add and subtract coplanar vectors

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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

QuantityDefinitionExample
Scalar A physical quantity described by a magnitude only (single number + unit). Mass \(m = 2\;\text{kg}\), Temperature \(T = 25^{\circ}\text{C}\)
Vector A physical quantity described by a magnitude **and** a direction. It can be written in terms of two perpendicular components. Displacement \(\mathbf{s}\), Force \(\mathbf{F}\)

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.

A simple vector drawn as an arrow on a set of x‑ and y‑axes
Typical vector drawn on a 2‑D coordinate system.

Algebraic (Component) Representation

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

\[ \mathbf{A}=A_x\hat{\mathbf i}+A_y\hat{\mathbf j} \]

where \(A_x\) and \(A_y\) 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:

\[ A_x = A\cos\theta,\qquad A_y = A\sin\theta \]

Conversely, given components \(A_x\) and \(A_y\):

\[ A = \sqrt{A_x^{2}+A_y^{2}},\qquad \theta = \tan^{-1}\!\left(\frac{A_y}{A_x}\right) \]

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

Adding Coplanar Vectors

1. Graphical Methods

MethodSteps (quick‑check)Illustration
Tip‑to‑Tail (Polygon)
  1. Draw the first vector \(\mathbf{A}\) with its tail at the origin.
  2. Place the tail of \(\mathbf{B}\) at the tip of \(\mathbf{A}\) (keep length and direction unchanged).
  3. Draw the resultant \(\mathbf{R}\) from the origin to the tip of \(\mathbf{B}\).
 Tip‑to‑tail addition: A followed by B, resultant from origin to final tip
Parallelogram
  1. Draw \(\mathbf{A}\) and \(\mathbf{B}\) with a common tail at the origin.
  2. Complete the parallelogram by drawing lines parallel to each vector through the tip of the other.
  3. The diagonal from the origin gives the resultant \(\mathbf{R}\).
 Parallelogram method: two vectors from common tail, diagonal is resultant

2. Component (Algebraic) Method

For \(\mathbf{A}=A_x\hat{\mathbf i}+A_y\hat{\mathbf j}\) and \(\mathbf{B}=B_x\hat{\mathbf i}+B_y\hat{\mathbf j}\):

\[ \boxed{\mathbf{R}= \mathbf{A}+\mathbf{B}= (A_x+B_x)\hat{\mathbf i}+ (A_y+B_y)\hat{\mathbf j}} \]

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).

StepDescriptionIllustration
1. Reverse \(\mathbf{B}\) Draw a vector \(-\mathbf{B}\) with the same magnitude as \(\mathbf{B}\) but pointing opposite to \(\mathbf{B}\). Original vector B and its reversed counterpart -B
2. Add \(\mathbf{A}\) and \(-\mathbf{B}\) Use either tip‑to‑tail or parallelogram method as for addition. Tip‑to‑tail addition of A and -B giving resultant A‑B

Component Approach

\[ \boxed{\mathbf{A}-\mathbf{B}= (A_x-B_x)\hat{\mathbf i}+ (A_y-B_y)\hat{\mathbf j}} \]

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: \[ F_R=\sqrt{F_{Rx}^{2}+F_{Ry}^{2}}=\sqrt{(5.98)^{2}+(49.64)^{2}}=50.0\;\text{N} \]
  4. Resultant direction (measured anticlockwise from east): \[ \theta_R=\tan^{-1}\!\left(\frac{F_{Ry}}{F_{Rx}}\right)=\tan^{-1}\!\left(\frac{49.64}{5.98}\right)=83.1^{\circ} \]

    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: \[ d_x = d_{1x}-d_{2x}=7.72-0=7.72\;\text{m} \] \[ d_y = d_{1y}-d_{2y}=9.19-(-5)=14.19\;\text{m} \]
  3. Resultant magnitude: \[ d=\sqrt{d_x^{2}+d_y^{2}}=\sqrt{7.72^{2}+14.19^{2}}=16.2\;\text{m} \]
  4. Resultant direction** (east of north): \[ \theta=\tan^{-1}\!\left(\frac{d_x}{d_y}\right)=\tan^{-1}\!\left(\frac{7.72}{14.19}\right)=28.6^{\circ} \]

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

Summary Table

Operation Graphical Method (steps) Component Formula Resultant Magnitude Resultant Direction
\(\mathbf{A}+\mathbf{B}\)
  1. Tip‑to‑tail or parallelogram.
  2. Draw resultant from origin (or from first tail).
 \((A_x+B_x)\hat{\mathbf i}+(A_y+B_y)\hat{\mathbf j}\) \(\sqrt{(A_x+B_x)^2+(A_y+B_y)^2}\) \(\tan^{-1}\!\bigl((A_y+B_y)/(A_x+B_x)\bigr)\) (adjust quadrant)
\(\mathbf{A}-\mathbf{B}\)
  1. Reverse \(\mathbf{B}\) to obtain \(-\mathbf{B}\).
  2. Add \(\mathbf{A}\) and \(-\mathbf{B}\) using tip‑to‑tail or parallelogram.
 \((A_x-B_x)\hat{\mathbf i}+(A_y-B_y)\hat{\mathbf j}\) \(\sqrt{(A_x-B_x)^2+(A_y-B_y)^2}\) \(\tan^{-1}\!\bigl((A_y-B_y)/(A_x-B_x)\bigr)\) (adjust quadrant)

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.

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