3 Vectors: Triple Products, Lecture notes of Mathematical Physics

Download 3 Vectors: Triple Products and more Lecture notes Mathematical Physics in PDF only on Docsity!

OLLSCOIL NA hEIREANN MA NUAD

THE NATIONAL UNIVERSITY OF IRELAND MAYNOOTH

MATHEMATICAL PHYSICS

EE

Engineering Mathematics II

Vector Triple Products

Prof. D. M. Heffernan and Mr. S. Pouryahya

3 Vectors: Triple Products

3.1 The Scalar Triple Product

The scalar triple product, as its name may suggest, results in a scalar as its result. It

is a means of combining three vectors via cross product and a dot product. Given the

vectors

A = A 1 i + A 2 j + A 3 k

B = B 1 i + B 2 j + B 3 k

C = C 1 i + C 2 j + C 3 k

a scalar triple product will involve a dot product and a cross product

A · (B × C)

It is necessary to perform the cross product before the dot product when computing a

scalar triple product,

B × C =

i j k

B 1 B 2 B 3

C 1 C 2 C 3

= i

B 2 B 3

C 2 C 3

− j

B 1 B 3

C 1 C 3

• k

B 1 B 2

C 1 C 2

since A = A 1 i + A 2 j + A 3 k one can take the dot product to find that

A · (B × C) = (A 1 )

B 2 B 3

C 2 C 3

− (A 2 )

B 1 B 3

C 1 C 3

+ (A 3 )

B 1 B 2

C 1 C 2

which is simply

Important Formula 3.1.

A · (B × C) =

A 1 A 2 A 3

B 1 B 2 B 3

C 1 C 2 C 3

The usefulness of being able to write the scalar triple product as a determinant is not only

due to convenience in calculation but also due to the following property of determinants

Note 3.1. Exchanging any two adjacent rows in a determinant changes the sign of the

original determinant.

... example continued

Take the dot product with A to find

A · (B × C) = (2)(0) + (3)(0) + (−1)(−4)

Method 2 :

Evaluate the determinant

A · (B × C) =

Example 3.1.2. Prove that

Important Formula 3.2.

A · B × C = A × B · C

Solution:

Notice that there are no brackets given here as the only way to evaluate the scalar triple

products is to perform the cross products before performing the dot products

a

. Let

A = A 1 i + A 2 j + A 3 k

B = B 1 i + B 2 j + B 3 k

C = C 1 i + C 2 j + C 3 k

now,

A · B × C =

A 1 A 2 A 3

B 1 B 2 B 3

C 1 C 2 C 3

C 1 C 2 C 3

B 1 B 2 B 3

A 1 A 2 A 3

C 1 C 2 C 3

A 1 A 2 A 3

B 1 B 2 B 3

= C · A × B = A × B · C

a This is due to the fact that if the dot product is evaluate first one would be left with a cross product

between a scalar and a vector which is not defined.

3.2 The Vector Triple Product

The vector triple product, as its name suggests, produces a vector. It is the result of

taking the cross product of one vector with the cross product of two other vectors.

Important Formula 3.3 (Vector Triple Product).

A × (B × C) = (A · C)B − (A · B)C

Proving the vector triple product formula can be done in a number of ways. The straight-

forward method is to assign

A = A 1 i + A 2 j + A 3 k

B = B 1 i + B 2 j + B 3 k

C = C 1 i + C 2 j + C 3 k

and work out the various dot and cross products to show that the result is the same.

Here we shall however go through a slightly more subtle but less calculation heavy proof.

Note 3.2. The vector A × (B × C) must be in the same plane as B and C. This is due

to fact that the vector that results from the cross product is perpendicular to both the

vectors whose product has just been taken. Since one can say that A × (B × C) is on

the same plane as B and C it follows that

A × (B × C) = αB + βC

where α and β are scalars.

We introduce a new coordinate system with the unit vector i

′ along the vector B, j

′ a

unit vector (orthogonal to i

′ ), which is on the same plane as the both the vectors B and

C, and k

′ a unit vector orthogonal to both i

′ and j

′ 1

. Using this basis allows one to write

the vectors B and C as

B = B 1 i

′

• 0j

′

• 0k

′

C = C 1 i

′

• C 2 j

′

• 0k

′

however there is no special reduction to the representation of the vector A in terms of

this new basis thus,

A = A 1 i

′

• A 2 j

′

• A 3 k

′ .

1 the unit vector k

′ will thus point in the same direction as the vector B × C.

This is the desired result as returning to our definitions of A, Band C in this basis,

A = A 1 i

′

• A 2 j

′

• A 3 k

′

B = B 1 i

′

C = C 1 i

′

• C 2 j

′

one finds that,

A · B = A 1 B 1

A · C = A 1 C 1 + A 2 C 2

Hence,

A × (B × C) = (A 2 C 2 + A 1 C 1 )B 1 i

′ − A 1 B 1 (C 1 i

′

• C 2 j

′ )

= (A · C)B − (A · B)C.

3.3 Area and Volume Using a Cross Product

3.3.1 The area of a parallelogram.

θ

B

A h

(a) The parallelogram

θ

B

A

h

(b) The triangle

Figure 2: Areas related to the cross product.

The are of a parallelogram is simply given by the product of the base and the height of

the parallelogram. Here this is given by

Area of parallelogram = h | B |

= (| A | sin(θ)) | B |

= | A | | B | sin(θ)

= |A × B|

3.3.2 The area of a triangle.

The area of a triangle is half the base times the height. From the figure we have

Area of triangle =

| B | h =

| B | (| A | sin(θ))

|A × B|

3.4 Summary of Vector Rules

Here we list most of the main results concerning vectors,

A · A = A

2 ≡ | A |

2 (3.4.1)

A · B = B · A (3.4.2)

A · (αB) = α(A · B) (3.4.3)

A · (B × C) = (A × B) · C (3.4.4)

A × A = 0 (3.4.5)

A × B = −B × A (3.4.6)

A × (αB) = α(A × B) (3.4.7)

A × (B × C) = (A · C)B − (A · B)C (3.4.8)

3.4.1 Worked problem

Example 3.4.1 (Manipulating vectors without evaluation).

Prove that

(i) (A × B) × (C × D) = C(A · B × D) − D(A · B × C)

(ii) (A × B) × (C × D) = B(A · C × D) − A(B · C × D)

Solution:

(i) let U = A × B

(A × B) × (C × D) = U × (C × D)

= (U · D)C − (U · C)D using Eq 3.4.

= (A × B · D)C − (A × B · C)D

= (A · B × D)C − (A · B × C)D using Eq 3.4.

= C(A · B × D) − D(A · B × C).

(ii) let V = (C × D)

(A × B) × (C × D) = −(C × D) × (A × B) using Eq 3.4.

= −V × (A × B)

= − {(V · B)A − (V · A)B} using Eq 3.4.

= (C × D · A)B − (C × D · B)A

= (C · D × A)B − (C · D × B)A using Eq 3.4.

= B(A · C × D) − A(B · C × D).