scale them, but multiplying two vectors turns out to mean two genuinely different things.
The scalar (dot) product combines two vectors into a single number that measures how much they point the same way.
The vector (cross) product combines two 3D vectors into a new vector at right angles to both.
1. In modelling, the dot product measures alignment, angles, and the work done by a force, while the cross product gives areas and the normal direction to a surface.
Both appear on every AI HL paper, and since a GDC is required on every paper, the calculator does the arithmetic while you supply the correct set-up.

Hint

The dot product returns a scalar (a plain number) and is written $\mathbf{v} \cdot \mathbf{w}$.
The cross product returns a vector and is written $\mathbf{v} \times \mathbf{w}$.

The Scalar (Dot) Product: Two Formulas, Both In The Booklet

The component form is the one you compute with: multiply matching components and add. $$\mathbf{v} \cdot \mathbf{w} = v_1 w_1 + v_2 w_2 + v_3 w_3$$
The geometric form is the one you reason with, linking the product to the angle $\theta$ between the vectors. $$\mathbf{v} \cdot \mathbf{w} = |\mathbf{v}|\,|\mathbf{w}|\cos\theta$$
A large positive value means the vectors are well aligned, a value near zero means they are close to perpendicular, and a negative value means they point in broadly opposite directions.

Note

Both forms are printed in the formula booklet under AHL 3.13, so the exam work is choosing the right one and reading the angle correctly.

Computing The Dot Product By Hand And On Your GDC

By hand, pair up matching components, multiply each pair, and add the results to one number.
On the GDC, store the two vectors and use the built-in dot-product command; stating that command is the method, and it is the fast, error-free route on Paper 2 and Paper 3.

Example question

Given $\mathbf{v} = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}$ and $\mathbf{w} = \begin{pmatrix} 4 \\ 5 \\ -2 \end{pmatrix}$, find $\mathbf{v} \cdot \mathbf{w}$.

Solution

Multiply matching components and add: $$\mathbf{v} \cdot \mathbf{w} = 3(4) + (-1)(5) + 2(-2)$$
Evaluate: $$12 - 5 - 4 = 3$$
On the GDC, entering both vectors and applying the dot-product command returns the same value, $3$.

Finding The Angle Between Two Vectors

Rearranging the geometric form makes the angle the subject. $$\cos\theta = \frac{\mathbf{v} \cdot \mathbf{w}}{|\mathbf{v}|\,|\mathbf{w}|} \qquad\Rightarrow\qquad \theta = \arccos\!\left(\frac{\mathbf{v} \cdot \mathbf{w}}{|\mathbf{v}|\,|\mathbf{w}|}\right)$$
1. This rearrangement is not printed separately in the booklet; it follows directly from the geometric form given under AHL 3.13.
The recipe is always the same: dot product on top, the two magnitudes on the bottom, then $\arccos$.
1. Set your GDC to the angle mode the question wants; angles between vectors are usually reported in degrees, so check the mode before you evaluate the $\arccos$.

Example question

Find the angle between $\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$, giving your answer to the nearest degree.

Solution

Dot product: $$\mathbf{a} \cdot \mathbf{b} = 1(2) + 2(-1) + 2(2) = 4$$
Magnitudes: $$|\mathbf{a}| = \sqrt{1+4+4} = 3$ and $|\mathbf{b}| = \sqrt{4+1+4} = 3$$
Substitute into the set-up: $$\cos\theta = \dfrac{4}{3 \times 3} = \dfrac{4}{9}$$
So $\theta = \arccos\tfrac{4}{9} \approx 64^\circ$

Perpendicular And Parallel Vectors: The Two Extreme Angles

Perpendicular: The Dot Product Is Zero

If $\theta = 90^\circ$ then $\cos\theta = 0$, so the whole product collapses to zero. $$\mathbf{v} \cdot \mathbf{w} = 0 \iff \mathbf{v} \perp \mathbf{w} \quad(\text{for non-zero vectors})$$
This is the fastest perpendicularity check in the course: no magnitudes and no $\arccos$, just one dot product.

