3D Vectors & The Scalar Product

Figure 3.3: 3D Vectors and Scalar Projection

1. Vectors in Three Dimensions ($\mathbb{R}^3$)

A vector in three dimensions represents a displacement across three mutually orthogonal axes. It can be written in Cartesian basis notation ($\mathbf{i}, \mathbf{j}, \mathbf{k}$) or in column vector form:

$$\mathbf{v} = v_1 \mathbf{i} + v_2 \mathbf{j} + v_3 \mathbf{k} = \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}$$

Magnitude (Length): $|\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}$. Geometrically, this is the spatial diagonal of a rectangular cuboid with side lengths $|v_1|, |v_2|, |v_3|$.
Unit Vector: A vector with identical direction but length equal to 1: $\hat{\mathbf{v}} = \frac{\mathbf{v}}{|\mathbf{v}|}$.
Displacement Between Points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$: $$\vec{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \\ z_2 - z_1 \end{pmatrix}, \quad |\vec{AB}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$$
Direction Cosines: If $\alpha, \beta, \gamma$ are the angles that $\mathbf{v}$ makes with the positive $x, y, z$-axes, then $\cos\alpha = \frac{v_1}{|\mathbf{v}|}$, $\cos\beta = \frac{v_2}{|\mathbf{v}|}$, and $\cos\gamma = \frac{v_3}{|\mathbf{v}|}$, satisfying the invariant: $$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$$

2. The Scalar (Dot) Product

The scalar product of two vectors $\mathbf{u}$ and $\mathbf{v}$ yields a real scalar quantity defined by two mathematically equivalent formulations:

$$\mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + u_3 v_3 = |\mathbf{u}| |\mathbf{v}| \cos\theta$$

where $\theta$ is the angle between the two vectors with $0 \le \theta \le \pi$ ($0^\circ \le \theta \le 180^\circ$).

Core Properties of the Scalar Product

• Angle Determination: $\cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|} = \frac{u_1 v_1 + u_2 v_2 + u_3 v_3}{\sqrt{u_1^2 + u_2^2 + u_3^2} \sqrt{v_1^2 + v_2^2 + v_3^2}}$.
If $\mathbf{u} \cdot \mathbf{v} > 0$, $\theta$ is acute ($0 \le \theta < \pi/2$).
If $\mathbf{u} \cdot \mathbf{v} < 0$, $\theta$ is obtuse ($\pi/2 < \theta \le \pi$).

• Orthogonality (Perpendicularity) Test: For two non-zero vectors $\mathbf{u}$ and $\mathbf{v}$, $$\mathbf{u} \perp \mathbf{v} \iff \mathbf{u} \cdot \mathbf{v} = 0$$

• Self-Product & Magnitude: $\mathbf{u} \cdot \mathbf{u} = |\mathbf{u}|^2 \iff |\mathbf{u}| = \sqrt{\mathbf{u} \cdot \mathbf{u}}$.

• Cauchy-Schwarz Inequality: $|\mathbf{u} \cdot \mathbf{v}| \le |\mathbf{u}| |\mathbf{v}|$, with equality if and only if $\mathbf{u}$ and $\mathbf{v}$ are collinear (parallel).

3. Vector Projections & Orthogonal Decomposition (Syllabus AHL 3.11)

The scalar product enables decomposing any vector $\mathbf{u}$ into two perpendicular vector components relative to a reference vector $\mathbf{v}$:

