Mathematical Physics Vol 1

Chapter 3. Examples

62

Proof Assume that vectors n , m , p are mutually orthogonal, namely that n · m = 0 , n · p = 0 and m · p = 0. Since the space is three-dimensional a fourth vector that would be mutually orthogonal with these three vectors does not exist. As per definition, vectors are linearly independent if the following is true α n + β m + γ p = 0 ⇒ α = β = γ = 0 . From here, it follows that α n + β m + γ p = 0 / · n ⇒ α n · n + β m · n + γ p · n = 0 ⇒ ⇒ α | n | 2 + 0 + 0 = 0 ⇒ α = 0 , α n + β m + γ p = 0 / · m ⇒ α n · m + β m · m + γ p · m = 0 ⇒ ⇒ 0 + β | m | 2 + 0 = 0 ⇒ β = 0 , α n + β m + γ p = 0 / · p ⇒ α n · p + β m · p + γ p · p = 0 ⇒ ⇒ 0 + 0 + γ | p | 2 = 0 ⇒ γ = 0 .

Problem10 Give a geometric interpretation of the vector product.

Answer a) The area of the parallelogram constructed over vectors a and b (Fig. 3.8), is P = | b |· h = | b |·| a | sin θ , that is P = | a × b | . Thus, the magnitude of a vector product of two vectors is equal to the area of the parallelogram formed by these two vectors. b) The area of a triangle whose two sides are vectors a and b is equal to half the intensity of their vector product (Fig. 3.8).

1 2 |

a × b | .

P ∆ =

Figure 3.8: Area of a triangle.