A drone moves in a straight line such that its position vector at time $t$ seconds is given by $\mathbf{r} = \begin{pmatrix} 5 \\ -2 \\ 10 \end{pmatrix} + t \begin{pmatrix} 3 \\ 4 \\ -1 \end{pmatrix}$ metres. Calculate the exact speed of the drone.

1. Identify the velocity vector from the kinematic equation $\mathbf{r} = \mathbf{r}_0 + t\mathbf{v}$, which is $\mathbf{v} = \begin{pmatrix} 3 \\ 4 \\ -1 \end{pmatrix}$. 2. Recall that speed is a scalar quantity representing the magnitude of the velocity vector: $|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}$. 3. Substitute the vector components into the magnitude formula: $\sqrt{3^2 + 4^2 + (-1)^2}$. 4. Calculate the exact value: $\sqrt{9 + 16 + 1} = \sqrt{26}$. 5. The exact speed of the drone is $\sqrt{26}\text{ m s}^{-1}$ .

Object A's position is given by $\mathbf{r}_A = \begin{pmatrix} 2 \\ 5 \end{pmatrix} + t \begin{pmatrix} 4 \\ -1 \end{pmatrix}$ and Object B's position is $\mathbf{r}_B = \begin{pmatrix} 10 \\ 0 \end{pmatrix} + t \begin{pmatrix} 2 \\ 1 \end{pmatrix}$, where $t \ge 0$ is the time in seconds. Determine if the objects collide, and explain your reasoning.

1. Set up the condition for a collision, which requires both position vectors to be equal at the same time: $\mathbf{r}_A = \mathbf{r}_B$. 2. Equate the $x$-components of the two objects: $2 + 4t = 10 + 2t$. 3. Solve this linear equation for $t$: $2t = 8 \implies t = 4\text{ s}$. 4. Substitute $t = 4$ into the $y$-component for Object A: $5 - 1(4) = 1$. 5. Substitute $t = 4$ into the $y$-component for Object B: $0 + 1(4) = 4$. 6. Conclude that since the $y$-coordinates ($1 \neq 4$) do not match at $t=4$, the objects cross paths at different times, meaning they do not collide .

Two ships, A and B, move such that their position vectors at time $t$ hours are $\mathbf{r}_A = \begin{pmatrix} 0 \\ 20 \end{pmatrix} + t \begin{pmatrix} 15 \\ 5 \end{pmatrix}$ and $\mathbf{r}_B = \begin{pmatrix} 10 \\ 5 \end{pmatrix} + t \begin{pmatrix} 5 \\ 15 \end{pmatrix}$ in km. Find the exact time $t$ when the ships are closest together.

1. Find the relative position vector representing the distance between them: $\vec{AB} = \mathbf{r}_B - \mathbf{r}_A$. 2. Subtract the components: $\vec{AB} = \begin{pmatrix} 10 - 0 \\ 5 - 20 \end{pmatrix} + t \begin{pmatrix} 5 - 15 \\ 15 - 5 \end{pmatrix} = \begin{pmatrix} 10 \\ -15 \end{pmatrix} + t \begin{pmatrix} -10 \\ 10 \end{pmatrix}$. 3. Set up the squared distance function $D^2 = x(t)^2 + y(t)^2$: $D^2 = (10 - 10t)^2 + (-15 + 10t)^2$. 4. Expand and simplify the expression: $D^2 = (100 - 200t + 100t^2) + (225 - 300t + 100t^2) = 200t^2 - 500t + 325$. 5. Minimize the quadratic $D^2$ using the vertex formula $t = \frac{-b}{2a}$: $t = \frac{500}{2(200)}$. 6. Evaluate the fraction: $t = \frac{500}{400} = 1.25$. The ships are closest together at exactly $1.25$ hours .

Two submarines, Alpha and Beta, are tracked on a naval radar. Their positions at time $t$ hours are: \[r_A = \begin{pmatrix} -4 \\ 10 \end{pmatrix} + t \begin{pmatrix} 6 \\ -2 \end{pmatrix} \quad \text{and} \quad r_B = \begin{pmatrix} 14 \\ 2 \end{pmatrix} + t \begin{pmatrix} -3 \\ 4 \end{pmatrix}\] Show algebraically that the paths of the submarines cross, but they do not collide.