Scalar Projection (Signed Magnitude): The signed length of the shadow cast by $\mathbf{u}$ along the direction of $\mathbf{v}$: $$\text{scal}_{\mathbf{v}}(\mathbf{u}) = |\mathbf{u}| \cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|}$$
Vector Projection (Parallel Component $\mathbf{u}_\parallel$): The actual vector shadow pointing along $\mathbf{v}$: $$\text{proj}_{\mathbf{v}}(\mathbf{u}) = \mathbf{u}_\parallel = \left(\frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|}\right) \hat{\mathbf{v}} = \left(\frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|^2}\right) \mathbf{v}$$
Perpendicular Component ($\mathbf{u}_\perp$): The component orthogonal to $\mathbf{v}$: $$\mathbf{u}_\perp = \mathbf{u} - \text{proj}_{\mathbf{v}}(\mathbf{u}) = \mathbf{u} - \left(\frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|^2}\right) \mathbf{v}$$ Notice that $\mathbf{u}_\perp \cdot \mathbf{v} = \mathbf{u}\cdot\mathbf{v} - \frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{v}|^2}(\mathbf{v}\cdot\mathbf{v}) = \mathbf{u}\cdot\mathbf{v} - \mathbf{u}\cdot\mathbf{v} = 0$, guaranteeing orthogonality!

Worked Examples

Worked Example 1: Angle Between Two 3D Vectors (Paper 1 Style · 6 marks)

Consider the vectors $\mathbf{u} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$.
(a) Find the scalar product $\mathbf{u} \cdot \mathbf{v}$. [2 marks]
(b) Calculate the magnitudes $|\mathbf{u}|$ and $|\mathbf{v}|$. [2 marks]
(c) Hence find the exact cosine of the angle $\theta$ between $\mathbf{u}$ and $\mathbf{v}$, and state whether $\theta$ is acute or obtuse. [2 marks]

Official Markscheme Solution:

(a) Scalar Product: [2 marks]
$\mathbf{u} \cdot \mathbf{v} = (2)(1) + (-1)(2) + (2)(2)$ [M1]
$= 2 - 2 + 4 = 4$ [A1].

(b) Magnitudes: [2 marks]
$|\mathbf{u}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$ [A1].
$|\mathbf{v}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$ [A1].

(c) Cosine & Nature: [2 marks]
$\cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|} = \frac{4}{3 \times 3} = \frac{4}{9}$ [A1].
Since $\cos\theta = \frac{4}{9} > 0$, $\theta$ is an acute angle ($\theta \approx 63.6^\circ$) [R1].

Worked Example 2: Orthogonality with Quadratic Parameter (Paper 1 Style · 6 marks)

The vectors $\mathbf{a} = \begin{pmatrix} k \\ 3 \\ -2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} k - 1 \\ k \\ 4 \end{pmatrix}$ are orthogonal.
(a) Show that $k^2 + 2k - 8 = 0$. [3 marks]
(b) Find the possible values of $k$. [3 marks]

Official Markscheme Solution:

(a) Quadratic Formation: [3 marks]
Orthogonal vectors satisfy $\mathbf{a} \cdot \mathbf{b} = 0$ [M1]:
$k(k - 1) + 3(k) + (-2)(4) = 0$ [M1]
$k^2 - k + 3k - 8 = 0 \implies k^2 + 2k - 8 = 0$ [A1][AG].

(b) Solving Parameter: [3 marks]
Factorize the quadratic: [M1]
$(k + 4)(k - 2) = 0$
Solutions: $k = 2$ or $k = -4$ [A1].

Worked Example 3: Vector Projection & Orthogonal Component (Paper 1 Style · 6 marks)

Let $\mathbf{u} = \begin{pmatrix} 4 \\ 1 \\ 3 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}$.
(a) Find the vector projection of $\mathbf{u}$ onto $\mathbf{v}$. [3 marks]
(b) Decompose $\mathbf{u}$ into two vectors $\mathbf{u}_\parallel$ and $\mathbf{u}_\perp$ such that $\mathbf{u} = \mathbf{u}_\parallel + \mathbf{u}_\perp$, where $\mathbf{u}_\parallel$ is parallel to $\mathbf{v}$ and $\mathbf{u}_\perp$ is perpendicular to $\mathbf{v}$. [3 marks]

Official Markscheme Solution:

(a) Vector Projection: [3 marks]
Calculate scalar product and magnitude squared of $\mathbf{v}$: [M1]
$\mathbf{u} \cdot \mathbf{v} = 4(2) + 1(-2) + 3(1) = 8 - 2 + 3 = 9$ [A1].
$|\mathbf{v}|^2 = 2^2 + (-2)^2 + 1^2 = 4 + 4 + 1 = 9$ [A1].
$\mathbf{u}_\parallel = \text{proj}_{\mathbf{v}}(\mathbf{u}) = \left(\frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|^2}\right) \mathbf{v} = \left(\frac{9}{9}\right) \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix}$ [A1].

