LECTURE NOTES ON TRIPLE PRODUCT

PROF. SHILPA PATRA

Department of Mathematics Narajole Raj College

1. Triple Product In vector , a branch of mathematics, the triple product is a product of three 3-dimensional vectors, usually Euclidean vectors. The name ”triple product” is used for two different products, the -valued scalar triple product and the vector-valued vector triple product.

2. Scalar triple product The scalar triple product (also called the mixed product, box product, or triple scalar product) is defined as the of one of the vectors with the of the other two.

Definition 2.1. For a given set of three vectors a, b, and c, the scalar (a×b)·c is called a scalar triple product of a, b, and c.

Remark: a · b is a scalar, and so (a · b) × c has no meaning.

Note: Given any three vectors a, b, and c, the following are scalar triple prod- ucts:

(a × b) · c, (b × c) · a, (c × a) · b, a · (b × c), b · (c × a), c · (a × b) (2.1) (b × a) · c, (c × b) · a, (a × c) · b, a · (c × b), b · (a × c), c · (b × a)

Geometric interpretation of scalar triple product: 1 Paper:C4T Unit-4 Topic- Triple product

Geometrically, the scalar triple product (a × b) · c is the (signed) volume of the defined by the three vectors given. Indeed, the magnitude of the vector a × b is the area of the parallelogram formed by using a and b; and the direction of the vector a × b is perpendicular to the plane parallel to both a and b.

Therefore, |(a × b) · c| is |a × b||c|| cos(θ)|, where θ is the angle between a × b and c. From the above figure, we observe that |c|| cos(θ)| is the height of the paral- lelepiped formed by using the three vectors as adjacent vectors. Thus, |(a × b) · c| is the volume of the parallelepiped.

Theorem 2.2. If a = a1i+a2j+a3k, b = b1i+b2j+b3k, and c = c1i+c2j+c3k, then

a1 a2 a3

(a × b) · c = b1 b2 b3 .

c1 c2 c3 Paper:C4T Unit-4 Topic- Triple product

Proof. By definition, we have

i j k

(a × b) · c = a1 a2 a3 · c

b1 b2 b3

= [(a2b3 − a3b2)i + (a3b1 − a1b3)j + (a1b2 − a2b1)k] · (c1i + c2j + c3k)

= (a2b3 − a3b2)c1 + (a3b1 − a1b3)c2 + (a1b2 − a2b1)c3

a1 a2 a3

= b1 b2 b3 .

c1 c2 c3  Theorem 2.3. For a given set of three vectors a, b, and c, (a × b) · c = a · (b × c). Proof.

b1 b2 b3

a · (b × c) = (b × c) · a = c1 c2 c3

a1 a2 a3

a1 a2 a3

= − c1 c2 c3 by R1 ↔ R3

b1 b2 b3

a1 a2 a3

= b1 b2 b3 by R2 ↔ R3

c1 c2 c3 = (a × b) · c

 Note: By Theorem 2.3, it follows that, in a scalar triple product, dot and cross can be interchanged without altering the order of occurrences of the vectors. For instance, we have (a × b) · c = a · (b × c), Since dot and cross can be interchanged. = (b × c) · a, Since dot product is commutative. = b · (c × a), Since dot and cross can be interchanged. = (c × a) · b, Since dot product is commutative. = c · (a × b), Since dot and cross can be interchanged. Notation: For any three vectors a, b, and c, the scalar triple product (a × b) · c is denoted by [a, b, c] is read as box a, b, c . For this reason and also because the absolute value of a scalar triple product represents the volume of a box (rectangular parallelepiped),a scalar triple product is also called a box product.

Note: By the above, we have [a, b, c] = (a × b) · c = a · (b × c) = (b × c) · a = b · (c × a) = [b, c, a] [b, c, a] = (b × c) · a = b · (c × a) = (c × a) · b = c · (a × b = [c, a, b]. Paper:C4T Unit-4 Topic- Triple product

In other words, [a, b, c] = [b, c, a] = [c, a, b] ; that is, if the three vectors are per- muted in the same cyclic order, the value of the scalar triple product remains the same.

If any two vectors are interchanged in their position in a scalar triple product, then the value of the scalar triple product is (-1) times the original value. More explicitly, [a, b, c] = [b, c, a] = [c, a, b] = −[a, c, b] = −[c, b, a] = −[b, a, c]. Theorem 2.4. The scalar triple product preserves addition and scalar multipli- cation. That is, [a + b, c, d] = [a, c, d] + [b, c, d]; [αa, b, c] = α[a, b, c]; ∀α ∈ R [a, b + c, d] = [a, b, d] + [a, c, d]; [a, αb, c] = α[a, b, c]; ∀α ∈ R [a, b, c + d] = [a, b, c] + [a, b, d]; [a, b, αc] = α[a, b, c]; ∀α ∈ R Theorem 2.5. The scalar triple product of three non-zero vectors is zero if, and only if, the three vectors are coplanar. Proof. Leta, b, c be any three non-zero vectors. Then, (a × b) · c = 0 ⇐⇒ a × b = 0 or (c and a × b are perpendicular) ⇐⇒ (a and b are parallel) or (a, b, and c are coplanar) ⇐⇒ a, b, and c are coplanar.  Example 2.6. If 2i − j + 3k, 3i + 2j + k, i + mj + 4k are coplanar, find the value of m . Proof. Since the given three vectors are coplanar, we have

