From Column Form to Magnitude and Bearing (and Back)

1. Magnitude of a column vector

The column vector $\left(\begin{array}{c}x\\ y\end{array}\right)$ describes a journey of x units East/West and y units North/South. These two movements are at right angles, so with the vector itself they form a right-angled triangle. The vector is the longest side (hypotenuse). Pythagoras' theorem therefore gives the magnitude:

|v| = √(x² + y²)

The sign of x or y does not matter, because squares are always positive. For $\left(\begin{array}{c}3\\ 4\end{array}\right)$, |v| = √(9 + 16) = √25 = 5. For $\left(\begin{array}{c}-5\\ 12\end{array}\right)$, |v| = √(25 + 144) = √169 = 13.

▶ Follow the triangle: components first, then the hypotenuse, then the bearing angle.

2. Finding the bearing

We use the tangent ratio: tan α = opposite ÷ adjacent. Here α is the acute angle between the vector and the North–South line (not the East–West line). Its opposite side is the East–West component |x| and its adjacent side is the North–South component |y|.

α = tan⁻¹( |x| ÷ |y| )

Then we adjust α to a full bearing depending on the compass area of the vector. Always sketch a small diagram first to see which area it is in.

▶ A vector into the South-West: find α from the South line, then add 180°.

Special cases: if y = 0 the vector lies along the East–West line, so the bearing is 090° (x > 0) or 270° (x < 0). If x = 0 it lies along the North–South line, so the bearing is 000° (y > 0) or 180° (y < 0). For $\left(\begin{array}{c}0\\ -7\end{array}\right)$ the magnitude is 7 and the bearing is 180°.

3. A complete routine

1. 1Sketch the vector from its signs: which compass area?
2. 2Magnitude: |v| = √(x² + y²).
3. 3Find α = tan⁻¹(|x| ÷ |y|), using positive values.
4. 4Use the area to find the bearing: NE α; SE 180° − α; SW 180° + α; NW 360° − α.
5. 5Write the result as (magnitude, bearing) with a three-figure bearing, and check against the sketch.

Let's try $\left(\begin{array}{c}-5\\ 12\end{array}\right)$. It points to the North-West (x negative, y positive). |v| = 13. α = tan⁻¹(5 ÷ 12) = 22.6°. North-West: bearing = 360° − 22.6° = 337.4°. The answer (13, 337.4°) is in the North-West area (270° to 360°) ✓.

Another: $\left(\begin{array}{c}6\\ -8\end{array}\right)$. It points South-East. |v| = 10, α = tan⁻¹(6 ÷ 8) = 36.9°, bearing = 180° − 36.9° = 143.1°. Check: 143.1° is between 090° and 180° ✓.

4. Going back: from (magnitude, bearing) to column form

The reverse conversion uses the formulae from the last lesson: x = d sin θ and y = d cos θ, where d is the magnitude and θ the bearing. For (20 km, 310°): x = 20 sin 310° = −15.32 and y = 20 cos 310° = 12.86, so the column vector is $\left(\begin{array}{c}-15.32\\ 12.86\end{array}\right)$. The signs say 15.32 km West and 12.86 km North, which is the North-West, as 310° should be.

To test both conversions together, convert your answer back. For (13, 337.4°) we get x = 13 sin 337.4° = −5.00 and y = 13 cos 337.4° = 12.00, which is the original $\left(\begin{array}{c}-5\\ 12\end{array}\right)$ ✓.

Worked examples

Try each one yourself first, then open the solution and compare every step.