Example question

Find the value of $t$ for which $\begin{pmatrix} 1 \\ t \\ 2 \end{pmatrix}$ and $\begin{pmatrix} 3 \\ -1 \\ t \end{pmatrix}$ are perpendicular.

Solution

Perpendicular means the dot product is zero.
Compute it: $1(3) + t(-1) + 2(t) = 3 + t$
Set equal to zero and solve: $3 + t = 0 \Rightarrow t = -3$

Parallel: One Vector Is A Multiple Of The Other

Two non-zero vectors are parallel when one is a scalar multiple of the other, $\mathbf{v} = k\mathbf{w}$.
Then $\theta = 0^\circ$ or $180^\circ$, so $\cos\theta = \pm 1$ and the size of the dot product is the full product of the magnitudes. $$|\mathbf{v} \cdot \mathbf{w}| = |\mathbf{v}|\,|\mathbf{w}| \iff \mathbf{v} \parallel \mathbf{w}$$

The Angle Between Two Lines: Use Their Direction Vectors

A straight line's tilt in space is carried entirely by its direction vector, not by the point it passes through.
So the angle between two lines is found by applying the dot-product angle formula to their two direction vectors.
If the formula returns an obtuse angle, subtract from $180^\circ$ to report the acute angle between the lines.

Example question

Solution

Use the direction vectors $\mathbf{d}_1 = (1,2,2)$ and $\mathbf{d}_2 = (2,2,1)$.
Dot product: $$\mathbf{d}_1 \cdot \mathbf{d}_2 = 1(2) + 2(2) + 2(1) = 8$$
Magnitudes: $$|\mathbf{d}_1| = 3$ and $|\mathbf{d}_2| = 3$$
So $\cos\theta = \dfrac{8}{9}$, giving $\theta \approx 27^\circ$, which is already acute.

The Component (Scalar Projection) Of One Vector On Another

The component of $\mathbf{v}$ in the direction of $\mathbf{w}$ is the length of the shadow $\mathbf{v}$ casts along $\mathbf{w}$, which is $|\mathbf{v}|\cos\theta$.
Replacing $\cos\theta$ using the geometric form turns this into a formula you can compute directly. $$|\mathbf{v}|\cos\theta = \frac{\mathbf{v} \cdot \mathbf{w}}{|\mathbf{w}|}$$
This projection formula is not printed in the booklet, so you derive it from the geometric dot-product form on the day.

Example question

A force $\mathbf{F} = \begin{pmatrix} 6 \\ 8 \end{pmatrix}\,\mathrm{N}$ acts on a crate. Find the component of $\mathbf{F}$ along the direction $\mathbf{d} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}$.

Solution

Apply the scalar projection $\dfrac{\mathbf{F} \cdot \mathbf{d}}{|\mathbf{d}|}$.
Dot product: $$\mathbf{F} \cdot \mathbf{d} = 6(4) + 8(3) = 48$$
Magnitude: $$|\mathbf{d}| = \sqrt{16 + 9} = 5$$
Component $= \dfrac{48}{5} = 9.6\,\mathrm{N}$

The Vector (Cross) Product: A Product That Returns A Vector

The component form is the one you compute with. $$\mathbf{v} \times \mathbf{w} = \begin{pmatrix} v_2 w_3 - v_3 w_2 \\ v_3 w_1 - v_1 w_3 \\ v_1 w_2 - v_2 w_1 \end{pmatrix}$$
The geometric form gives the length and direction of the output vector. $$|\mathbf{v} \times \mathbf{w}| = |\mathbf{v}|\,|\mathbf{w}|\sin\theta$$
The direction is perpendicular to both $\mathbf{v}$ and $\mathbf{w}$, fixed by the right-hand rule: curl the fingers of your right hand from $\mathbf{v}$ to $\mathbf{w}$ and your thumb points along $\mathbf{v} \times \mathbf{w}$.

Note

Both the component form and $|\mathbf{v} \times \mathbf{w}| = |\mathbf{v}|\,|\mathbf{w}|\sin\theta$ are given in the formula booklet under AHL 3.13.
Each output component skips its own row and cross-multiplies the other two, and the middle component carries a sign flip, so write it carefully.