(b) Orthogonal Decomposition: [3 marks]
$\mathbf{u}_\perp = \mathbf{u} - \mathbf{u}_\parallel$ [M1]
$= \begin{pmatrix} 4 \\ 1 \\ 3 \end{pmatrix} - \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ 2 \end{pmatrix}$ [A1].
Check orthogonality: $\mathbf{u}_\perp \cdot \mathbf{v} = 2(2) + 3(-2) + 2(1) = 4 - 6 + 2 = 0$ [R1].
Decomposition: $\mathbf{u} = \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix} + \begin{pmatrix} 2 \\ 3 \\ 2 \end{pmatrix}$ [AG].

Worked Example 4: Geometric Proof of Rhombus Diagonals (Paper 1 Style · 5 marks)

Let $ABCD$ be a parallelogram defined by vectors $\vec{AB} = \mathbf{a}$ and $\vec{AD} = \mathbf{b}$.
(a) Express the diagonals $\vec{AC}$ and $\vec{BD}$ in terms of $\mathbf{a}$ and $\mathbf{b}$. [2 marks]
(b) Prove that the diagonals are perpendicular if and only if $ABCD$ is a rhombus. [3 marks]

Official Markscheme Solution:

(a) Diagonals: [2 marks]
• $\vec{AC} = \vec{AB} + \vec{BC} = \mathbf{a} + \mathbf{b}$ [A1].
• $\vec{BD} = \vec{BA} + \vec{AD} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}$ [A1].

(b) Proof of Equivalence: [3 marks]
The diagonals are perpendicular if and only if their scalar product is zero: [M1]
$\vec{AC} \cdot \vec{BD} = 0 \iff (\mathbf{a} + \mathbf{b}) \cdot (\mathbf{b} - \mathbf{a}) = 0$.
Expand using the distributive property: [M1]
$(\mathbf{b} + \mathbf{a}) \cdot (\mathbf{b} - \mathbf{a}) = \mathbf{b} \cdot \mathbf{b} - \mathbf{a} \cdot \mathbf{a} = |\mathbf{b}|^2 - |\mathbf{a}|^2 = 0$ [A1].
$|\mathbf{b}|^2 - |\mathbf{a}|^2 = 0 \iff |\mathbf{a}| = |\mathbf{b}|$ [R1].
Since adjacent sides $AB$ and $AD$ have equal lengths, all four sides of the parallelogram are equal, which is the definition of a rhombus [AG].

Worked Example 5: Triangle Geometry & Altitude via Vectors (Paper 2 Style · 8 marks)

The points $A(1, 2, 3)$, $B(3, 0, 2)$, and $C(2, 4, 1)$ form a triangle in $\mathbb{R}^3$.
(a) Find the vectors $\vec{AB}$ and $\vec{AC}$. [2 marks]
(b) Calculate the size of $\angle BAC$ in degrees. [3 marks]
(c) Find the exact area of triangle $ABC$. [3 marks]

Official Markscheme Solution:

(a) Vectors: [2 marks]
$\vec{AB} = \begin{pmatrix} 3 - 1 \\ 0 - 2 \\ 2 - 3 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \\ -1 \end{pmatrix}$ [A1].
$\vec{AC} = \begin{pmatrix} 2 - 1 \\ 4 - 2 \\ 1 - 3 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}$ [A1].

(b) Angle $\angle BAC$: [3 marks]
• $\vec{AB} \cdot \vec{AC} = 2(1) + (-2)(2) + (-1)(-2) = 2 - 4 + 2 = 0$ [M1][A1]!
Since the dot product is exactly 0: [R1]
$\cos(\angle BAC) = 0 \implies \angle BAC = 90^\circ$ (or $\frac{\pi}{2}$ radians) [A1].
Triangle $ABC$ is a right-angled triangle at vertex $A$.

