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Introduction to Vectors: Components, Magnitude and Direction — Answers

Answer sheet with worked solutions for the Year 11 trigonometry worksheet.

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Success TutoringWorksheetsAnswer sheetYear 11 · Mathematics · Trigonometry

Introduction to Vectors: Components, Magnitude and Direction — Answers

Part A: Components, magnitude and direction

A vector has both size (magnitude) and direction; a scalar has size only. In two dimensions we write a vector with the unit vectors i (1 unit in the x-direction) and j (1 unit in the y-direction): a = xi + yj.

Magnitude: |a| = √(x2 + y2). Direction: the angle θ from the positive x-axis, found from tan θ = and the quadrant the vector points into. Going the other way, a vector of magnitude r at angle θ is r cos θ i + r sin θ j.

Worked example. a = 3i − 4j. |a| = √(9 + 16) = 5. It points right and down (fourth quadrant); the reference angle is tan−1() = 53.1°, so θ = 360° − 53.1° = 306.9°.

1.Find the magnitude of u = 6i + 8j.

Answer: 10

  1. |u| = √(62 + 82) = √(36 + 64) = √100 = 10
2.Find the exact magnitude of v = −2i + 5j.

Answer: √29

  1. |v| = √((−2)2 + 52) = √(4 + 25) = √29
3.Find the direction of w = −3i + 3j, as an angle measured from the positive x-axis (0° ≤ θ < 360°).

Answer: 135°

  1. w points left and up, so it is in the second quadrant.
  2. Reference angle: tan−1() = 45°
  3. θ = 180° − 45° = 135°
4.Which vector has magnitude 13?
  1. A5i + 12j
  2. B7i + 6j
  3. C10i + 3j
  4. D13i + 13j

Answer: 5i + 12j

√(52 + 122) = √169 = 13. The others give √85, √109 and 13√2.

5.A vector has magnitude 10 and makes an angle of 30° with the positive x-axis (pointing into the first quadrant). Write it in component form, exactly.

Answer: 5√3i + 5j

  1. x-component: 10 cos 30° = 10 × = 5√3
  2. y-component: 10 sin 30° = 10 × = 5
  3. Vector = 5√3i + 5j (about 8.66i + 5j)

Part B: Operations on vectors

Add or subtract vectors component by component, and multiply by a scalar k by multiplying both components: k(xi + yj) = kxi + kyj. Two vectors are parallel when one is a scalar multiple of the other.

The vector from point A to point B is b − a (position of B minus position of A). A unit vector has magnitude 1: divide a vector by its magnitude to get the unit vector in its direction.

Worked example. a = 2i + j and b = −i + 4j. Then a + b = i + 5j, and 3a − b = (6 + 1)i + (3 − 4)j = 7i − j.

6.Let a = 4i − 2j and b = i + 5j. Find a + b.

Answer: 5i + 3j

  1. Add the components: (4 + 1)i + (−2 + 5)j = 5i + 3j
7.Using the same vectors, find 2a − 3b.

Answer: 5i − 19j

  1. 2a = 8i − 4j and 3b = 3i + 15j
  2. 2a − 3b = (8 − 3)i + (−4 − 15)j = 5i − 19j
8.For the points A(1, 2) and B(7, −6), the vector from A to B is (answer space) and its magnitude is (answer space).

Answer: 6i − 8j, 10

The vector from A to B is (position of B) − (position of A) = (7 − 1)i + (−6 − 2)j = 6i − 8j. Its magnitude is √(36 + 64) = 10.

9.Find the unit vector in the direction of 3i + 4j, writing the components as decimals.

Answer: 0.6i + 0.8j

  1. Magnitude: √(9 + 16) = 5
  2. Unit vector = (3i + 4j) = i + j = 0.6i + 0.8j
  3. Check: √(0.36 + 0.64) = 1
10.The vector 2i + kj is parallel to 6i − 9j when k = (answer space).

Answer: −3

6i − 9j = 3(2i − 3j), so the parallel vector with x-component 2 is 2i − 3j, giving k = −3.

11.For any two vectors, |a + b| = |a| + |b|.

