Vectors - Properties Of Scalar Product

Vectors - Properties of Scalar Product

The scalar product of two vectors is denoted by a dot (·) between the vectors.
The scalar product of two vectors is commutative: A · B = B · A.
The scalar product of two parallel vectors is equal to the product of their magnitudes: A · B = |A||B|.
The scalar product of two perpendicular vectors is zero: A · B = 0.
The scalar product of a vector with itself is equal to the square of its magnitude: A · A = |A|^2.

Vectors - Properties of Scalar Product (contd.)

The scalar product of two vectors can be expressed in terms of their components:
- A · B = AₓBₓ + AᵧBᵧ + A₂B₂.
The scalar product can also be written using the magnitudes and angle between the vectors:
- A · B = |A||B| cosθ.
If the value of the scalar product is positive, the angle between the vectors is acute.
If the value of the scalar product is negative, the angle between the vectors is obtuse.

Vectors - Properties of Scalar Product (contd.)

The scalar product of two vectors can be used to find the angle between them:
- cosθ = (A · B) / (|A||B|).
The scalar product can also be written using the unit vectors i, j, and k:
- A · B = AₓBₓ + AᵧBᵧ + A₂B₂ = |A||B| cosθ.
The angle between two vectors can be calculated as:
- cosθ = (AₓBₓ + AᵧBᵧ + A₂B₂) / (|A||B|).

Example 1

Find the scalar product of two vectors A = 3i + 2j and B = 5i + 4j. Solution: A · B = (3)(5) + (2)(4) = 15 + 8 = 23.

Example 2

Calculate the angle between two vectors C = 4i + 3j and D = 2i - 2j. Solution: C · D = (4)(2) + (3)(-2) = 8 - 6 = 2. |C| = √(4² + 3²) = √(16 + 9) = √25 = 5. |D| = √(2² + (-2)²) = √(4 + 4) = √8. cosθ = C · D / (|C||D|) = 2 / (5√8) ≈ 0.1788. θ ≈ arccos(0.1788) ≈ 80.43 degrees.

Equation for Projection

The scalar product A · B can be used to find the projection of a vector A onto another vector B. Projection of A onto B = |A| cosθ = (A · B) / |B|.

Slide 11: Vectors - Properties of Scalar Product (contd.)

The scalar product of two vectors can also be negative when the vectors are in opposite directions.
The scalar product of two vectors can be used to find the work done by a force when it is applied along a certain direction.
The scalar product can be used to solve problems related to the equilibrium of a particle.
The scalar product can be used to calculate the distance between a point and a line.
The scalar product can be used to find the area of a parallelogram.

Slide 12: Example 3

Find the work done by a force of 10 N in displacing an object along a vector A = 3i + 4j. Solution: Work done = Force × Displacement × cosθ. Since the force and displacement are parallel, θ = 0 degrees. Work done = 10 × |A| × cos0 = 10 × √(3² + 4²) × 1 = 10 × 5 = 50 J.

Slide 13: Example 4

Calculate the area of a parallelogram with two adjacent sides A = 2i + j and B = 4i + 3j. Solution: Area of parallelogram = |A × B|. |A × B| = |(2)(3) - (1)(4)| = |6 - 4| = 2.

Slide 14: Unit Vectors

Unit vectors have a magnitude of 1.
The unit vectors in the x, y, and z directions are represented as i, j, and k, respectively.
Any vector can be expressed as a linear combination of the unit vectors.

Slide 15: Example 5

Express the vector C = 3i - 4j + 2k as a linear combination of the unit vectors i, j, and k. Solution: C = 3i - 4j + 2k.

Slide 16: Orthogonal Vectors

Orthogonal vectors are perpendicular to each other.
The dot product of orthogonal vectors is zero.
Two vectors A and B are orthogonal if their scalar product, A · B, is equal to zero.

Slide 17: Example 6

Check whether the vectors E = 3i - 2j + k and F = 4i + 6j - 5k are orthogonal. Solution: E · F = (3)(4) + (-2)(6) + (1)(-5) = 12 - 12 - 5 = -5. Since the scalar product is not equal to zero, the vectors E and F are not orthogonal.

Slide 18: Cross Product

The cross product of two vectors is denoted by the symbol × between the vectors.
The cross product of two vectors is not commutative: A × B = -B × A.
The cross product of two parallel vectors is equal to zero: A × B = -B × A = 0.

