Lecture 15: Dot Product and Angles Between Vectors✅

Overview of Dot Product Properties and Its Applications

  • The dot product (also called scalar product) plays a crucial role in physics and engineering.

  • It is used to calculate work done by a force when the displacement is not directly along the force vector.

Concept of Work Done by a Force

  • Work is defined as the product of the component of force acting in the direction of displacement.

  • Work formula: W=∣F∣×∣A∣×cos(θ)W = |F| \times |A| \times \text{cos}(\theta) Where:

    • ∣F∣|F| = magnitude of the force

    • ∣A∣|A| = magnitude of displacement

    • θ\theta = angle between the force and displacement vector

  • Unit of work: Joules (1 Joule = 1 Newton-meter)

Example of Work Done

  • Scenario: A person pulls a door at an angle leading to displacement.

  • Given:

    • Force: ∣F∣=20 N|F| = 20 \text{ N}

    • Displacement: ∣A∣=3 m|A| = 3 \text{ m}

    • Angle: θ=60∘=π3 radians\theta = 60^\circ = \frac{\pi}{3} \text{ radians}

  • Calculation:
    W=20×3×cos(60∘)=20×3×12=30 JW = 20 \times 3 \times \text{cos}(60^\circ) = 20 \times 3 \times \frac{1}{2} = 30 \text{ J}

  • This example illustrates the importance of the dot product in defining work done.

Definition of Dot Product

  • The dot product of two vectors A and B is defined as: A∙B=∣A∣×∣B∣×cos(θ)A \bullet B = |A| \times |B| \times \text{cos}(\theta)

    • θ\theta is the angle between A and B, typically within the range [0, π\pi].

Finding the Dot Product

  • Example Vectors:

    • A=3iA = 3 \mathbf{i}

    • B=2i+2jB = 2 \mathbf{i} + 2 \mathbf{j}

  • Magnitudes:

    • ∣A∣=3|A| = 3

    • ∣B∣=22|B| = 2\sqrt{2}

  • Angle formed: θ=π4\theta = \frac{\pi}{4}

  • Dot product calculation:
    A∙B=3×(22)×cos(π4)A \bullet B = 3 \times (2\sqrt{2}) \times \text{cos}(\frac{\pi}{4})

  • Result: 626\sqrt{2}

Perpendicular Vectors

  • Vectors A and B are perpendicular if θ=π2\theta = \frac{\pi}{2} (90 degrees).

  • In this case, cos(π2)=0\text{cos}(\frac{\pi}{2}) = 0, leading to:
    A∙B=0A \bullet B = 0.

Example of Perpendicular Vectors

  • Vectors:

    • A=i+jA = \mathbf{i} + \mathbf{j}

    • B=−i−jB = -\mathbf{i} - \mathbf{j}

  • Verification:

    • Dot product:
      A∙B=0A \bullet B = 0

Calculating the Angle Between Vectors

  • Given Vectors:

    • A=i+3jA = \mathbf{i} + \sqrt{3}\mathbf{j}

    • B=−i+j+kB = -\mathbf{i} + \mathbf{j} + \mathbf{k}

  • To find the angle:

    • Use the dot product definition.

    • Calculate magnitudes and dot product.

    • Use formula: θ=arc cos(A∙B∣A∣∣B∣)\theta = \text{arc cos} \left(\frac{A \bullet B}{|A||B|}\right)

Scalar Product in Component Form

  • Vectors can be represented in 2D or 3D as:
    A=(a<em>1,a</em>2,a<em>3)A = (a<em>1, a</em>2, a<em>3) B=(b</em>1,b<em>2,b</em>3)B = (b</em>1, b<em>2, b</em>3)

  • Expanded dot product:
    A∙B=a<em>1b</em>1+a<em>2b</em>2+a<em>3b</em>3A \bullet B = a<em>1b</em>1 + a<em>2b</em>2 + a<em>3b</em>3

  • Important Property: The result of zero confirms that vectors are perpendicular.

Properties of Dot Product

  1. A∙0=0A \bullet 0 = 0 (Dot product with a zero vector is zero)

  2. Commutative: A∙B=B∙AA \bullet B = B \bullet A

  3. Distributive: A∙(B+C)=A∙B+A∙CA \bullet (B + C) = A \bullet B + A \bullet C

  4. Scalar multiplication: k(A∙B)=(kA)∙B=A∙(kB)k(A \bullet B) = (kA) \bullet B = A \bullet (kB)

  5. A and B are perpendicular if A∙B=0A \bullet B = 0 (and vice versa)

Conclusion

  • The dot product is essential for calculations in mathematics, physics, and engineering.

  • Understanding both definitions (cosine and component form) enhances versatility in problem-solving.

  • Future discussions will explore various applications of the dot product in deeper contexts such as projections and optimization problems.