IB · MATH AI HL

Mathematics: Applications & Interpretation HL

Geometry and trigonometry: vectors — Topic 3 HL

Name: ____________________Date: October 2, 2026

1.
A line has vector equation r = (1, 2, 3) + t(2, −1, 1). Find the coordinates of the point on the line when t = 2.
[2 marks]
Marking points
- Substitutes t = 2 into each component: x = 1 + 2(2), y = 2 − 2, z = 3 + 2.
- Obtains the point (5, 0, 5).
Examiner tip: A vector equation of a line packages a known point and a direction vector together — substituting any value of the parameter t generates a specific point on the line.
2.
Marking analysis: A learner attempts the following task: “A line has vector equation r = (1, 2, 3) + t(2, −1, 1). Find the coordinates of the point on the line when t = 2.” Their response addresses only this point: “Substitutes t = 2 into each component: x = 1 + 2(2), y = 2 − 2, z = 3 + 2.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[2 marks]
Marking points
- Recognises credit for the stated point: Substitutes t = 2 into each component: x = 1 + 2(2), y = 2 − 2, z = 3 + 2.
- Identifies the missing requirement: Obtains the point (5, 0, 5).
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
3.
Find the angle between the vectors a = (3, 4, 0) and b = (1, 2, 2), using the dot product.
[4 marks]
Marking points
- Calculates a·b = (3)(1) + (4)(2) + (0)(2) = 11.
- Calculates |a| = √(9+16+0) = 5 and |b| = √(1+4+4) = 3.
- Uses cos(θ) = (a·b)/(|a||b|) = 11/15.
- Obtains θ ≈ 42.8°.
Examiner tip: The dot product formula for the angle between two vectors works in any number of dimensions — the same method applies whether the vectors are 2D or 3D.
4.
Marking analysis: A learner attempts the following task: “Find the angle between the vectors a = (3, 4, 0) and b = (1, 2, 2), using the dot product.” Their response addresses only this point: “Calculates a·b = (3)(1) + (4)(2) + (0)(2) = 11.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Calculates a·b = (3)(1) + (4)(2) + (0)(2) = 11.
- Identifies the missing requirement: Calculates |a| = √(9+16+0) = 5 and |b| = √(1+4+4) = 3.
- Identifies the missing requirement: Uses cos(θ) = (a·b)/(|a||b|) = 11/15.
- Identifies the missing requirement: Obtains θ ≈ 42.8°.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
5.
Find the cross product a × b for a = (2, 0, 0) and b = (1, 3, 0), and hence find the area of the parallelogram formed by these two vectors.
[4 marks]
Marking points
- Uses the cross product formula a × b = (a2b3−a3b2, a3b1−a1b3, a1b2−a2b1).
- Obtains a × b = (0, 0, 6).
- States that the area of the parallelogram equals |a × b|.
- Obtains an area of 6 square units.
Examiner tip: The magnitude of the cross product of two vectors always gives the area of the parallelogram they span — halving it gives the area of the triangle formed by the same two vectors.
6.
Marking analysis: A learner attempts the following task: “Find the cross product a × b for a = (2, 0, 0) and b = (1, 3, 0), and hence find the area of the parallelogram formed by these two vectors.” Their response addresses only this point: “Uses the cross product formula a × b = (a2b3−a3b2, a3b1−a1b3, a1b2−a2b1).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Uses the cross product formula a × b = (a2b3−a3b2, a3b1−a1b3, a1b2−a2b1).
- Identifies the missing requirement: Obtains a × b = (0, 0, 6).
- Identifies the missing requirement: States that the area of the parallelogram equals |a × b|.
- Identifies the missing requirement: Obtains an area of 6 square units.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
7.
Triangle ABC has vertices A(0, 0, 0), B(4, 0, 0), C(0, 3, 0). Use the cross product of vectors AB and AC to find the area of the triangle.
[4 marks]
Marking points
- Finds AB = (4, 0, 0) and AC = (0, 3, 0).
- Calculates AB × AC = (0, 0, 12).
- Uses the triangle area formula Area = (1/2)|AB × AC|.
- Obtains an area of 6 square units.
Examiner tip: Even for a triangle lying flat in the z = 0 plane, using the full 3D cross product method works correctly and generalises immediately to triangles that are not flat.