(c) Area of Triangle: [3 marks]
Because $\angle BAC = 90^\circ$, the area is $\frac{1}{2} |\vec{AB}| |\vec{AC}|$ [M1].
$|\vec{AB}| = \sqrt{2^2 + (-2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$ [A1].
$|\vec{AC}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$ [A1].
$\text{Area} = \frac{1}{2} \times 3 \times 3 = \frac{9}{2} = 4.5$ square units [A1].

Practice Exercises (10 Tiered Questions)

Work through these 10 examination-calibrated exercises ranging from core foundational skills to advanced Paper 1 and Paper 2 problems. Expand each exercise to view the full IB markscheme and examiner advice.

Exercise 1 · Magnitude and Unit Vector

Given $\mathbf{a} = 2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}$, find $|\mathbf{a}|$ and the unit vector in the direction of $\mathbf{a}$.

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$|\mathbf{a}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$ [A1].
Unit vector: $\hat{\mathbf{a}} = \frac{\mathbf{a}}{|\mathbf{a}|} = \frac{1}{7}(2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}) = \begin{pmatrix} 2/7 \\ -3/7 \\ 6/7 \end{pmatrix}$ [A1].

Exercise 2 · Direct Scalar Product

Calculate $\mathbf{u}\cdot\mathbf{v}$ where $\mathbf{u} = \begin{pmatrix} 5 \\ -1 \\ 4 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix}$.

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$\mathbf{u}\cdot\mathbf{v} = (5)(-2) + (-1)(3) + (4)(1)$ [M1]
$= -10 - 3 + 4 = -9$ [A1].
Examiner Advice: A negative scalar product immediately indicates that the angle between the vectors is obtuse.

Exercise 3 · Orthogonality Test

Show that $\mathbf{p} = \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix}$ and $\mathbf{q} = \begin{pmatrix} 6 \\ -1 \\ 1 \end{pmatrix}$ are perpendicular.

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$\mathbf{p}\cdot\mathbf{q} = 1(6) + 4(-1) + (-2)(1) = 6 - 4 - 2 = 0$ [M1].
Since $\mathbf{p}\cdot\mathbf{q} = 0$ and neither vector is the zero vector, $\mathbf{p}$ and $\mathbf{q}$ are orthogonal (perpendicular) [A1][R1].

Exercise 4 · Acute Angle Between Vectors

Find the acute angle between $\mathbf{a} = \mathbf{i} + \mathbf{j}$ and $\mathbf{b} = \mathbf{j} + \mathbf{k}$ in radians.

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• $\mathbf{a}\cdot\mathbf{b} = 1(0) + 1(1) + 0(1) = 1$ [M1].
• $|\mathbf{a}| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}$, $|\mathbf{b}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$ [A1].
• $\cos\theta = \frac{1}{\sqrt{2} \times \sqrt{2}} = \frac{1}{2}$ [A1].
$\theta = \arccos\left(\frac{1}{2}\right) = \frac{\pi}{3}$ radians (or $60^\circ$) [A1].

Exercise 5 · Parameter for Given Angle

Find the value of $p$ such that the angle between $\mathbf{u} = \begin{pmatrix} 1 \\ p \\ 0 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$ is $\frac{\pi}{3}$.

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$\mathbf{u}\cdot\mathbf{v} = 1(0) + p(1) + 0(1) = p$ [M1].
$|\mathbf{u}| = \sqrt{1 + p^2}$, $|\mathbf{v}| = \sqrt{0 + 1 + 1} = \sqrt{2}$ [A1].
$\cos\left(\frac{\pi}{3}\right) = \frac{1}{2} = \frac{p}{\sqrt{2(1 + p^2)}}$ [M1].
Square both sides: $\frac{1}{4} = \frac{p^2}{2(1 + p^2)} \implies 2(1 + p^2) = 4p^2 \implies 2 + 2p^2 = 4p^2 \implies 2p^2 = 2 \implies p^2 = 1$ [M1].
Since $\cos(\pi/3) > 0$, $p$ must be positive: $p = 1$ [A1].

