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MCV4U Geometric Vectors
Grade 12 Unit 5 — practice questions with full solutions

Magnitude and direction, resultants, the cosine law for vector sums, and resolving vectors into components — vectors before coordinates.

Free MCV4U geometric vectors practice for Grade 12 students in Ontario. These 5 questions cover the same material as a typical unit test or exam review on this topic, and every one comes with a complete worked solution — not just an answer key. Useful whether you are preparing for a MCV4U unit test, catching up on a lesson, or reviewing before the final exam.
MCV4U · Grade 12 Geometry and Algebra of Vectors 4 core 1 challenge
1Core
Two forces act on an object at right angles: 40 N east and 30 N north. Find the magnitude and direction of the resultant force.
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  1. Because the forces are perpendicular, the resultant is the hypotenuse of a right triangle.
  2. Magnitude = √(402 + 302) = √(1600 + 900) = √2500 = 50 N.
  3. Direction: the angle north of east satisfies tan θ = 30 ÷ 40 = 0.75.
  4. θ = arctan(0.75) ≈ 36.9°.
Final answer50 N at about 36.9° north of east
2Core
Vectors a and b have magnitudes |a| = 5 and |b| = 8, with an angle of 60° between them. Find |a + b|, to two decimal places.
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  1. Placing the vectors tip to tail, the triangle containing a + b has an interior angle of 180° − 60° = 120°.
  2. Using the cosine law in the form |a + b|2 = |a|2 + |b|2 + 2|a||b|cos 60°.
  3. = 25 + 64 + 2(5)(8)(0.5) = 89 + 40 = 129.
  4. |a + b| = √129 ≈ 11.36.
Final answer√129 ≈ 11.36
3Core
An aircraft flies due north with an airspeed of 300 km/h while a 60 km/h wind blows due east. Find the aircraft's ground speed and its direction, to one decimal place.
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  1. The two velocities are perpendicular, so add them as components: 60 east and 300 north.
  2. Ground speed = √(3002 + 602) = √(90000 + 3600) = √93600.
  3. √93600 ≈ 305.9 km/h.
  4. Direction: tan θ = 60 ÷ 300 = 0.2, so θ = arctan(0.2) ≈ 11.3° east of north.
Final answer≈ 305.9 km/h, about 11.3° east of north
4Core
A force of 100 N is applied at 30° above the horizontal. Resolve it into horizontal and vertical components, to one decimal place.
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  1. The horizontal component is |F|cos θ.
  2. 100 cos 30° = 100(0.8660) ≈ 86.6 N.
  3. The vertical component is |F|sin θ.
  4. 100 sin 30° = 100(0.5) = 50 N.
Final answerHorizontal ≈ 86.6 N, vertical = 50 N
5Challenge
Vectors a and b satisfy |a| = 6, |b| = 10, and the angle between them is 120°. Find |a − b|.
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  1. Use |a − b|2 = |a|2 + |b|2 − 2|a||b|cos θ.
  2. cos 120° = −0.5.
  3. = 36 + 100 − 2(6)(10)(−0.5) = 136 + 60 = 196.
  4. |a − b| = √196 = 14.
Final answer14

MCV4U Geometric Vectors — common questions

Short answers to the things students ask most about this unit.

How do I add two vectors that are not at right angles?

Place them tip to tail and use the cosine law on the resulting triangle. Only when the vectors are perpendicular can you use the Pythagorean theorem directly.

What does resolving a vector into components mean?

It means splitting one vector into perpendicular parts, usually horizontal and vertical, using magnitude times cosine for the horizontal and magnitude times sine for the vertical.

More MCV4U practice — grade 12 calculus and vectors

MCV4U Rates of Change & LimitsMCV4U DerivativesMCV4U Applications of Derivatives & Curve SketchingMCV4U Derivatives of Exponential & Trigonometric FunctionsAll MCV4U units
← Unit 4: Derivatives of Exponential & Trigonometric FunctionsUnit 6: Cartesian Vectors, Dot & Cross Products →
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