8.
Marking analysis: A learner attempts the following task: “Triangle ABC has vertices A(0, 0, 0), B(4, 0, 0), C(0, 3, 0). Use the cross product of vectors AB and AC to find the area of the triangle.” Their response addresses only this point: “Finds AB = (4, 0, 0) and AC = (0, 3, 0).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Finds AB = (4, 0, 0) and AC = (0, 3, 0).
- Identifies the missing requirement: Calculates AB × AC = (0, 0, 12).
- Identifies the missing requirement: Uses the triangle area formula Area = (1/2)|AB × AC|.
- Identifies the missing requirement: Obtains an area of 6 square units.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
9.
Find the scalar projection of vector a = (3, 4) onto vector b = (1, 0).
[3 marks]
Marking points
- Uses the scalar projection formula: (a·b)/|b|.
- Calculates a·b = (3)(1) + (4)(0) = 3, and |b| = 1.
- Obtains a scalar projection of 3.
Examiner tip: The scalar projection of a onto b measures how far a extends in the direction of b — it equals the length of the shadow a would cast on the line through b.
10.
Marking analysis: A learner attempts the following task: “Find the scalar projection of vector a = (3, 4) onto vector b = (1, 0).” Their response addresses only this point: “Uses the scalar projection formula: (a·b)/|b|.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Marking points
- Recognises credit for the stated point: Uses the scalar projection formula: (a·b)/|b|.
- Identifies the missing requirement: Calculates a·b = (3)(1) + (4)(0) = 3, and |b| = 1.
- Identifies the missing requirement: Obtains a scalar projection of 3.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
11.
Find the distance between the points A(1, 2, 3) and B(4, 6, 3) in three-dimensional space.
[2 marks]
Marking points
- Uses the 3D distance formula √[(x2−x1)² + (y2−y1)² + (z2−z1)²].
- Substitutes to obtain √(3² + 4² + 0²) = √25 = 5.
Examiner tip: The 3D distance formula is simply Pythagoras' theorem applied one dimension further — when the z-coordinates are equal, it reduces exactly to the familiar 2D formula.
12.
Marking analysis: A learner attempts the following task: “Find the distance between the points A(1, 2, 3) and B(4, 6, 3) in three-dimensional space.” Their response addresses only this point: “Uses the 3D distance formula √[(x2−x1)² + (y2−y1)² + (z2−z1)²].” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[2 marks]
Marking points
- Recognises credit for the stated point: Uses the 3D distance formula √[(x2−x1)² + (y2−y1)² + (z2−z1)²].
- Identifies the missing requirement: Substitutes to obtain √(3² + 4² + 0²) = √25 = 5.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
13.
Line 1 has vector equation r = (1, 1, 0) + t(1, 2, 0). Line 2 has vector equation r = (0, 3, 0) + s(1, −1, 0). Determine whether the two lines intersect, and if so, find the point of intersection.
[5 marks]
Marking points
- Sets the x-components equal: 1 + t = s.
- Sets the y-components equal: 1 + 2t = 3 − s.
- Substitutes s = 1 + t into the second equation and solves to obtain t = 1/3.
- Finds s = 4/3, and confirms both parameters give the same point (since the z-components are both 0, no contradiction arises).
- States the point of intersection as (4/3, 5/3, 0).
Examiner tip: In three dimensions, two lines might not intersect even if their equations look similar — always verify that all three coordinate equations are satisfied by the same parameter values before concluding the lines meet.
14.
Marking analysis: A learner attempts the following task: “Line 1 has vector equation r = (1, 1, 0) + t(1, 2, 0). Line 2 has vector equation r = (0, 3, 0) + s(1, −1, 0). Determine whether the two lines intersect, and if so, find the point of intersection.” Their response addresses only this point: “Sets the x-components equal: 1 + t = s.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks]
Marking points
- Recognises credit for the stated point: Sets the x-components equal: 1 + t = s.
- Identifies the missing requirement: Sets the y-components equal: 1 + 2t = 3 − s.
- Identifies the missing requirement: Substitutes s = 1 + t into the second equation and solves to obtain t = 1/3.
- Identifies the missing requirement: Finds s = 4/3, and confirms both parameters give the same point (since the z-components are both 0, no contradiction arises).