For paths to cross, set the $x$ and $y$ lines equal using DIFFERENT parameters ( $\lambda$ and $\mu$ ). [M1] $x: -4 + 6\lambda = 14 - 3\mu \implies 6\lambda + 3\mu = 18 \implies 2\lambda + \mu = 6$ . [M1] $y: 10 - 2\lambda = 2 + 4\mu \implies -2\lambda - 4\mu = -8 \implies \lambda + 2\mu = 4$ . [M1] Solving simultaneously gives $\lambda = 8/3$ and $\mu = 2/3$ . [A1] Because a solution exists, the geometric paths cross. However, because $\lambda \neq \mu$ ( $8/3 \neq 2/3$ ), the submarines reach the intersection point at different times, so they do not collide. [R1]

Vectors and Kinematics Questions

What you need to know

The dot product for angles, cross product for perpendicular vectors. HL AI applies vectors to force decomposition and structural engineering. 3D vectors — dot product and cross product overview →

r = a + λb for a line, and r = a + tv for an object moving with constant velocity. HL AI asks where two paths meet or how close two moving objects get. (Equations of planes are AA HL only.) Vector equations of lines in 2D and 3D overview →

What's examined in AI HL vectors and kinematics

The question bank covers these vectors and kinematics question types (number of questions in brackets):

Vector Geometry & Algebra (19)
Constant Velocity Kinematics (16)
Calculus-based Kinematics (9)

Key formulas

Vector dot product: $\vec{a} \cdot \vec{b} = | \vec{a} | | \vec{b} | \cos θ = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}$
Vector magnitude: $| \vec{a} | = \sqrt{a_{1}^{2} + a_{2}^{2} + a_{3}^{2}}$
Angle between two vectors: $\cos θ = \frac{\vec{a} \cdot \vec{b}}{| \vec{a} | | \vec{b} |}$
Vector equation of a line: $\vec{r} = \vec{a} + t \vec{d}$

In the same notation as the IB formula booklet. All AI HL formulas →

Explained, with a worked example: Magnitude of a vector

Vectors and Kinematics worked examples

Worked example 1: Speed from a Kinematic Vector Equation · easy

A drone moves in a straight line such that its position vector at time $t$ seconds is given by $r = (\begin{matrix} 5 \\ - 2 \\ 10 \end{matrix}) + t (\begin{matrix} 3 \\ 4 \\ - 1 \end{matrix})$ metres. Calculate the exact speed of the drone.

Solution

5 steps

Examiner tip: In the parametric or vector equation of a line, the vector multiplied by the parameter

t

is always the velocity (direction) vector, not the initial position vector.

Worked example 2: Determining if Moving Objects Collide · medium

Object A's position is given by $r_{A} = (\begin{matrix} 2 \\ 5 \end{matrix}) + t (\begin{matrix} 4 \\ - 1 \end{matrix})$ and Object B's position is $r_{B} = (\begin{matrix} 10 \\ 0 \end{matrix}) + t (\begin{matrix} 2 \\ 1 \end{matrix})$ , where $t \geq 0$ is the time in seconds. Determine if the objects collide, and explain your reasoning.

Solution

6 steps · 1 to fill in

Examiner tip: For a true collision, both objects must share the exact same coordinates at the exact same time

t

. If the times differ when solving

x

and

y

simultaneously, their paths cross, but no collision occurs.

Worked example 3: Minimizing Distance Between Objects · hard

Two ships, A and B, move such that their position vectors at time $t$ hours are $r_{A} = (\begin{matrix} 0 \\ 20 \end{matrix}) + t (\begin{matrix} 15 \\ 5 \end{matrix})$ and $r_{B} = (\begin{matrix} 10 \\ 5 \end{matrix}) + t (\begin{matrix} 5 \\ 15 \end{matrix})$ in km. Find the exact time $t$ when the ships are closest together.

