Grade by Grade
C. Geometry and Algebra of Vectors · Vectors in Two-Space and Three-Space

Converting Between Cartesian Form and Magnitude-Direction Form

C1.3 C1.2About 70 minutes
Learning goals
  • I can use x = r cos θ and y = r sin θ to write a vector given by magnitude and direction in Cartesian form.
  • I can find the magnitude and direction of a vector given in Cartesian form, in any quadrant.
  • I can work with directions given as standard angles, true bearings and quadrant bearings.
  • I can avoid the tan⁻¹ quadrant trap by sketching and checking signs.

Words to know

reference angle
The acute angle α between a vector and the x-axis.
magnitude-direction form
A vector given by its length and a direction (standard angle, bearing or quadrant bearing); polar form (r, θ) is one example.
resolve
To split a vector into its x- and y-components.
tan⁻¹ (inverse tangent)
The calculator function that gives an angle from a tangent ratio. It only returns angles from −90° to 90°.

Two languages, one vector

A navigation app knows your velocity as "95 km/h on a bearing of 060°" because that's what a driver understands. But to add that velocity to a wind or current, or to animate a moving dot on a screen, the computer needs the components [x, y]. Translating back and forth between magnitude-direction (polar) form and Cartesian form is one of the most-used skills in this unit, and it is pure trigonometry from Grade 11 and MHF4U.

From magnitude and direction to components

Put the vector's tail at the origin. If its magnitude is r and its standard angle is θ, the head is at the point (x, y) where

x = r cos θ and y = r sin θ

In quadrant I this is just SOH CAH TOA in the right triangle formed by the vector and its components. In MCR3U (the unit Trigonometric Ratios, Identities and Laws) you defined cos θ = x/r and sin θ = y/r for every angle, so the same formulas work in every quadrant: the signs of cos θ and sin θ automatically give the signs of the components.

xyθrx = r cos θy = r sin θr = 8, θ = 30°x = 8 cos 30° = 4√3 ≈ 6.93y = 8 sin 30° = 4[4√3, 4]
The vector and its components form a right triangle. The adjacent side is x = r cos θ and the opposite side is y = r sin θ.
Worked example

Example 1: The curriculum's sample problem

Represent the vector with a magnitude of 8 and a direction of 30° anticlockwise from the positive x-axis in Cartesian form.

  1. x = 8 cos 30° = 8(√3/2) = 4√3.
  2. y = 8 sin 30° = 8(1/2) = 4.
  3. Check: √((4√3)² + 4²) = √(48 + 16) = √64 = 8 ✓

[4√3, 4], about [6.93, 4].

Worked example

Example 2: An angle in quadrant III

Write the polar vector (15, 250°) in Cartesian form, to two decimal places.

  1. x = 15 cos 250° ≈ 15(−0.3420) ≈ −5.13.
  2. y = 15 sin 250° ≈ 15(−0.9397) ≈ −14.10.
  3. Both components are negative, which matches quadrant III (250° is between 180° and 270°). ✓

[−5.13, −14.10]

From components to magnitude and direction

Given [x, y]:

  1. Magnitude: r = √(x² + y²).
  2. Reference angle α (the acute angle between the vector and the x-axis): α = tan⁻¹(|y| ÷ |x|).
  3. Standard angle θ: look at the signs of x and y to find the quadrant, then use the chart below.

Calculator warning. The tan⁻¹ key only returns angles from −90° to 90°. If you type tan⁻¹(y ÷ x) directly, you must add 180° when x is negative, and add 360° if the answer is still negative. This is the rule the curriculum gives as θ = tan⁻¹(y/x) or tan⁻¹(y/x) + 180°. Drawing a quick sketch first is the safest check.

quadrant Iθ = ααquadrant IIθ = 180° − ααquadrant IIIθ = 180° + ααquadrant IVθ = 360° − αα
The reference angle α is always measured to the x-axis. The quadrant decides how α turns into the standard angle θ.
Worked example

Example 3: Drawing the vector (8, 6)

Represent the vector [8, 6] as a directed line segment: give its magnitude, its standard angle and its true bearing.

  1. r = √(8² + 6²) = √100 = 10.
  2. Both components are positive, so the vector is in quadrant I and θ = α = tan⁻¹(6/8) ≈ 36.9°.
  3. True bearing: β = 90° − 36.9° = 53.1°, which is N 53.1° E.
  4. To draw it: choose a scale (say 1 cm per unit), draw a 10 cm arrow at 36.9° above the positive x-axis.

A directed segment of length 10 at θ ≈ 36.9° (bearing 053.1°, N 53.1° E).

Worked example

Example 4: A vector in quadrant II

Write [−5, 12] in magnitude-direction form, giving the standard angle and the true bearing.

  1. r = √(25 + 144) = √169 = 13.
  2. x < 0 and y > 0, so quadrant II. α = tan⁻¹(12/5) ≈ 67.4°.
  3. θ = 180° − 67.4° = 112.6°. (Calculator check: tan⁻¹(12 ÷ −5) ≈ −67.4°, and adding 180° gives 112.6°. ✓)
  4. Bearing: β = 90° − 112.6° = −22.6°, plus 360° gives 337.4°. That is 22.6° west of north.