- Identifies the missing requirement: States the point of intersection as (4/3, 5/3, 0).
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
15.
Two sites A(2, 3) and B(8, 7) are used to generate part of a Voronoi diagram. Find the equation of the perpendicular bisector of AB, which forms the boundary between the two sites' regions.
[5 marks]
Marking points
- Finds the midpoint of AB: ((2+8)/2, (3+7)/2) = (5, 5).
- Finds the gradient of AB: (7−3)/(8−2) = 2/3.
- Finds the perpendicular gradient as the negative reciprocal: −3/2.
- Uses the point-gradient form through (5, 5) with gradient −3/2.
- States the perpendicular bisector equation y − 5 = −1.5(x − 5).
Examiner tip: Every edge of a Voronoi diagram lies on the perpendicular bisector of the segment joining two neighbouring sites — points on this line are equally close to both sites.
16.
Marking analysis: A learner attempts the following task: “Two sites A(2, 3) and B(8, 7) are used to generate part of a Voronoi diagram. Find the equation of the perpendicular bisector of AB, which forms the boundary between the two sites' regions.” Their response addresses only this point: “Finds the midpoint of AB: ((2+8)/2, (3+7)/2) = (5, 5).” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks]
Marking points
- Recognises credit for the stated point: Finds the midpoint of AB: ((2+8)/2, (3+7)/2) = (5, 5).
- Identifies the missing requirement: Finds the gradient of AB: (7−3)/(8−2) = 2/3.
- Identifies the missing requirement: Finds the perpendicular gradient as the negative reciprocal: −3/2.
- Identifies the missing requirement: Uses the point-gradient form through (5, 5) with gradient −3/2.
- Identifies the missing requirement: States the perpendicular bisector equation y − 5 = −1.5(x − 5).
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
17.
A support cable has direction vector (1, 2, 2) and is attached to a horizontal floor, whose normal vector is (0, 0, 1). Find the angle the cable makes with the floor.
[4 marks]
Marking points
- Finds the angle between the cable's direction and the normal vector using the dot product: cos(φ) = |d·n|/(|d||n|).
- Calculates |d·n| = 2, |d| = 3, |n| = 1, giving cos(φ) = 2/3.
- Uses the relationship between the angle to the normal and the angle to the plane: angle to floor = 90° − φ, equivalently sin(angle) = 2/3.
- Obtains an angle of approximately 41.8° with the floor.
Examiner tip: The angle between a line and a plane is always measured from the plane itself, not from the plane's normal — remember to subtract the angle found with the normal from 90°, or equivalently use sine instead of cosine directly.
18.
Marking analysis: A learner attempts the following task: “A support cable has direction vector (1, 2, 2) and is attached to a horizontal floor, whose normal vector is (0, 0, 1). Find the angle the cable makes with the floor.” Their response addresses only this point: “Finds the angle between the cable's direction and the normal vector using the dot product: cos(φ) = |d·n|/(|d||n|).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Finds the angle between the cable's direction and the normal vector using the dot product: cos(φ) = |d·n|/(|d||n|).
- Identifies the missing requirement: Calculates |d·n| = 2, |d| = 3, |n| = 1, giving cos(φ) = 2/3.
- Identifies the missing requirement: Uses the relationship between the angle to the normal and the angle to the plane: angle to floor = 90° − φ, equivalently sin(angle) = 2/3.
- Identifies the missing requirement: Obtains an angle of approximately 41.8° with the floor.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
19.
Find the shortest distance from the point P(1, 1, 1) to the line through A(0, 0, 0) with direction vector d = (1, 1, 0).
[5 marks]
Marking points
- Finds the vector AP = (1, 1, 1).
- Uses the formula distance = |AP × d|/|d|.
- Calculates AP × d = (1×0−1×1, 1×1−1×0, 1×1−1×1) = (−1, 1, 0).
- Calculates |AP × d| = √2 and |d| = √2.
- Obtains a shortest distance of 1.
Examiner tip: The formula |AP × d|/|d| works because the magnitude of the cross product gives the area of the parallelogram spanned by AP and d, and dividing by the base length |d| converts that area into a perpendicular height.
20.
