Scalar Product of Vectors

What is the result of the scalar product of two vectors?

Which law does the scalar product follow?

If the scalar product of two vectors A and B is zero, what can be inferred about the two vectors?

What does the projection of vector B onto vector A represent?

What is the dot product of a vector A with itself?

How is the scalar product A.B defined mathematically?

What does the expression A.(B + C) equal in terms of scalar products?

Given two vectors A and B in component form, how is their scalar product calculated?

Scalar Product of Vectors

The scalar product or dot product of two vectors A and B is a scalar quantity.
Defined as: A.B = AB cos θ, where θ is the angle between the vectors A and B.
Geometric interpretation: A.B is the product of the magnitude of A and the component of B along A.
Alternatively: It's the product of the magnitude of B and the component of A along B.
Commutative law: A.B = B.A
Distributive law: A.(B + C) = A.B + A.C
Scalar multiplication: λA.(λB) = λ(A.B), where λ is a real number.
For unit vectors i, j, k:
- i.i = j.j = k.k = 1
- i.j = j.k = k.i = 0
Given two vectors A = Ax i + Ay j + Az k and B = Bx i + By j + Bz k, their scalar product is: A.B = AxBx + AyBy + AzBz
Special cases:
- A.A = A² = Ax² + Ay² + Az²
- A.B = 0 if A and B are perpendicular (θ = 90°).

Example: Finding Angle and Projection

Calculate the angle between force F = (3i + 4j - 5k) and displacement d = (5i + 4j + 3k).
- Calculate F.d = 16
- Calculate F.F = 50
- Calculate d.d = 50
- Use the formula: cos θ = F.d / (|F| |d|) = 16 / 50 = 0.32
- Find θ = cos⁻¹(0.32)
Projection of F on d:
- The projection of F onto d is F.d / |d| = 16 / √(50) units.

This quiz covers the scalar (dot) product of vectors, including its definition, geometric interpretation, and fundamental properties such as commutative and distributive laws. Additionally, it includes special cases and calculations involving angles and projections between vectors. Test your understanding of these concepts with this engaging quiz!