Solution

6 steps

Examiner tip: Minimizing the square of the distance (

D^{2}

) is mathematically equivalent to minimizing the distance (

D

) itself, and avoids dealing with complex square root derivatives or solver errors.

Try these IB Maths AI HL vectors and kinematics questions

Three questions from the bank, easiest first. Mark schemes and AI marking of your written working are in the practice area. One of them has its full worked solution here, open to everyone.

Question 1 · easy · 4 marks · Paper 1

Consider the vector $v = (\begin{matrix} 2 \\ - 4 \\ 4 \end{matrix})$ .
(a) Calculate the exact magnitude of $v$ . [2 marks]
(b) Find the unit vector in the direction of $v$ . [2 marks]

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Question 2 · medium · 5 marks · Paper 1

Two submarines, Alpha and Beta, are tracked on a naval radar. Their positions at time $t$ hours are: $r_{A} = (\begin{matrix} - 4 \\ 10 \end{matrix}) + t (\begin{matrix} 6 \\ - 2 \end{matrix}) and r_{B} = (\begin{matrix} 14 \\ 2 \end{matrix}) + t (\begin{matrix} - 3 \\ 4 \end{matrix})$ Show algebraically that the paths of the submarines cross, but they do not collide.

Full worked solution Free, no sign-in

For paths to cross, set the $x$ and $y$ lines equal using DIFFERENT parameters ( $λ$ and $μ$ ). [M1]
$x : - 4 + 6 λ = 14 - 3 μ ⟹ 6 λ + 3 μ = 18 ⟹ 2 λ + μ = 6$ . [M1]
$y : 10 - 2 λ = 2 + 4 μ ⟹ - 2 λ - 4 μ = - 8 ⟹ λ + 2 μ = 4$ . [M1]
Solving simultaneously gives $λ = 8 / 3$ and $μ = 2 / 3$ . [A1]
Because a solution exists, the geometric paths cross. However, because $λ \neq μ$ ( $8 / 3 \neq 2 / 3$ ), the submarines reach the intersection point at different times, so they do not collide. [R1]

Every question here has a full worked solution. See them all →

Question 3 · hard · 7 marks · Paper 1

An asteroid is moving through space with position vector $r_{A} = (\begin{matrix} 50 \\ 20 \\ - 10 \end{matrix}) + t (\begin{matrix} - 8 \\ 4 \\ 2 \end{matrix})$ . A space station is located at $S (10, 30, - 5)$ .
Let $D (t)$ be the distance between the asteroid and the space station at time $t$ .
(a) Find an algebraic expression for $D (t)^{2}$ in terms of $t$ . [3 marks]
(b) By graphing $D (t)^{2}$ on your GDC or using calculus, find the exact time $t$ when the asteroid is closest to the space station, and state this minimum distance. [4 marks]

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All 44 vectors and kinematics questions with mark schemes →

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FAQ

How many IB Maths AI HL vectors and kinematics questions are there?

There are 44 exam-style vectors and kinematics questions in the AI HL question bank (Paper 1: 37 · Paper 2: 7), graded 13 easy, 16 medium, 14 hard, 1 very hard. Every question has a full IB-style mark scheme (M, A and R marks).

Is vectors and kinematics on Paper 1 or Paper 2?

Both. In the bank, Paper 1: 37 · Paper 2: 7. Practise with your GDC — AI papers expect calculator methods throughout.

Where can I get the mark schemes?

Open the AI HL Unit 3 practice page: every question has a step-by-step IB-style mark scheme, and you can photograph your working for instant AI marking. The worked examples on this page are free.

IB Maths AI HL Vectors and Kinematics Questions

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What you need to know

What's examined in AI HL vectors and kinematics

Key formulas

Vectors and Kinematics worked examples

Worked example 1: Speed from a Kinematic Vector Equation · easy

Solution

Worked example 2: Determining if Moving Objects Collide · medium

Solution

Worked example 3: Minimizing Distance Between Objects · hard

Solution

Try these IB Maths AI HL vectors and kinematics questions

Question 1 · easy · 4 marks · Paper 1

Question 2 · medium · 5 marks · Paper 1

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Question 3 · hard · 7 marks · Paper 1

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