Slide 19: Cross Product (contd.)

The cross product of two vectors can be expressed in terms of their components:
- A × B = (AᵧB₂ - A₂Bᵧ)i + (A₂Bₓ - AₓB₂)j + (AₓBᵧ - AᵧBₓ)k.
The cross product can also be written using the magnitudes and angle between the vectors:
- |A × B| = |A||B| sinθ.
The cross product of two vectors gives a result that is orthogonal to both vectors.

Slide 20: Example 7

Find the cross product of two vectors G = 2i + j - 3k and H = 3i + 4j + 2k. Solution: G × H = [(1)(2) - (-3)(4)]i + [(-3)(3) - (2)(2)]j + [(2)(4) - (1)(3)]k = [2 - (-12)]i + [(-9) - 4]j + [8 - 3]k = 14i - 13j + 5k.

Slide 21: Application of Cross Product

The cross product of two vectors can be used to find the moment of a force.
The moment of a force is a measure of its tendency to rotate an object about an axis.
The magnitude of the moment of a force F about a point P is given by: |M| = |F||r| sinθ, where |F| is the magnitude of the force, |r| is the perpendicular distance from the point P to the line of action of the force, and θ is the angle between the force and the vector r.

Slide 22: Example 8

Find the moment of a force of 8 N about a point P, located at a distance of 5 m from the line of action of the force. The angle between the force and the vector r is 60 degrees. Solution: |M| = (8)(5) sin60 = (8)(5)(√3/2) = 20√3 Nm.

Slide 23: Scalar Triple Product

The scalar triple product of three vectors A, B, and C is denoted by (A × B) · C.
The scalar triple product can be used to find the volume of a parallelepiped.
The volume V of a parallelepiped formed by three vectors A, B, and C is given by: V = |(A × B) · C|.

Slide 24: Example 9

Find the volume of a parallelepiped with sides A = 3i + j - 2k, B = 2i + 4j + 5k, and C = -i + 2j + 3k. Solution: V = |(A × B) · C| = |(14i + 11j + 10k) · (-i + 2j + 3k)| = |(-14) + 22 + 30| = 38 cubic units.

Slide 25: Vector Triple Product

The vector triple product of three vectors A, B, and C is denoted by A × (B × C).
The vector triple product can be used to find the moment of a force about an axis.
The moment of a force F about an axis defined by a unit vector ĥ is given by: M = F × (r × ĥ), where r is the position vector of the point of application of the force relative to the axis.

Slide 26: Example 10

Find the moment of a force of 10 N about an axis defined by the unit vector ĥ = i + 2j - k, with the force applied at a point located at r = 2i - 3j + 4k. Solution: M = (10) × [(2i - 3j + 4k) × (i + 2j - k)] = (10) × [(6i + 2j - 3k) × (i + 2j - k)] = (10) × (-3i + 15j + 14k) = -30i + 150j + 140k Nm.

Slide 27: Distributive Property of Cross Product

The cross product of two vectors satisfies the distributive property over addition: A × (B + C) = (A × B) + (A × C).
Similarly, the cross product of a scalar and a vector satisfies the distributive property: k(A × B) = (kA) × B = A × (kB).

Slide 28: Example 11

Simplify the expression A × (B + C), where A = i + 2j - k, B = 4i - 3j + 2k, and C = 2i + j - 4k. Solution: A × (B + C) = (i + 2j - k) × [(4i - 3j + 2k) + (2i + j - 4k)] = (i + 2j - k) × (6i - 2j - 2k) = (i + 2j - k) × (-6i + 2j + 2k) = -10i + 10j + 10k.

Slide 29: Applications of Vectors in Physics

Vectors play a crucial role in various branches of physics, including mechanics, electromagnetism, and quantum mechanics.
They are used to represent physical quantities such as force, velocity, acceleration, electric field, magnetic field, and more.
Vectors help in understanding the direction, magnitude, and interactions of these physical quantities.
They are essential in solving problems related to motion, forces, energy, and other concepts in physics.

Slide 30: Summary

The scalar product of two vectors allows us to find the angle between them and solve problems related to work done, equilibrium, and distance calculations.
The cross product of two vectors helps in finding the moment of a force, volume of a parallelepiped, and moment of a force about an axis.
Both scalar and vector triple products have applications in physics.
Vectors have widespread applications in various branches of physics and are crucial for understanding and solving problems in the subject.