Lecture 16: Vectors and Their Applications

Introduction to Vectors

  • Continuation of discussion on vectors.
  • Focus on scalar resolute, vector resolute, and applications of the dot product.

Decomposing Vectors

  • Given two vectors a and b forming an angle ( \theta ):
    • Component of b in direction of a = ( |b| \cos \theta )
    • Component of b perpendicular to a = ( |b| \sin \theta )
  • Term "component of b in direction a" corresponds to ( |b| \cos \theta ).

Dot Product of Vectors

  • Definition of the dot product: ( \mathbf{a} \cdot \mathbf{b} = |a||b|\cos \theta )
  • Rearranging this gives: ( \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|a||b|} )
  • This implies the component of b in direction a is given by:
    • ( \frac{\mathbf{a} \cdot \mathbf{b}}{|a|} \hat{a} ) where ( \hat{a} = \frac{\mathbf{a}}{|a|} ) is the unit vector in direction of a.

Example 1: Vectors a and b

  • Let ( \mathbf{a} = 2\mathbf{i} ) and ( \mathbf{b} = 2\mathbf{i} + 3\mathbf{j} )
  • Calculate component of b in direction a:
    • Dot Product: ( \mathbf{a} \cdot \mathbf{b} = (2)(2) + (0)(3) = 4 )
    • Magnitude of ( \mathbf{a} = 2 )
    • Therefore, component = ( \frac{4}{2} = 2 )

Example 2: Vectors with Negative Components

  • Let ( \mathbf{a} = -3\mathbf{i} ) and ( \mathbf{b} = 2\mathbf{i} + 3\mathbf{j} )
  • Calculate component of b in direction a:
    • Dot Product: ( \mathbf{a} \cdot \mathbf{b} = (-3)(2) + (0)(3) = -6 )
    • Magnitude: ( |\mathbf{a}| = 3 )
    • Therefore, component = ( \frac{-6}{3} = -2 )
  • Interpretation: A negative component indicates the angle between a and b is greater than ( \frac{\pi}{2} ).

Vector Resolution

  • Scalar component leads to the vector projection defined as:
    • ( \text{PROJ}_{a} b = \left( \frac{b \cdot a}{|a|^2} \right) a )
  • Explains how to obtain vector projection by multiplying the scalar component by the unit vector in direction of a.

Example: Projection Calculation

  • Let ( \mathbf{a} = 2\mathbf{i} ) and ( \mathbf{b} = 2\mathbf{i} + 3\mathbf{j} )
  • Components calculated previously indicate the projection of b onto a is simply ( 2\mathbf{i} ).

Terminology

  • Scalar resolute = scalar component in a given direction.
  • Vector resolute = resolved part of the vector in the direction of another.
  • Important to understand ‘projection’ vs ‘rejection’ of vectors.

Vector Rejection

  • The vector rejection is defined as the component of b that is perpendicular to a:
    • ( ext{Rejection} = b - \text{projection of } b \text{ onto } a )
    • Denoted as ( b_{\perp} ) (perpendicular component).

Applications of Vectors

  • Work done = dot product of force and displacement vectors.
  • Finding angles between vectors using the dot product formula.

Practical Application

  1. Analyzing forces on a particle:
    • Calculating work done by forces acting on a particle on a surface (e.g. moving a door).
  2. Using displacements and forces in mechanics, including equilibrium conditions in systems with multiple vectors.

Final Example: Balancing Forces

  • Application involving three tensions attached to a pole: (100 N each).
  • Analyze their components to maintain vertical orientation of the pole.
  • Solve to find required conditions for balance (specific value for d).

Conclusion

  • Review essential points regarding vector components, projection, rejection, and their applications in mechanical contexts.
  • Open for questions and further clarifications during instructor's consultation hours.