Vectors, Arrows and Angles – Getting Directional in Maths​

Vectors, Arrows and Angles – Getting Directional in Maths

Today’s maths lessons were all about vectors—those handy arrows that tell us not just how far, but which way. Whether we were introducing GCSE students to vector notation or helping A-Level students break down 3D vector problems, the theme was clear: direction matters.

We explored how to:

• Represent vectors using arrows and coordinates

• Add and subtract vectors geometrically and algebraically

• Find magnitudes and directions

• Use vectors in geometric proofs and navigation problems

Finding Magnitudes and Directions in A-Level Maths
Hemel Private Tuition – A-Level Focus

Today’s A-Level Maths topic: Vectors – Magnitude and Direction
This is where geometry meets algebra, and we learn to turn coordinates into meaningful measurements.

🔹 Magnitude – How Long Is That Vector?

The magnitude of a vector tells us its length — essentially the distance between the start and end points.

For a 2D vector a = (x, y), the magnitude is:

$$|\mathbf{a}| = \sqrt{x^2 + y^2}$$In 3D, for a = (x, y, z):$$|\mathbf{a}| = \sqrt{x^2 + y^2 + z^2}$$

It’s just an application of Pythagoras – but in multiple dimensions!

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🔹 Direction – Where’s It Pointing?

To find the direction (angle θ) of a vector in 2D, we use trigonometry:

If a = (x, y), then

$$\theta = \tan^{-1} \left(\frac{y}{x}\right)$$

Make sure to consider the quadrant the vector lies in — the inverse tangent only gives angles from –90° to +90°, so adjust accordingly for vectors in the second or third quadrant.

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🧠 Why It Matters

Whether it’s physics, navigation, or mechanics, vectors give us control over motion and force. Knowing both how far (magnitude) and where (direction) a vector is pointing is essential in solving real-world problems — and plenty of exam ones too.

Today, our students practised:

• Converting between coordinate form and magnitude/direction

• Resolving vectors into components

• Applying vector direction to projectile motion and forces

For our A-Level students, we even took a deep dive into scalar products and solving vector equations—perfect preparation for mechanics modules.

There’s something satisfying about seeing a messy problem turn into a clean arrow pointing exactly where it should. And with plenty of diagrams, animations, and real-world examples (including sailing and drone paths!), it all started to make sense.

A projectile is fired upwards at 60 degrees to the horizontal at 45 m/s. Using vectors, resolve the velocity into its vertical and horizontal vectors, and then determine the maximum height it can achieve.

Given:

• Initial speed $u = 45\, \text{m/s}$

• Angle $\theta = 60^\circ$

• Acceleration due to gravity $g = 9.8\, \text{m/s}^2$

We’re going to:

1. Resolve the initial velocity into horizontal and vertical components

2. Use kinematic equations to calculate the maximum height reached

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Step 1: Resolve the velocity into components

Using vector resolution:

• Horizontal velocity:

  $$u_x = u \cos \theta = 45 \cos 60^\circ = 45 \times 0.5 = 22.5\, \text{m/s}$$

• Vertical velocity:

  $$u_y = u \sin \theta = 45 \sin 60^\circ = 45 \times 0.866 = 38.97\, \text{m/s}$$

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Step 2: Calculate Maximum Height

At maximum height, the vertical velocity becomes 0.

Use the kinematic equation:

$$v^2 = u^2 + 2as$$

Let’s solve for $s = h$, the maximum height, with:

• Final vertical velocity $v = 0$

• Initial vertical velocity $u = 38.97\, \text{m/s}$

• Acceleration $a = -9.8\, \text{m/s}^2$ (negative because it acts downward)

$$0 = (38.97)^2 + 2(-9.8)h$$ $$0 = 1517.2 - 19.6h$$ $$19.6h = 1517.2$$ $$h = \frac{1517.2}{19.6} \approx 77.4\, \text{m}$$

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✅ Final Answer:

• Horizontal velocity: $22.5\, \text{m/s}$

• Vertical velocity: $38.97\, \text{m/s}$

• Maximum height: $\boxed{77.4\, \text{m}}$

If your child is struggling with direction (literally or mathematically), we’re here to help.

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