13 at θ ≈ 112.6°, which is a true bearing of 337.4° (N 22.6° W).

x (E)y (N)(−5, 12)12−5θ ≈ 112.6°22.6°13magnitude 13standard angle 112.6°true bearing 337.4°N 22.6° W
[−5, 12] has magnitude 13. It is 112.6° anticlockwise from east, which is the same as 22.6° west of north.

Working straight from a bearing

If a vector has magnitude r and true bearing β, you can convert β to a standard angle first (θ = 90° − β). Or use the bearing directly. Because a bearing is measured from north (the y-axis), the roles of sine and cosine swap:

x = r sin β and y = r cos β

Both methods give the same answer, since cos(90° − β) = sin β and sin(90° − β) = cos β (the cofunction identities from MHF4U's Trigonometric Identities and Equations unit).

Worked example

Example 5: A velocity on a bearing

A snowmobile travels at 12 m/s on a bearing of 210°. Write its velocity in Cartesian form (x east, y north).

  1. Method 1: θ = 90° − 210° = −120°, plus 360° gives 240°. x = 12 cos 240° = −6 and y = 12 sin 240° = −6√3.
  2. Method 2: x = 12 sin 210° = 12(−1/2) = −6 and y = 12 cos 210° = 12(−√3/2) = −6√3.
  3. Both components negative: south-west, which matches S 30° W. ✓

[−6, −6√3] m/s, about [−6, −10.39] m/s.

Just for fun

Your calculator's tan⁻¹ key only knows about half of the plane (from −90° to 90°). The other half is your job. Think of it as the calculator politely asking you to draw a sketch.

Hands-on activity

Measure the components

You need: a ruler, a protractor, graph paper (or plain paper), a pencil and a calculator.

  1. Draw x- and y-axes. Using 1 cm per unit, draw a 10 cm vector at 35° above the positive x-axis.
  2. From the head, draw a line straight down to the x-axis. Measure the x-component and the y-component with your ruler.
  3. Calculate 10 cos 35° and 10 sin 35° and compare.
  4. Repeat with a 10 cm vector at 160° and one at 250°. Record the signs of the components as well as the sizes.
  5. Finally, choose a point such as (−6, 4) on the paper, draw the vector from the origin, and measure its length and angle with the ruler and protractor. Then calculate r and θ.

What to look for: at 35°, about 8.2 cm and 5.7 cm (10 cos 35° ≈ 8.19, 10 sin 35° ≈ 5.74). At 160°, x ≈ −9.4 and y ≈ 3.4. At 250°, x ≈ −3.4 and y ≈ −9.4. For (−6, 4): r = √52 ≈ 7.2 cm and θ = 180° − tan⁻¹(4/6) ≈ 146.3°.

Your turn

1

Write (10, 60°) in Cartesian form, exactly.

2

Write (20, 150°) in Cartesian form, exactly.

3

Write [6, 8] in polar form.

4

Write [−4, −4] in polar form, exactly.

5

Write [9, −12] in polar form.

6

A wind blows at 50 km/h on a true bearing of 300°. Write its velocity in Cartesian form (x east, y north).

7

Your calculator says tan⁻¹(−3 ÷ −4) = 36.87°. What is the standard angle of [−4, −3]? Why is the calculator's answer wrong?

More practice: warm-up

1

Write (4, 0°) in Cartesian form.

2

Write (2, 90°) in Cartesian form.

3

Find the magnitude of [5, 12].

4

In which quadrant does [−3, 5] point?

5

Write (6, 30°) in Cartesian form, exactly.

More practice: on your own

1

A plane flies at 600 km/h on a heading of N 20° W. Write its velocity in Cartesian form.

2

Write [−2, 7] in polar form.

3

A force is [120, −50] N. Find its magnitude and its direction as a true bearing and a quadrant bearing.

4

Write (5, −30°) with a standard angle between 0° and 360°, then in Cartesian form.

5

Write 40 N at a true bearing of 135° in Cartesian form.

Challenge

1

A vector has magnitude 25 and x-component −7. Find every possible standard angle.

2

A vector of magnitude 10 has an x-component exactly twice its y-component. Find both possible vectors exactly, and their standard angles.

3

Prove that (r, θ) and (r, θ + 180°) are opposite vectors.

Check your understanding

Quiz

1.

What is the Cartesian form of a vector with magnitude 8 at 30° anticlockwise from the positive x-axis?

2.

What is the polar form of [−6, 6]?

3.

A calculator gives tan⁻¹(−3 ÷ −3) = 45°. What is the standard angle of [−3, −3]?

4.

What is the x-component of a 40 N force at a standard angle of 120°?

5.

A vector has magnitude r and true bearing β (x east, y north). What is its x-component?

6.

What is the standard angle of [5, −12], to the nearest tenth of a degree?

For parents and teachers

Big idea (C1.3): a 2-D vector can be written as components [x, y] or as a magnitude and direction, and trigonometry converts between them: x = r cos θ, y = r sin θ, r = √(x² + y²), with the quadrant chosen from the signs.

How to help: Before your teen presses any calculator keys, ask: "Which quadrant is it in? Should x be positive or negative?" That one question prevents most errors. Make sure the calculator is in degree mode (DEG on the screen).

Common difficulties:

  • Trusting tan⁻¹(y ÷ x) without adjusting for quadrants II and III.
  • Using cos for the y-component when the angle is a bearing (from north, x = r sin β).
  • Rounding the angle early, then getting components that are slightly off.
  • Calculator in radian mode (cos 30 in radians is about 0.154, not 0.866).