Exercise 6 · Scalar Projection

Find the scalar projection of $\mathbf{a} = 2\mathbf{i} + 3\mathbf{j} - \mathbf{k}$ in the direction of $\mathbf{b} = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k}$.

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Scalar projection is given by $\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}$ [M1]:
$\mathbf{a}\cdot\mathbf{b} = 2(1) + 3(-2) + (-1)(2) = 2 - 6 - 2 = -6$ [A1].
$|\mathbf{b}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = 3$ [A1].
$\text{scal}_{\mathbf{b}}(\mathbf{a}) = \frac{-6}{3} = -2$ [A1].
Examiner Advice: A negative scalar projection means the projection points in the direction opposite to $\mathbf{b}$.

Exercise 7 · Vector Proof of Pythagoras' Theorem

Let $\mathbf{u}$ and $\mathbf{v}$ be perpendicular vectors. Prove using dot products that $|\mathbf{u} + \mathbf{v}|^2 = |\mathbf{u}|^2 + |\mathbf{v}|^2$.

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Expand using the self-product property: [M1]
$|\mathbf{u} + \mathbf{v}|^2 = (\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v})$
$= \mathbf{u}\cdot\mathbf{u} + 2(\mathbf{u}\cdot\mathbf{v}) + \mathbf{v}\cdot\mathbf{v}$ [M1]
$= |\mathbf{u}|^2 + 2(\mathbf{u}\cdot\mathbf{v}) + |\mathbf{v}|^2$ [A1].
Since $\mathbf{u} \perp \mathbf{v}$, $\mathbf{u}\cdot\mathbf{v} = 0$ [R1].
Therefore, $|\mathbf{u} + \mathbf{v}|^2 = |\mathbf{u}|^2 + 0 + |\mathbf{v}|^2 = |\mathbf{u}|^2 + |\mathbf{v}|^2$ [AG].

Exercise 8 · Direction Cosines Invariant

A vector $\mathbf{v}$ in $\mathbb{R}^3$ makes angles $\alpha = 45^\circ$ with the $x$-axis and $\beta = 60^\circ$ with the $y$-axis. Find all possible values for the angle $\gamma$ that $\mathbf{v}$ makes with the $z$-axis.

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Apply the direction cosine identity: $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$ [M1].
$\cos(45^\circ) = \frac{1}{\sqrt{2}} \implies \cos^2(45^\circ) = \frac{1}{2}$ [A1].
$\cos(60^\circ) = \frac{1}{2} \implies \cos^2(60^\circ) = \frac{1}{4}$ [A1].
$\frac{1}{2} + \frac{1}{4} + \cos^2\gamma = 1 \implies \frac{3}{4} + \cos^2\gamma = 1 \implies \cos^2\gamma = \frac{1}{4}$ [M1].
$\cos\gamma = \pm\frac{1}{2}$ [A1].
• If $\cos\gamma = \frac{1}{2} \implies \gamma = 60^\circ$ (or $\frac{\pi}{3}$) [A1].
• If $\cos\gamma = -\frac{1}{2} \implies \gamma = 120^\circ$ (or $\frac{2\pi}{3}$) [A1].
Values: $\gamma = 60^\circ$ or $120^\circ$.

Exercise 9 · Finding Shortest Distance to a Line Segment

Let $A(0, 0, 0)$, $B(2, 4, 4)$, and $P(1, 1, 5)$. Find the coordinates of the foot of the perpendicular from $P$ to the line $AB$, and determine the shortest distance from $P$ to the line.

