Vectors in Two Dimensions

Lesson goal: build accurate physics fluency for vectors in two dimensions and use that fluency to support clear HSC-style scientific writing.

This page is materialised into the MentorMind course shell from existing teaching, textbook, and eduKG material. Use it as the main lesson surface; use the tutor for targeted repair, worked examples, and concise writing feedback.

• How is the motion of an object moving in a straight line described and predicted?

From The Feynman Lectures on Physics, Vol I, Chapter 11:

Vector addition is not a trick; it is a statement about how nature combines directions. Treating velocities like scalars loses the geometry of motion.

• Represent vectors using components and diagrams.
• Resolve vectors into perpendicular components.
• Add and subtract vectors in two dimensions.
• Apply vector methods to velocity and displacement.

Vector components

A vector $\vec{R}$ at angle $\theta$ from the horizontal has components:

$$R_x = R\cos\theta, \quad R_y = R\sin\theta$$

The angle is always measured from the positive x-axis (east direction) unless otherwise specified.

Interactive: Vector Resolution

The diagram below shows a vector being resolved into its x and y components:

Vector addition

Graphical methods (head-to-tail) and component methods both yield the resultant. The component method is preferred for calculations.

Interactive: Vector Addition (Tail-to-Head)

Add two vectors using the tail-to-head method. The resultant (dashed) connects the start to the end.

Method using components:

1. Resolve each vector into x and y components
2. Add all x components: $R_x = A_x + B_x$
3. Add all y components: $R_y = A_y + B_y$
4. Find magnitude: $R = \sqrt{R_x^2 + R_y^2}$
5. Find direction: $\theta = \tan^{-1}\left(\frac{R_y}{R_x}\right)$

Resultant magnitude and direction

$$R = \sqrt{R_x^2 + R_y^2}, \quad \theta = \tan^{-1}\left(\frac{R_y}{R_x}\right)$$

Watch the quadrant! If $R_x < 0$, the angle from $\tan^{-1}$ needs adjustment (add 180 degrees).

Interactive: Three-Vector Addition

When adding more than two vectors, the component method becomes essential:

Example 1: Resolve a vector

A displacement of 50 m is directed 30 degrees north of east.

Solution:

1. $R_x = 50\cos(30 degrees) = 50 \times 0.866 = 43.3$ m (east)
2. $R_y = 50\sin(30 degrees) = 50 \times 0.5 = 25.0$ m (north)
3. Components describe the east and north parts of the motion.

Example 2: Add two vectors

A walker goes 3.0 m east, then 4.0 m north.

Solution:

1. Components: $R_x = 3.0$ m, $R_y = 4.0$ m
2. Resultant magnitude: $R = \sqrt{3.0^2 + 4.0^2} = \sqrt{25} = 5.0$ m
3. Direction: $\theta = \tan^{-1}(4.0/3.0) = 53 degrees$ north of east

Example 3: Subtract vectors (relative motion)

A boat's velocity is 6.0 m/s east. The current is 2.0 m/s north. Find the boat's velocity relative to the water.

Solution:

1. $\vec{v}{bw} = \vec{v}{be} - \vec{v}_{we}$
2. Components: $v_x = 6.0$ m/s, $v_y = -2.0$ m/s (subtract current)
3. Speed: $\sqrt{6.0^2 + 2.0^2} = 6.3$ m/s
4. Direction: $\tan^{-1}(2.0/6.0) = 18 degrees$ south of east

Example 4: Find components from magnitude and direction

A force of 25 N acts at 60 degrees above the horizontal.

Solution:

1. $F_x = 25\cos(60 degrees) = 12.5$ N (horizontal)
2. $F_y = 25\sin(60 degrees) = 21.7$ N (vertical)

• Misconception: Components are always equal to the magnitude. Correction: Components depend on direction; only at 45 degrees are they equal.

• Misconception: Vector addition is the same as adding magnitudes. Correction: Directions change the result; 3 m east + 4 m north ≠ 7 m.

• Misconception: A negative component means the vector is negative. Correction: It only indicates direction along the axis.

• Misconception: $\tan^{-1}$ always gives the correct angle. Correction: You must check the quadrant based on component signs.

Easy (2 marks)

Resolve a 10 m displacement at 60 degrees above the horizontal into components.

• Correct cosine and sine components (2)

Answer:

• $x = 10\cos(60 degrees) = 5.0$ m
• $y = 10\sin(60 degrees) = 8.7$ m

Medium (4 marks)

Two forces act on a point: 8 N east and 6 N north. Find the resultant magnitude and direction.

• Correct components and resultant magnitude (2)
• Correct direction (2)

Answer:

• $R = \sqrt{8^2 + 6^2} = \sqrt{100} = 10$ N
• $\theta = \tan^{-1}(6/8) = 37 degrees$ north of east

Hard (5 marks)

An aircraft travels 200 km/h north relative to the air. A wind of 60 km/h blows east. Find the ground velocity and its direction.

• Correct vector model (1)
• Components and magnitude (2)
• Correct direction statement (2)

Solution:

The ground velocity is the sum of air velocity and wind velocity:

• $v_x = 0 + 60 = 60$ km/h (east)
• $v_y = 200 + 0 = 200$ km/h (north)

Ground speed:

• $v = \sqrt{60^2 + 200^2} = \sqrt{43600} = 209$ km/h

Direction:

• $\theta = \tan^{-1}(60/200) = 17 degrees$ east of north

Answer: 209 km/h at 17 degrees east of north

Test your understanding with these interactive questions:

• Components encode vector direction in x and y axes.
• Vector addition can be solved with components and Pythagoras.
• Direction is found using inverse tangent with correct quadrant.
• Vector methods underpin two-dimensional motion.

Check your understanding:

After studying this section, you should be able to:

• Resolve a vector into perpendicular components
• Add two or more vectors using components
• Calculate resultant magnitude using Pythagoras
• Determine direction using inverse tangent
• Check which quadrant the resultant lies in

When answering questions from this lesson, separate:

• the physical quantity being discussed,
• the model or law being applied,
• the mathematical relationship, including units,
• the conclusion in words.

For explanation questions, write in the pattern: claim -> physics reason -> consequence. For calculation questions, state the formula, substitute with units, calculate, then interpret the answer.