Marking analysis: A learner attempts the following task: “Find the shortest distance from the point P(1, 1, 1) to the line through A(0, 0, 0) with direction vector d = (1, 1, 0).” Their response addresses only this point: “Finds the vector AP = (1, 1, 1).” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks]
Marking points
- Recognises credit for the stated point: Finds the vector AP = (1, 1, 1).
- Identifies the missing requirement: Uses the formula distance = |AP × d|/|d|.
- Identifies the missing requirement: Calculates AP × d = (1×0−1×1, 1×1−1×0, 1×1−1×1) = (−1, 1, 0).
- Identifies the missing requirement: Calculates |AP × d| = √2 and |d| = √2.
- Identifies the missing requirement: Obtains a shortest distance of 1.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
21.
A cuboid has dimensions 6 cm × 8 cm × 10 cm. Find the angle between the space diagonal of the cuboid and its base.
[4 marks]
Marking points
- Finds the base diagonal: √(6² + 8²) = 10 cm.
- Finds the space diagonal: √(6² + 8² + 10²) ≈ 14.14 cm.
- Uses tan(angle) = height/base diagonal = 10/10.
- Obtains an angle of 45°.
Examiner tip: When the vertical height equals the base diagonal, as happens here, the angle with the base is always exactly 45° — a useful pattern to recognise quickly.
22.
Marking analysis: A learner attempts the following task: “A cuboid has dimensions 6 cm × 8 cm × 10 cm. Find the angle between the space diagonal of the cuboid and its base.” Their response addresses only this point: “Finds the base diagonal: √(6² + 8²) = 10 cm.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Finds the base diagonal: √(6² + 8²) = 10 cm.
- Identifies the missing requirement: Finds the space diagonal: √(6² + 8² + 10²) ≈ 14.14 cm.
- Identifies the missing requirement: Uses tan(angle) = height/base diagonal = 10/10.
- Identifies the missing requirement: Obtains an angle of 45°.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
23.
Three mobile phone towers are located at A(0, 0), B(6, 0), and C(3, 5). A phone at point P(2, 2) connects to whichever tower is closest. Calculate the distance from P to each tower, and state which tower the phone connects to.
[4 marks]
Marking points
- Calculates PA = √(2² + 2²) ≈ 2.83.
- Calculates PB = √(4² + 2²) ≈ 4.47.
- Calculates PC = √(1² + 3²) ≈ 3.16.
- States that since PA is the smallest distance, the phone connects to tower A.
Examiner tip: This 'closest site' logic is exactly what a Voronoi diagram encodes visually — each region in the diagram shows every point that is closer to one particular site than to any other.
24.
Marking analysis: A learner attempts the following task: “Three mobile phone towers are located at A(0, 0), B(6, 0), and C(3, 5). A phone at point P(2, 2) connects to whichever tower is closest. Calculate the distance from P to each tower, and state which tower the phone connects to.” Their response addresses only this point: “Calculates PA = √(2² + 2²) ≈ 2.83.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Calculates PA = √(2² + 2²) ≈ 2.83.
- Identifies the missing requirement: Calculates PB = √(4² + 2²) ≈ 4.47.
- Identifies the missing requirement: Calculates PC = √(1² + 3²) ≈ 3.16.
- Identifies the missing requirement: States that since PA is the smallest distance, the phone connects to tower A.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
25.
A line has vector equation r = (1, 2, 3) + t(2, −1, 1). Determine whether the point (7, −1, 7) lies on this line.
[4 marks]
Marking points
- Sets up three equations: 1 + 2t = 7, 2 − t = −1, 3 + t = 7.
- Solves the first equation to obtain t = 3, and the second to obtain t = 3.
- Solves the third equation to obtain t = 4, which is inconsistent with the other two.
- Concludes that since no single value of t satisfies all three equations, the point does not lie on the line.
Examiner tip: To check whether a point lies on a 3D line, all three coordinate equations must give the same value of the parameter — agreement in just one or two coordinates is not sufficient.
26.
Marking analysis: A learner attempts the following task: “A line has vector equation r = (1, 2, 3) + t(2, −1, 1). Determine whether the point (7, −1, 7) lies on this line.” Their response addresses only this point: “Sets up three equations: 1 + 2t = 7, 2 − t = −1, 3 + t = 7.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Sets up three equations: 1 + 2t = 7, 2 − t = −1, 3 + t = 7.