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$\vec{AB} = \begin{pmatrix} 2 \\ 4 \\ 4 \end{pmatrix}$, $\vec{AP} = \begin{pmatrix} 1 \\ 1 \\ 5 \end{pmatrix}$ [M1].
The foot of the perpendicular $F$ is the vector projection of $\vec{AP}$ onto $\vec{AB}$ [M1]:
$\vec{AP}\cdot\vec{AB} = 1(2) + 1(4) + 5(4) = 2 + 4 + 20 = 26$ [A1].
$|\vec{AB}|^2 = 2^2 + 4^2 + 4^2 = 4 + 16 + 16 = 36$ [A1].
$\vec{AF} = \frac{26}{36} \begin{pmatrix} 2 \\ 4 \\ 4 \end{pmatrix} = \frac{13}{18} \begin{pmatrix} 2 \\ 4 \\ 4 \end{pmatrix} = \begin{pmatrix} 13/9 \\ 26/9 \\ 26/9 \end{pmatrix}$ [A1].
Coordinates of foot: $F\left(\frac{13}{9}, \frac{26}{9}, \frac{26}{9}\right)$.
Shortest distance is $|\vec{PF}|$:
$\vec{PF} = \vec{AF} - \vec{AP} = \begin{pmatrix} 13/9 - 1 \\ 26/9 - 1 \\ 26/9 - 5 \end{pmatrix} = \begin{pmatrix} 4/9 \\ 17/9 \\ -19/9 \end{pmatrix}$ [M1].
$|\vec{PF}| = \frac{1}{9}\sqrt{16 + 289 + 361} = \frac{1}{9}\sqrt{666} = \frac{3\sqrt{74}}{9} = \frac{\sqrt{74}}{3} \approx 2.87$ units [A1].

Exercise 10 · Angle Bisector Vector

Let $\mathbf{u}$ and $\mathbf{v}$ be non-zero vectors. Prove that the vector $\mathbf{w} = |\mathbf{v}|\mathbf{u} + |\mathbf{u}|\mathbf{v}$ bisects the angle between $\mathbf{u}$ and $\mathbf{v}$.

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Notice that $\mathbf{w} = |\mathbf{u}||\mathbf{v}| \left( \frac{\mathbf{u}}{|\mathbf{u}|} + \frac{\mathbf{v}}{|\mathbf{v}|} \right) = |\mathbf{u}||\mathbf{v}| (\hat{\mathbf{u}} + \hat{\mathbf{v}})$ [M1].
Consider the angle $\theta_1$ between $\mathbf{w}$ and $\mathbf{u}$, and $\theta_2$ between $\mathbf{w}$ and $\mathbf{v}$ [M1]:
$\mathbf{w}\cdot\mathbf{u} = |\mathbf{v}|(\mathbf{u}\cdot\mathbf{u}) + |\mathbf{u}|(\mathbf{v}\cdot\mathbf{u}) = |\mathbf{v}||\mathbf{u}|^2 + |\mathbf{u}|(\mathbf{u}\cdot\mathbf{v}) = |\mathbf{u}| [ |\mathbf{u}||\mathbf{v}| + \mathbf{u}\cdot\mathbf{v} ]$ [A1].
$\cos\theta_1 = \frac{\mathbf{w}\cdot\mathbf{u}}{|\mathbf{w}||\mathbf{u}|} = \frac{|\mathbf{u}||\mathbf{v}| + \mathbf{u}\cdot\mathbf{v}}{|\mathbf{w}|}$ [A1].
Similarly, $\mathbf{w}\cdot\mathbf{v} = |\mathbf{v}|(\mathbf{u}\cdot\mathbf{v}) + |\mathbf{u}||\mathbf{v}|^2 = |\mathbf{v}| [ \mathbf{u}\cdot\mathbf{v} + |\mathbf{u}||\mathbf{v}| ]$.
$\cos\theta_2 = \frac{\mathbf{w}\cdot\mathbf{v}}{|\mathbf{w}||\mathbf{v}|} = \frac{|\mathbf{u}||\mathbf{v}| + \mathbf{u}\cdot\mathbf{v}}{|\mathbf{w}|}$ [A1].
Since $\cos\theta_1 = \cos\theta_2$ and both angles lie in $[0, \pi]$, $\theta_1 = \theta_2$ [R1].
Therefore, $\mathbf{w}$ exactly bisects the angle between $\mathbf{u}$ and $\mathbf{v}$ [AG].

