Physics Study Notes: Vectors, Scalar Quantities, and Vector Operations

Fundamental Concepts of Vectors and Scalars

  • Vector Quantities:

    • Defined as physical quantities that possess both a magnitude (numerical value with appropriate units) and a specific spatial direction.

    • Visually modeled as directional arrows where:

    • The orientation and arrowhead specify the vector direction.

    • The length of the arrow is directly proportional to the vector magnitude.

    • Key examples of vector quantities include:

    • Position (r\mathbf{r})

    • Displacement (Δx\Delta \mathbf{x})

    • Velocity (v\mathbf{v})

    • Acceleration (a\mathbf{a})

    • Force (F\mathbf{F})

  • Scalar Quantities:

    • Defined as physical quantities that are completely specified by magnitude alone, lacking directional orientation.

    • Key examples of scalar quantities include:

    • Distance (dd)

    • Speed (vv)

    • Time (tt)

Methods of Vector Representation

  • Primary Representations of Vectors:

    • Magnitude and direction format (e.g., 5 m5\,m at 60∘60^\circ North of East).

    • Unit vector notation.

    • Graphical arrow diagrams.

  • Unit Vector Notation:

    • Represents a vector as the sum of its orthogonal constituent components in the x-, y-, and z-directions.

    • Uses dimensionless unit vectors of unit length 11 along coordinate axes:

    • i^\hat{i} denotes the unit vector along the x-direction.

    • j^\hat{j} denotes the unit vector along the y-direction.

    • k^\hat{k} denotes the unit vector along the z-direction.

    • Standard position vector formula in unit vector notation:     r=Ai^+Bj^+Ck^\mathbf{r} = A\hat{i} + B\hat{j} + C\hat{k}     where AA, BB, and CC represent the scalar component magnitudes along the x-, y-, and z-axes respectively.

  • Decomposing Vector Magnitude and Direction into Unit Vector Form:

    • To convert a vector with magnitude AA and direction angle θ\theta (relative to the positive x-axis) into unit vector notation:     Ax=A×cos⁡(θ)A_x = A \times \cos(\theta)     Ay=A×sin⁡(θ)A_y = A \times \sin(\theta)

    • Standard two-dimensional unit vector expression:     A=Axi^+Ayj^\mathbf{A} = A_x\hat{i} + A_y\hat{j}

    • Worked Example: Express vector A=5 m\mathbf{A} = 5\,m at 60∘60^\circ North of East in unit vector notation:

    • Ax=5 m×cos⁡(60∘)=+2.5 mA_x = 5\,m \times \cos(60^\circ) = +2.5\,m

    • Ay=5 m×sin⁡(60∘)=+4.33 m≈+4.3 mA_y = 5\,m \times \sin(60^\circ) = +4.33\,m \approx +4.3\,m

    • Unit vector expression: A=(2.5i^+4.3j^) m\mathbf{A} = (2.5\hat{i} + 4.3\hat{j})\,m

  • Resultant Vector Definition and Formula:

    • A resultant vector C\mathbf{C} is defined as the total vector sum of two or more constituent vectors.

    • Component-wise addition formula for C=A+B\mathbf{C} = \mathbf{A} + \mathbf{B}:     C=(Ax+Bx)i^+(Ay+By)j^\mathbf{C} = (A_x + B_x)\hat{i} + (A_y + B_y)\hat{j}

Graphical Vector Operations and the Tip-to-Tail Method

  • Tip-to-Tail Addition Method:

    • Vectors are added graphically by placing them sequentially head-to-tail.

    • Step 1: Draw vector A\mathbf{A} with proper scale and direction starting at the origin.

    • Step 2: Place the tail of vector B\mathbf{B} directly at the tip (arrowhead) of vector A\mathbf{A}.

    • Step 3: Draw the resultant vector C\mathbf{C} from the tail of vector A\mathbf{A} to the tip of vector B\mathbf{B}.   

      Vector addition diagram

  • Graphical Subtraction Method:

    • Vector subtraction C=A−B\mathbf{C} = \mathbf{A} - \mathbf{B} is executed as the addition of a negative vector:     C=A+(−B)\mathbf{C} = \mathbf{A} + (-\mathbf{B})

    • The vector −B-\mathbf{B} has equal magnitude to B\mathbf{B} but points in the exactly opposite direction (180∘180^\circ reversal).

    • Step 1: Draw vector A\mathbf{A}.

    • Step 2: Invert vector B\mathbf{B} to obtain −B-\mathbf{B}, placing its tail at the tip of A\mathbf{A}.

    • Step 3: Draw the resultant vector C\mathbf{C} from the tail of A\mathbf{A} to the tip of −B-\mathbf{B}.   

      Vector subtraction diagram

Comprehensive Vector Calculation Problems

  • Problem 1: Graphical Addition of Two Vectors

    • Objective: Find A+B\mathbf{A} + \mathbf{B} graphically and draw resultant vector C\mathbf{C}.

    • Method: Position the tail of B\mathbf{B} at the tip of A\mathbf{A}. The resultant C\mathbf{C} connects the origin of A\mathbf{A} to the arrowhead of B\mathbf{B}.

    • Vector Equation: C=A+B\mathbf{C} = \mathbf{A} + \mathbf{B}

  • Problem 2: Graphical Subtraction of Two Vectors

    • Objective: Find A−B\mathbf{A} - \mathbf{B} graphically and draw resultant vector C\mathbf{C}.

