represent a vector as two perpendicular components

Scalars and Vectors (Cambridge A‑Level Physics 9702 – Sub‑topic 1.4)

Learning Objectives

By the end of this lesson you will be able to:

• Define scalars and vectors and give appropriate physics examples.
• Represent any coplanar vector uniquely as two perpendicular components.
• Add and subtract coplanar vectors using both the graphical (tip‑to‑tail) method and the component method.
• Estimate the magnitude and direction of a vector from a sketch (AO2 skill).

Key Definitions

Properties of Vectors

1. Direction is shown by an arrow or by a unit‑vector notation.
2. Magnitude is always a non‑negative number.
3. Vectors can be added, subtracted, and multiplied by scalars.

Unit‑Vector Notation in the Cartesian Plane

The standard unit vectors are:

• \(\hat{\mathbf i}\) – along the positive \(x\)‑axis.
• \(\hat{\mathbf j}\) – along the positive \(y\)‑axis.

Any coplanar vector \(\vec{A}\) can be written uniquely as

\[

\boxed{\;\vec{A}=Ax\hat{\mathbf i}+Ay\hat{\mathbf j}\;}

\]

where \(Ax\) and \(Ay\) are the perpendicular (horizontal and vertical) components.

Decomposing a Vector into Perpendicular Components

For a vector \(\vec{A}\) of magnitude \(A\) making an angle \(\theta\) with the positive \(x\)-axis:

\[

\begin{aligned}

A_x &= A\cos\theta ,\\[2mm]

A_y &= A\sin\theta .

\end{aligned}

\]

These formulas follow directly from right‑triangle trigonometry. The representation is unique for a given \(\theta\) (i.e. a single pair \((Ax,Ay)\) corresponds to one vector).

Recovering magnitude and direction from components

\[

\begin{aligned}

A &= \sqrt{Ax^{2}+Ay^{2}},\\[2mm]

\theta &= \operatorname{atan2}(Ay,\,Ax)\;,

\end{aligned}

\]

where atan2 automatically selects the correct quadrant for \(\theta\). If a calculator does not have atan2, use the signs of \(Ax\) and \(Ay\) to adjust the result of \(\tan^{-1}(Ay/Ax)\).

Worked Example – Component Decomposition

A force \(\vec{F}\) has magnitude \(50\;\text{N}\) and acts \(30^{\circ}\) above the positive \(x\)-axis.

\[

\begin{aligned}

F_x &= 50\cos30^{\circ}=50\times0.866 = 43.3\;\text{N},\\[2mm]

F_y &= 50\sin30^{\circ}=50\times0.500 = 25.0\;\text{N}.

\end{aligned}

\]

\[

\boxed{\;\vec{F}= 43.3\hat{\mathbf i}+25.0\hat{\mathbf j}\;\text{N}\;}

\]

Vector Addition and Subtraction

1. Graphical (Tip‑to‑Tail) Method

1. Place the tail of the second vector at the head of the first.
2. Continue this process for any further vectors.
3. The resultant \(\vec{R}\) is drawn from the tail of the first vector to the head of the last.
4. Measure the length of \(\vec{R}\) and read its angle using the diagram’s scale.

When to use: Quick estimates, AO2 sketch questions, or when only a visual answer is required.

2. Component (Algebraic) Method

For \(\vec{A}=Ax\hat{\mathbf i}+Ay\hat{\mathbf j}\) and \(\vec{B}=Bx\hat{\mathbf i}+By\hat{\mathbf j}\):

\[

\begin{aligned}

\vec{A}+\vec{B} &= (Ax+Bx)\hat{\mathbf i}+(Ay+By)\hat{\mathbf j},\\[2mm]

\vec{A}-\vec{B} &= (Ax-Bx)\hat{\mathbf i}+(Ay-By)\hat{\mathbf j}.

\end{aligned}

\]

After obtaining the resultant components, convert back to magnitude and direction using the formulas in the previous section.

Summary Table – Graphical vs. Component Methods

Worked Example – Adding Two Forces (Component Method)

Force \(\vec{F}_1\): 30 N at \(30^{\circ}\) above the \(x\)-axis.

Force \(\vec{F}_2\): 40 N at \(120^{\circ}\) anticlockwise from the \(x\)-axis.

Step 1 – Resolve into components

\[

\begin{aligned}

F_{1x} &= 30\cos30^{\circ}=25.98\;\text{N}, &

F_{1y} &= 30\sin30^{\circ}=15.00\;\text{N},\\[2mm]

F_{2x} &= 40\cos120^{\circ}=40(-0.5)=-20.0\;\text{N}, &

F_{2y} &= 40\sin120^{\circ}=40(0.866)=34.64\;\text{N}.

\end{aligned}

\]

Step 2 – Add components

\[

\begin{aligned}

Rx &= F{1x}+F_{2x}=25.98-20.0=5.98\;\text{N},\\

Ry &= F{1y}+F_{2y}=15.00+34.64=49.64\;\text{N}.

\end{aligned}

\]

Step 3 – Resultant magnitude & direction

\[

\begin{aligned}

R &= \sqrt{Rx^{2}+Ry^{2}}=\sqrt{5.98^{2}+49.64^{2}}=50.0\;\text{N},\\[2mm]

\theta &= \operatorname{atan2}(49.64,\,5.98)=83.1^{\circ}\;\text{above the }x\text{-axis}.

\end{aligned}

\]

\[

\boxed{\;\vec{R}=5.98\hat{\mathbf i}+49.64\hat{\mathbf j}\;\text{N}\;\approx\;50\;\text{N at }83^{\circ}\;}

\]

Estimating Vector Size & Direction from a Sketch (AO2)

1. Identify the scale (e.g. 1 cm = 5 N).
2. Measure the vector length with a ruler.
3. Multiply by the scale factor → approximate magnitude.
4. Read the angle with a protractor; round to a sensible number of significant figures.

This technique is often sufficient for multiple‑choice or short‑answer questions where an exact calculation is unnecessary.

Common Vector Quantities in Physics

Summary Checklist

• Scalars = magnitude only; vectors = magnitude + direction.
• Any coplanar vector can be expressed uniquely as \(\vec{A}=Ax\hat{\mathbf i}+Ay\hat{\mathbf j}\).
• Component formulas: \(Ax=A\cos\theta\), \(Ay=A\sin\theta\).
  

  Reverse: \(A=\sqrt{Ax^{2}+Ay^{2}}\), \(\theta=\operatorname{atan2}(Ay,Ax)\).

• Vector addition/subtraction – choose the method that best fits the question (graphical for quick estimates, component for precise results).
• AO2 sketch skill: use scale, ruler, and protractor to obtain a reasonable magnitude and direction.