add and subtract coplanar vectors

Learning Objectives

• Distinguish between scalar and vector quantities.
• Express a vector as two perpendicular components using unit‑vector notation.
• Add and subtract coplanar vectors:

  • Graphically – tip‑to‑tail and parallelogram methods.
  • Analytically – component (algebraic) method.

• Convert between magnitude‑direction form and component form, and interpret the resulting angle correctly.

Scalars and Vectors

Reminder – Unit Vectors

\(\hat{\mathbf i}\) and \(\hat{\mathbf j}\) are unit vectors along the chosen x‑ and y‑axes:

• \(|\hat{\mathbf i}| = |\hat{\mathbf j}| = 1\)
• \(\hat{\mathbf i}\perp\hat{\mathbf j}\) (they are orthogonal)

Representing Vectors

Graphical Representation

An arrow whose length is proportional to the magnitude and whose orientation shows the direction.

Algebraic (Component) Representation

For a vector \(\mathbf{A}\) in the plane:

\[

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

\]

where \(Ax\) and \(Ay\) are the components along the x‑ and y-axes respectively.

Sign Conventions (2‑D Plane)

• East (right) = + \(x\)
• West (left) = – \(x\)
• North (up) = + \(y\)
• South (down) = – \(y\)

Conversion Between Forms

If a vector has magnitude \(A\) and makes an angle \(\theta\) measured anticlockwise from the positive x-axis:

\[

Ax = A\cos\theta,\qquad Ay = A\sin\theta

\]

Conversely, given components \(Ax\) and \(Ay\):

\[

A = \sqrt{Ax^{2}+Ay^{2}},\qquad

\theta = \tan^{-1}\!\left(\frac{Ay}{Ax}\right)

\]

Use the signs of \(Ax\) and \(Ay\) (or a calculator’s “atan2” function) to place \(\theta\) in the correct quadrant.

Adding Coplanar Vectors

1. Graphical Methods

Method	Steps (quick‑check)	Illustration
Tip‑to‑Tail (Polygon)	Draw the first vector \(\mathbf{A}\) with its tail at the origin. Place the tail of \(\mathbf{B}\) at the tip of \(\mathbf{A}\) (keep length and direction unchanged). Draw the resultant \(\mathbf{R}\) from the origin to the tip of \(\mathbf{B}\).
Parallelogram	Draw \(\mathbf{A}\) and \(\mathbf{B}\) with a common tail at the origin. Complete the parallelogram by drawing lines parallel to each vector through the tip of the other. The diagonal from the origin gives the resultant \(\mathbf{R}\).

2. Component (Algebraic) Method

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

\[

\boxed{\mathbf{R}= \mathbf{A}+\mathbf{B}= (Ax+Bx)\hat{\mathbf i}+ (Ay+By)\hat{\mathbf j}}

\]

Find the magnitude and direction of \(\mathbf{R}\) with the conversion formulas given above.

Subtracting Coplanar Vectors

Graphical Approach

Subtracting \(\mathbf{B}\) from \(\mathbf{A}\) is equivalent to adding the negative of \(\mathbf{B}\) (same length, opposite direction).

Step	Description	Illustration
1. Reverse \(\mathbf{B}\)	Draw a vector \(-\mathbf{B}\) with the same magnitude as \(\mathbf{B}\) but pointing opposite to \(\mathbf{B}\).
2. Add \(\mathbf{A}\) and \(-\mathbf{B}\)	Use either tip‑to‑tail or parallelogram method as for addition.

Component Approach

\[

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

\]

Again use the magnitude‑direction formulas to obtain the resultant’s size and angle.

Worked Example – Vector Addition

Problem: Two forces act in the horizontal plane.

• \(\mathbf{F}_1 = 30\;\text{N}\) at \(30^{\circ}\) north of east.
• \(\mathbf{F}_2 = 40\;\text{N}\) at \(120^{\circ}\) measured anticlockwise from east.

Find the resultant \(\mathbf{F}R = \mathbf{F}1+\mathbf{F}_2\).

1. Resolve each force into components (east = + \(x\), north = + \(y\)):

   \[

   \begin{aligned}

   F_{1x}&=30\cos30^{\circ}=30(0.866)=25.98\;\text{N}\\

   F_{1y}&=30\sin30^{\circ}=30(0.5)=15.00\;\text{N}\\[4pt]

   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}

   \]

2. Add the components:

   \[

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

   \]

   \[

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

   \]

3. Resultant magnitude:

   \[

   FR=\sqrt{F{Rx}^{2}+F_{Ry}^{2}}=\sqrt{(5.98)^{2}+(49.64)^{2}}=50.0\;\text{N}

   \]

4. Resultant direction (measured anticlockwise from east):

   \[

   \thetaR=\tan^{-1}\!\left(\frac{F{Ry}}{F_{Rx}}\right)=\tan^{-1}\!\left(\frac{49.64}{5.98}\right)=83.1^{\circ}

   \]

   Both components are positive → first quadrant → “\(83.1^{\circ}\) north of east”.

Worked Example – Vector Subtraction

Problem: A displacement \(\mathbf{d}1 = 12\;\text{m}\) at \(40^{\circ}\) east of north is followed by a displacement \(\mathbf{d}2 = 5\;\text{m}\) due south. Find the net displacement \(\mathbf{d} = \mathbf{d}1-\mathbf{d}2\).

1. Choose north = + \(y\), east = + \(x\):

   \[

   \begin{aligned}

   d_{1x}&=12\sin40^{\circ}=12(0.643)=7.72\;\text{m}\\

   d_{1y}&=12\cos40^{\circ}=12(0.766)=9.19\;\text{m}\\

   d_{2x}&=0\\

   d_{2y}&=-5\;\text{m}\quad(\text{south is –\(y\)})

   \end{aligned}

   \]

2. Subtract components:

   \[

   dx = d{1x}-d_{2x}=7.72-0=7.72\;\text{m}

   \]

   \[

   dy = d{1y}-d_{2y}=9.19-(-5)=14.19\;\text{m}

   \]

3. Resultant magnitude:

   \[

   d=\sqrt{dx^{2}+dy^{2}}=\sqrt{7.72^{2}+14.19^{2}}=16.2\;\text{m}

   \]

4. Resultant direction** (east of north):

   \[

   \theta=\tan^{-1}\!\left(\frac{dx}{dy}\right)=\tan^{-1}\!\left(\frac{7.72}{14.19}\right)=28.6^{\circ}

   \]

   Net displacement: \(16.2\;\text{m}\) at \(28.6^{\circ}\) east of north.

Summary Table

Key Points to Remember

• Scalars have magnitude only; vectors have magnitude + direction.
• State your coordinate axes and sign convention before resolving any vector.
• \(\hat{\mathbf i}\) and \(\hat{\mathbf j}\) are unit vectors of length 1 and are perpendicular.
• When converting components to an angle, always check the signs of the components to place the angle in the correct quadrant.
• Graphical methods help visualise the result; the component method is faster for calculations and is required in exam questions.
• Both addition and subtraction obey the same algebraic rules: add the components, then recombine into magnitude‑direction form.