Computing A Cross Product By Hand And On Your GDC

For each output component, cover its own row and cross-multiply the remaining pairs, remembering the minus sign on the middle entry.
On the GDC, storing the two vectors and calling the cross-product command is the reliable route, and quoting that command counts as the method.

Example question

Solution

Apply the component form: $$\mathbf{v} \times \mathbf{w} = \begin{pmatrix} 2(6)-3(5) \\ 3(4)-1(6) \\ 1(5)-2(4) \end{pmatrix}$$
Evaluate: $$\mathbf{v} \times \mathbf{w} = \begin{pmatrix} -3 \\ 6 \\ -3 \end{pmatrix}$$
Check with the dot product: $1(-3) + 2(6) + 3(-3) = 0$, so it is perpendicular to $\mathbf{v}$.

The Magnitude Is An Area: Parallelograms And Triangles

Take $\mathbf{v}$ as the base of a parallelogram; the height is the part of $\mathbf{w}$ at right angles to it, which is $|\mathbf{w}|\sin\theta$.
Base times height is exactly the magnitude of the cross product, and this area result is given in the formula booklet. $$\text{Area of parallelogram} = |\mathbf{v}|\,(|\mathbf{w}|\sin\theta) = |\mathbf{v} \times \mathbf{w}|$$
A triangle is half a parallelogram, so its area is half the magnitude. $$\text{Area of triangle} = \tfrac{1}{2}|\mathbf{v} \times \mathbf{w}|$$

Example question

Find the area of the triangle with vertices $A(1, 0, 0)$, $B(0, 2, 0)$ and $C(0, 0, 3)$.

Solution

Form two edge vectors from $A$: $$\vec{AB} = (-1, 2, 0)$ and $\vec{AC} = (-1, 0, 3)$$
Cross product: $$\vec{AB} \times \vec{AC} = (6, 3, 2)$$
Magnitude: $$|\vec{AB} \times \vec{AC}| = \sqrt{6^2 + 3^2 + 2^2} = \sqrt{49} = 7$$
Triangle area is half: $\tfrac{1}{2}(7) = 3.5$ square units

The Direction Of The Cross Product: Perpendicular To Both, Zero When Parallel

The single most useful fact for the exam is that $\mathbf{v} \times \mathbf{w}$ is perpendicular to both $\mathbf{v}$ and $\mathbf{w}$.
1. Given two vectors lying in a plane, their cross product is therefore a normal to that plane, which is the bridge to equations of planes.
If the vectors are parallel they span no area, so the magnitude is zero. $$\mathbf{v} \times \mathbf{w} = \mathbf{0} \iff \mathbf{v} \parallel \mathbf{w} \quad(\text{for non-zero vectors})$$

Common Mistake

Do not mix the two products: the dot product returns a scalar, the cross product returns a vector.
The dot product is zero for perpendicular vectors, but the cross product is zero for parallel vectors, so the zero conditions are opposite.
The dot product is commutative, but the cross product is anticommutative, $\mathbf{v} \times \mathbf{w} = -(\mathbf{w} \times \mathbf{v})$.

Dot Versus Cross: A Side-By-Side Summary

Feature	Dot product	Cross product
Output	Scalar (a number)	Vector (3D)
Component form	v1w1 + v2w2 + v3w3	(v2w3-v3w2, v3w1-v1w3, v1w2-v2w1)
Size formula	\|v\|\|w\|cos(theta)	\|v\|\|w\|sin(theta)
Geometric meaning	Alignment and projection	Area of parallelogram
Zero when	Vectors perpendicular	Vectors parallel
Order	Commutative	Anticommutative

Active recall

Write both booklet forms of $\mathbf{v} \cdot \mathbf{w}$ and state whether the output is a scalar or a vector.
State the quick test that tells you two non-zero vectors are perpendicular.
How do you find the angle between two lines from their equations?
What does $|\mathbf{v} \times \mathbf{w}|$ represent geometrically, and how do you get a triangle's area from it?
For non-zero vectors, when is $\mathbf{v} \cdot \mathbf{w} = 0$ and when is $\mathbf{v} \times \mathbf{w} = \mathbf{0}$?