    • Method: Reverse direction of B\mathbf{B} to form −B-\mathbf{B}, place tail of −B-\mathbf{B} at the tip of A\mathbf{A}, and draw C\mathbf{C} from tail of A\mathbf{A} to tip of −B-\mathbf{B}.

    • Vector Equation: C=A−B\mathbf{C} = \mathbf{A} - \mathbf{B}

  • Problem 3: Addition of Perpendicular Vectors (A=30 m\mathbf{A} = 30\,m East, B=50 m\mathbf{B} = 50\,m North)

    • Unit Vector Notation:

    • A=30i^ m\mathbf{A} = 30\hat{i}\,m

    • B=50j^ m\mathbf{B} = 50\hat{j}\,m

    • C=(30i^+50j^) m\mathbf{C} = (30\hat{i} + 50\hat{j})\,m

    • Magnitude Calculation:     ∣C∣=(30)2+(50)2=900+2500=3400≈58.3 m|\mathbf{C}| = \sqrt{(30)^2 + (50)^2} = \sqrt{900 + 2500} = \sqrt{3400} \approx 58.3\,m

    • Direction Calculation:     θ=tan⁡−1(5030)≈59∘ North of East\theta = \tan^{-1}\left(\frac{50}{30}\right) \approx 59^\circ\,\text{North of East}

  • Problem 4: Subtraction of Perpendicular Vectors (A=40 m\mathbf{A} = 40\,m South, B=50 m\mathbf{B} = 50\,m East)

    • Unit Vector Notation:

    • A=−40j^ m\mathbf{A} = -40\hat{j}\,m

    • B=50i^ m\mathbf{B} = 50\hat{i}\,m

    • C=A−B=−40j^−50i^=(−50i^−40j^) m\mathbf{C} = \mathbf{A} - \mathbf{B} = -40\hat{j} - 50\hat{i} = (-50\hat{i} - 40\hat{j})\,m

    • Magnitude Calculation:     ∣C∣=(−50)2+(−40)2=2500+1600=4100≈64.0 m|\mathbf{C}| = \sqrt{(-50)^2 + (-40)^2} = \sqrt{2500 + 1600} = \sqrt{4100} \approx 64.0\,m

    • Direction Calculation:     θ=tan⁡−1(4050)≈38.7∘ South of West\theta = \tan^{-1}\left(\frac{40}{50}\right) \approx 38.7^\circ\,\text{South of West}

  • Problem 5: Addition of Non-Perpendicular Vectors (A=25 m\mathbf{A} = 25\,m at 30∘30^\circ North of East, B=40 m\mathbf{B} = 40\,m North)

    • Decomposing Vector A\mathbf{A}:

    • Ax=25 m×cos⁡(30∘)≈21.65 m≈21.7 mA_x = 25\,m \times \cos(30^\circ) \approx 21.65\,m \approx 21.7\,m

    • Ay=25 m×sin⁡(30∘)=12.5 mA_y = 25\,m \times \sin(30^\circ) = 12.5\,m

    • A=(21.7i^+12.5j^) m\mathbf{A} = (21.7\hat{i} + 12.5\hat{j})\,m

    • Decomposing Vector B\mathbf{B}:

    • Bx=0i^ mB_x = 0\hat{i}\,m

    • By=40j^ mB_y = 40\hat{j}\,m

    • B=40j^ m\mathbf{B} = 40\hat{j}\,m

    • Summing Components for Resultant Vector C=A+B\mathbf{C} = \mathbf{A} + \mathbf{B}:

    • Cx=Ax+Bx=21.7 m+0 m=21.7 mC_x = A_x + B_x = 21.7\,m + 0\,m = 21.7\,m

    • Cy=Ay+By=12.5 m+40 m=52.5 mC_y = A_y + B_y = 12.5\,m + 40\,m = 52.5\,m

    • Unit Vector Notation: C=(21.7i^+52.5j^) m\mathbf{C} = (21.7\hat{i} + 52.5\hat{j})\,m

    • Magnitude Calculation:     ∣C∣=(21.7)2+(52.5)2=470.89+2756.25=3227.14≈56.8 m|\mathbf{C}| = \sqrt{(21.7)^2 + (52.5)^2} = \sqrt{470.89 + 2756.25} = \sqrt{3227.14} \approx 56.8\,m

    • Direction Calculation:     θ=tan⁡−1(52.521.7)≈67.5∘ North of East\theta = \tan^{-1}\left(\frac{52.5}{21.7}\right) \approx 67.5^\circ\,\text{North of East}

  • Problem 6: Algebraic Vector Addition in Unit Vector Notation (A=5i^−2j^\mathbf{A} = 5\hat{i} - 2\hat{j}, B=−2i^+6j^\mathbf{B} = -2\hat{i} + 6\hat{j})

    • Summing Component-by-Component (C=A+B\mathbf{C} = \mathbf{A} + \mathbf{B}):

    • Cx=5+(−2)=3C_x = 5 + (-2) = 3

    • Cy=−2+6=4C_y = -2 + 6 = 4

    • Unit Vector Notation: C=3i^+4j^\mathbf{C} = 3\hat{i} + 4\hat{j}

    • Magnitude Calculation:     ∣C∣=32+42=9+16=25=5|\mathbf{C}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5

    • Direction Calculation:     θ=tan⁡−1(43)≈53.1∘ above the positive x-axis\theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.1^\circ\,\text{above the positive x-axis}