Answer: False

Counterexample: a = 3i and b = 4j give |a + b| = |3i + 4j| = 5, but |a| + |b| = 7. Equality holds only when a and b point the same way.

Part C: The scalar (dot) product

For u = ai + bj and v = ci + dj, the scalar product (dot product) is u · v = ac + bd: multiply matching components and add. The result is a number, not a vector.

It also equals |u||v| cos θ, where θ is the angle between the vectors, so cos θ = . Two non-zero vectors are perpendicular exactly when their scalar product is 0.

Worked example. (i + 2j) · (4i − 2j) = 1 × 4 + 2 × (−2) = 0, so the two vectors are perpendicular.

12.Find (3i − 2j) · (5i + 4j).

Answer: 7

  1. Multiply matching components and add: 3 × 5 + (−2) × 4 = 15 − 8 = 7
13.Find the angle between a = 2i + j and b = i + 3j.

Answer: 45°

  1. a · b = 2 × 1 + 1 × 3 = 5
  2. |a| = √5 and |b| = √10
  3. cos θ = = = , so θ = 45°
14.The vectors 4i + mj and 3i − 6j are perpendicular when m = (answer space).

Answer: 2

Perpendicular vectors have a scalar product of 0: 4 × 3 + m × (−6) = 12 − 6m = 0, so m = 2.

Part D: Resultant velocities and forces

The combined effect of several velocities or forces is their resultant: add the vectors, then find the magnitude and direction of the sum. Set up axes first (for example, east as i and north as j).

Worked example. A plane flies east at 300 km/h in a wind blowing north at 40 km/h. Resultant = 300i + 40j, with speed √(90 000 + 1600) = 302.7 km/h, at tan−1() = 7.6° north of east.

15.A boat points due north at 12 km/h through still water while a current flows due east at 5 km/h. Its resultant speed is (answer space) km/h, on a course (answer space)° east of north (to 1 decimal place).

Answer: 13, 22.6

Velocity = 5i + 12j (east is i, north is j). Speed = √(25 + 144) = 13 km/h. The angle from north is tan−1() = 22.6°.

16.Two forces act on an object: F₁ = 30i + 10j newtons and F₂ = −10i + 40j newtons. The resultant force has magnitude (answer space) N (2 d.p.) and direction (answer space)° from the positive x-axis (1 d.p.).

Answer: 53.85, 68.2

Resultant = 20i + 50j. Magnitude = √(400 + 2500) = √2900 = 53.85 N. Direction = tan−1() = 68.2° (first quadrant).

Part E: Extended problem

A drone takes off from a point O. With i pointing east and j pointing north (distances in m), it first flies the displacement p = 120i + 50j and then the displacement q = −40i + 90j.

(a) Find the drone's final position vector r relative to O.

(b) How far is the drone from O, correct to 1 decimal place?

(c) Find the direction of r as an angle from the positive x-axis, correct to 1 decimal place.

(d) Find the angle between the two flight paths p and q, correct to the nearest degree.

(e) Write the vector the drone must fly to return directly to O. If it flies at 8 m/s, how long does the return take, correct to 1 decimal place?

17.Answer parts (a) to (e), showing full working.[10 marks]

Model answer: (a) r = p + q = (120 − 40)i + (50 + 90)j = 80i + 140j. (b) |r| = √(6400 + 19 600) = √26 000 ≈ 161.2 m. (c) tan−1() ≈ 60.3° from the positive x-axis (first quadrant). (d) p · q = 120 × (−40) + 50 × 90 = −4800 + 4500 = −300; |p| = √(14 400 + 2500) = 130 and |q| = √(1600 + 8100) = √9700 ≈ 98.49; cos θ = ≈ −0.0234, so θ ≈ 91°. (e) The return vector is −r = −80i − 140j, with length 161.245… m; time = 161.245… ÷ 8 ≈ 20.2 s.

Success criteria:

  • (a) adds components to get 80i + 140j (1 mark)
  • (b) magnitude ≈ 161.2 m (2 marks)
  • (c) direction ≈ 60.3° (2 marks)
  • (d) scalar product −300, both magnitudes and the angle ≈ 91° (3 marks)
  • (e) return vector −80i − 140j and time ≈ 20.2 s (2 marks)
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