IB Formula Booklet References (Topic 3)

Syllabus 3.11: Vectors in 3D & Scalar Product

The official IB DP Mathematics AA Formula Booklet provides:

Magnitude: $|\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}$
Scalar product: $\mathbf{v}\cdot\mathbf{w} = v_1 w_1 + v_2 w_2 + v_3 w_3$
Angle between two vectors: $\cos\theta = \frac{\mathbf{v}\cdot\mathbf{w}}{|\mathbf{v}||\mathbf{w}|}$
Perpendicular vectors: $\mathbf{v}\cdot\mathbf{w} = 0$

Key Formulas & Identities

1. Vector Projections & Orthogonal Components

Scalar projection of $\mathbf{u}$ on $\mathbf{v}$: $\text{scal}_{\mathbf{v}}(\mathbf{u}) = \frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{v}|}$
Vector projection of $\mathbf{u}$ on $\mathbf{v}$: $\text{proj}_{\mathbf{v}}(\mathbf{u}) = \left(\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{v}|^2}\right)\mathbf{v}$
Orthogonal component: $\mathbf{u}_\perp = \mathbf{u} - \text{proj}_{\mathbf{v}}(\mathbf{u})$

2. Direction Cosines Invariant

$\cos\alpha = \frac{v_1}{|\mathbf{v}|}, \quad \cos\beta = \frac{v_2}{|\mathbf{v}|}, \quad \cos\gamma = \frac{v_3}{|\mathbf{v}|}$
Fundamental identity: $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$
Unit vector in terms of direction cosines: $\hat{\mathbf{v}} = \cos\alpha\,\mathbf{i} + \cos\beta\,\mathbf{j} + \cos\gamma\,\mathbf{k}$

Common Exam Pitfalls

Pitfall 1: Scalar vs Vector Projection Formula

Dividing by the wrong magnitude is the single most common projection error! To project $\mathbf{u}$ onto $\mathbf{v}$, you MUST divide by the magnitude of the base vector $\mathbf{v}$ ($|\mathbf{v}|^2$), NEVER by $|\mathbf{u}|^2$.

Pitfall 2: Finding Angle Between Directed Segments

When finding $\angle BAC$, you must use vectors radiating FROM $A$: $\vec{AB}$ and $\vec{AC}$. If you mistakenly use $\vec{BA}$ and $\vec{AC}$, the vectors point in opposite directions and you will compute the supplementary exterior angle ($180^\circ - \theta$).

Pitfall 3: Scalar Product Output is a Scalar

The scalar product of two vectors is a SINGLE NUMBER, not a vector! Never write $\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix} = \begin{pmatrix} 4 \\ 10 \\ 18 \end{pmatrix}$. The answer is $4 + 10 + 18 = 32$.

Figure 3.3: 3D Vectors and Scalar Projection

1. Vectors in Three Dimensions ($\mathbb{R}^3$)

2. The Scalar (Dot) Product

Core Properties of the Scalar Product

3. Vector Projections & Orthogonal Decomposition (Syllabus AHL 3.11)

Worked Examples

Worked Example 1: Angle Between Two 3D Vectors (Paper 1 Style · 6 marks)

Worked Example 2: Orthogonality with Quadratic Parameter (Paper 1 Style · 6 marks)

Worked Example 3: Vector Projection & Orthogonal Component (Paper 1 Style · 6 marks)

Worked Example 4: Geometric Proof of Rhombus Diagonals (Paper 1 Style · 5 marks)

Worked Example 5: Triangle Geometry & Altitude via Vectors (Paper 2 Style · 8 marks)

Practice Exercises (10 Tiered Questions)

IB Formula Booklet References (Topic 3)

Syllabus 3.11: Vectors in 3D & Scalar Product

Key Formulas & Identities

1. Vector Projections & Orthogonal Components

2. Direction Cosines Invariant

Common Exam Pitfalls

Pitfall 1: Scalar vs Vector Projection Formula

Pitfall 2: Finding Angle Between Directed Segments

Pitfall 3: Scalar Product Output is a Scalar

Topic Diagnostic Test

Section A: Diagnostic Multiple Choice Questions

Section B: Examination-Style Free Response Questions