Vectors, Components & Trigonometry — Practice · Grade 12 Science

Chapter: Measurement & Vector Foundations

What this practice covers

By the end of Vectors, Components & Trigonometry you should be able to:

8 questions, numeric and written answer, each with a worked explanation.

Vocabulary used: scalar, vector, component.

Problem 1

Find resultant magnitude of 6N east and 8N north.
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

10 N

√(36+64)=10.

Problem 2

Two 12N opposite horizontal forces act. Find net force.
Answer and explanation

0N; opposite signed components cancel.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 3

A vector has components 5 units east and 12 units north. Find magnitude.
Hints

List the given quantities, select a relationship, and check units before substituting.

Answer and explanation

13 units

√(25+144)=13.

Problem 4

Why might atan(Ry/Rx) give a wrong direction?
Answer and explanation

The ratio alone can hide the quadrant; use component signs or atan 2.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 5

A displacement is 3 m west and 4 m north. Find magnitude. Show the reasoning, not just the conclusion.
Hints

Components sum vectorially, not as an ordinary total of lengths.

Answer and explanation

√(3²+4²)=5 m; direction is northwest.

Problem 6

Why can two equal vectors cancel?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

They cancel when their directions are opposite, giving zero vector sum despite nonzero individual magnitudes.

Problem 7

Add matching components before taking the magnitudeAdd matching components before taking the magnitude3 m east2 m northTip-to-tail route and direct displacementR = √13 m; θ = atan(2/3) ≈ 33.7°
Original exact component proportions for the 3 m east/2 m north example.
Interpret the supplied visual. Why is the resultant shorter than the 5 m route?
Answer and explanation

The 5 m is path length. The direct displacement is √(3²+2²)=√13≈3.61 m, the straight-line endpoint separation.

For vector addition, add x components to x components and y components to y components. The resultant is the single vector with those summed components. A 3 m east displacement followed by 2 m north gives (3,2) m, with magnitude √13 m and direction about 33.7° north of east. Adding magnitudes instead gives traveled distance along this route, not displacement magnitude. Use a quadrant-aware angle check when a component is negative; an inverse tangent ratio alone can select the wrong direction.

The 5 m is path length. The direct displacement is √(3²+2²)=√13≈3.61 m, the straight-line endpoint separation.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the physical or environmental relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 8

A walker moves 6 m east, then 8 m south. Find distance, displacement components, magnitude and direction; explain why 14 m is not the displacement magnitude.
Answer and explanation

Distance=14 m. Displacement=(6,−8) m with east/north positive; magnitude=10 m and direction≈53.1° south of east. The bent path is longer than its direct endpoint separation.

Distance=14 m.

Displacement=(6,−8) m with east/north positive; magnitude=10 m and direction≈53.1° south of east.

The bent path is longer than its direct endpoint separation.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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