- Identifies the missing requirement: Solves the first equation to obtain t = 3, and the second to obtain t = 3.
- Identifies the missing requirement: Solves the third equation to obtain t = 4, which is inconsistent with the other two.
- Identifies the missing requirement: Concludes that since no single value of t satisfies all three equations, the point does not lie on the line.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
27.
Find the unit vector in the direction of v = (3, −4, 12).
[3 marks]
Marking points
- Calculates the magnitude |v| = √(3² + (−4)² + 12²) = √169 = 13.
- Divides each component by the magnitude.
- Obtains the unit vector (3/13, −4/13, 12/13).
Examiner tip: A unit vector always has magnitude 1 and points in the same direction as the original vector — dividing by the magnitude is the only step needed, no change of direction occurs.
28.
Marking analysis: A learner attempts the following task: “Find the unit vector in the direction of v = (3, −4, 12).” Their response addresses only this point: “Calculates the magnitude |v| = √(3² + (−4)² + 12²) = √169 = 13.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Marking points
- Recognises credit for the stated point: Calculates the magnitude |v| = √(3² + (−4)² + 12²) = √169 = 13.
- Identifies the missing requirement: Divides each component by the magnitude.
- Identifies the missing requirement: Obtains the unit vector (3/13, −4/13, 12/13).
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
29.
Find the angle between two lines with direction vectors d1 = (1, 1, 1) and d2 = (2, −1, 1).
[4 marks]
Marking points
- Calculates d1·d2 = (1)(2) + (1)(−1) + (1)(1) = 2.
- Calculates |d1| = √3 and |d2| = √6.
- Uses cos(θ) = 2/(√3 × √6) ≈ 0.471.
- Obtains θ ≈ 61.9°.
Examiner tip: The angle between two lines in space is found from their direction vectors alone — the specific points the lines pass through do not affect this angle at all.
30.
Marking analysis: A learner attempts the following task: “Find the angle between two lines with direction vectors d1 = (1, 1, 1) and d2 = (2, −1, 1).” Their response addresses only this point: “Calculates d1·d2 = (1)(2) + (1)(−1) + (1)(1) = 2.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: Calculates d1·d2 = (1)(2) + (1)(−1) + (1)(1) = 2.
- Identifies the missing requirement: Calculates |d1| = √3 and |d2| = √6.
- Identifies the missing requirement: Uses cos(θ) = 2/(√3 × √6) ≈ 0.471.
- Identifies the missing requirement: Obtains θ ≈ 61.9°.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
31.
A boat has a velocity of (6, 0) km/h relative to the water (heading due east), while a current has velocity (1, 2) km/h. Find the boat's resultant velocity vector relative to the ground, its speed, and its bearing.
[5 marks]
Marking points
- Adds the velocity vectors: (6, 0) + (1, 2) = (7, 2).
- States the resultant velocity vector as (7, 2) km/h.
- Calculates the speed as √(7² + 2²) ≈ 7.28 km/h.
- Calculates the bearing using the angle from north: bearing = 90° − arctan(2/7).
- Obtains a bearing of approximately 074°.
Examiner tip: Relative velocity problems always add the object's velocity relative to the medium to the medium's velocity relative to the ground — treat each velocity as a vector and add component by component.
32.
Marking analysis: A learner attempts the following task: “A boat has a velocity of (6, 0) km/h relative to the water (heading due east), while a current has velocity (1, 2) km/h. Find the boat's resultant velocity vector relative to the ground, its speed, and its bearing.” Their response addresses only this point: “Adds the velocity vectors: (6, 0) + (1, 2) = (7, 2).” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks]
Marking points
- Recognises credit for the stated point: Adds the velocity vectors: (6, 0) + (1, 2) = (7, 2).
- Identifies the missing requirement: States the resultant velocity vector as (7, 2) km/h.
- Identifies the missing requirement: Calculates the speed as √(7² + 2²) ≈ 7.28 km/h.
- Identifies the missing requirement: Calculates the bearing using the angle from north: bearing = 90° − arctan(2/7).
- Identifies the missing requirement: Obtains a bearing of approximately 074°.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.