IB Mathematics Applications & Interpretation AHL 3.13—scalar and Vector Products Notes

IB Mathematics Applications & Interpretation AHL 3.13—scalar and Vector Products Notes

Why Multiply Two Vectors: Two Products For Two Different Jobs

The Scalar (Dot) Product: Two Formulas, Both In The Booklet

Computing The Dot Product By Hand And On Your GDC

Finding The Angle Between Two Vectors

Perpendicular And Parallel Vectors: The Two Extreme Angles

Perpendicular: The Dot Product Is Zero

Parallel: One Vector Is A Multiple Of The Other

The Angle Between Two Lines: Use Their Direction Vectors

The Component (Scalar Projection) Of One Vector On Another

The Vector (Cross) Product: A Product That Returns A Vector

Computing A Cross Product By Hand And On Your GDC

The Magnitude Is An Area: Parallelograms And Triangles

The Direction Of The Cross Product: Perpendicular To Both, Zero When Parallel

Dot Versus Cross: A Side-By-Side Summary

Why Multiply Two Vectors: Two Products For Two Different Jobs

The Scalar (Dot) Product: Two Formulas, Both In The Booklet

Computing The Dot Product By Hand And On Your GDC

Finding The Angle Between Two Vectors

Perpendicular And Parallel Vectors: The Two Extreme Angles

Perpendicular: The Dot Product Is Zero

Parallel: One Vector Is A Multiple Of The Other

The Angle Between Two Lines: Use Their Direction Vectors

The Component (Scalar Projection) Of One Vector On Another

The Vector (Cross) Product: A Product That Returns A Vector

Computing A Cross Product By Hand And On Your GDC

The Magnitude Is An Area: Parallelograms And Triangles

The Direction Of The Cross Product: Perpendicular To Both, Zero When Parallel

Dot Versus Cross: A Side-By-Side Summary

IB Mathematics Applications & Interpretation AHL 3.13—scalar and Vector Products Notes

Number and Algebra

Functions

Geometry and Trigonometry

Statistics and Probability

Calculus

IB Mathematics Applications & Interpretation AHL 3.13—scalar and Vector Products Notes

Number and Algebra

Functions

Geometry and Trigonometry

Statistics and Probability

Calculus

Why Multiply Two Vectors: Two Products For Two Different Jobs

The Scalar (Dot) Product: Two Formulas, Both In The Booklet

Computing The Dot Product By Hand And On Your GDC

Finding The Angle Between Two Vectors

Perpendicular And Parallel Vectors: The Two Extreme Angles

Perpendicular: The Dot Product Is Zero

Parallel: One Vector Is A Multiple Of The Other

The Angle Between Two Lines: Use Their Direction Vectors

The Component (Scalar Projection) Of One Vector On Another

The Vector (Cross) Product: A Product That Returns A Vector

Computing A Cross Product By Hand And On Your GDC

The Magnitude Is An Area: Parallelograms And Triangles

The Direction Of The Cross Product: Perpendicular To Both, Zero When Parallel

Dot Versus Cross: A Side-By-Side Summary

Why Multiply Two Vectors: Two Products For Two Different Jobs

The Scalar (Dot) Product: Two Formulas, Both In The Booklet

Computing The Dot Product By Hand And On Your GDC

Finding The Angle Between Two Vectors

Perpendicular And Parallel Vectors: The Two Extreme Angles

Perpendicular: The Dot Product Is Zero

Parallel: One Vector Is A Multiple Of The Other

The Angle Between Two Lines: Use Their Direction Vectors

The Component (Scalar Projection) Of One Vector On Another

The Vector (Cross) Product: A Product That Returns A Vector

Computing A Cross Product By Hand And On Your GDC

The Magnitude Is An Area: Parallelograms And Triangles

The Direction Of The Cross Product: Perpendicular To Both, Zero When Parallel

Dot Versus Cross: A Side-By-Side Summary