2 −1 3

3 2 1 = 0 =⇒ m = −3

1 m 4  Example 2.7. Show that the four points (6, 7, 0), (16, 19, 4), (0, 3, 6), (2, 5,10) lie on a same plane. Proof. Let A = (6, 7, 0), B = (16, 19, 4), C = (0, 3, 6), D = (2, 5,10) . To show that the four points A, B, C, D lie on a plane, we have to prove that the three −−→ −→ −−→ vectors AB, AC and AD are coplanar. Now, −−→ −−→ −→ AB = OB − OA = (16i − 19j − 4k) − (6i − 7j) = (10i − 12j − 4k) −→ −−→ −→ −−→ −−→ −→ AC = OC − OA = (−6i + 10j − 6k) and AD = OD − OA = (−4i + 2j + 10k). we have

10 −12 −4 −−→ −→ −−→ [AB, AC, AD] = −6 10 −6 = 0.

−4 2 10 Paper:C4T Unit-4 Topic- Triple product

−−→ −→ −−→ Therefore, the three vectors AB, AC and AD are coplanar and hence the four points A, B, C, and D lie on a plane. 

Example 2.8. If the vectors a, b, c are coplanar, then prove that the vectors a + b, b + c, c + a are also coplanar.

Proof. Since the vectors a, b, c are coplanar, we have [a, b, c] = 0 Using the prop- erties of the scalar triple product, we get [a + b, b + c, c + a] = [a, b + c, c + a] + [b, b + c, c + a] = [a, b, c + a] + [a, c, c + a] + [b, b, c + a] + [b, c, c + a] = [a, b, c] + [a, b, a] + [a, c, c] + [a, c, a] + [b, b, c] + [b, b, a] + [b, c, c] + [b, c, a] = [a, b, c] + [a, b, c] = 2[a, b, c] = 0.

Hence the vectors a + b, b + c, c + a are coplanar. 

Theorem 2.9. If a, b, c and p, q, r be any systems of three vectors, and if p = x1a + y1b + z1c, q = x2a + y2b + z2c, and r = x3a + y3b + z3c, then

x1 y1 z1

[p, q, r] = x2 y2 z2 [a, b, c].

x3 y3 z3 Proof. Applying the distributive law of cross product and using a × a = b × b = c × c = 0

b × a = −a × b, a × c = −c × a, c × b = −b × c. We get

p × q = (x1a + y1b + z1c) × (x2a + y2b + z2c)

= (x1y2 − x2y1)(a × b) + (y1z2 − y2z1)(b × c) + (z1x2 − z2x1)(c × a)

x1 x2 y1 y2 z1 z2 = (a × b) + (b × c) + (c × a). y1 y2 z1 z2 x1 x2 Hence, we get

[p, q, r] = (p × q) · (x3a + y3b + z3c)   y1 y2 z1 z2 x1 x2 = x3 + y3 + z3 [a, b, c] z1 z2 x1 x2 y1 y2   y1 z1 z1 x1 x1 y1 = x3 + y3 + z3 [a, b, c] y2 z2 z2 x2 x2 y2

x1 y1 z1

= x2 y2 z2 [a, b, c].

x3 y3 z3



Example 2.10. If a, b, c are three vectors, prove that [a + c, a + b, a + b + c] = −[a, b, c]. Paper:C4T Unit-4 Topic- Triple product

Proof. Using theorem 2.9, we get

1 0 1

[a + c, a + b, a + b + c] = 1 1 0 [a, b, c]

1 1 1 = −[a, b, c].  Properties(Summary): • The scalar triple product is unchanged under a circular shift of its three operands (a, b, c): a · (b × c) = b · (c × a) = c · (a × b). • Swapping the positions of the operators without re-ordering the operands leaves the triple product unchanged. This follows from the preceding prop- erty and the commutative property of the dot product. a · (b × c) = (a × b) · c. • Swapping any two of the three operands negates the triple product. This follows from the circular-shift property and the anticommutativity of the cross product. a · (b × c) = −a · (c × b) = −b · (a × c) = −c · (b × a). • The scalar triple product can also be understood as the of the 3 × 3 that has the three vectors either as its rows or its columns   a1 a2 a3 a · (b × c) = det b1 b2 b3  = det (a, b, c) . c1 c2 c3 • If the scalar triple product is equal to zero, then the three vectors a, b, and c are coplanar, since the parallelepiped defined by them would be flat and have no volume. • If any two vectors in the scalar triple product are equal, then its value is zero: a · (a × b) = (a × b) · a = a · (b × b) = (b × a) · a = 0.

3. Vector triple product The vector triple product is defined as the cross product of one vector with the cross product of the other two.

Definition 3.1. The vector triple product of u, v, w is u × (v × w). Caution: The vector triple product is not associative, i.e. in general u × (v × w) 6= (u × v) × w. To see this, we note that (u × v) × w is perpendicular to u × v which is normal to a plane determined by u and v. So, (u × v) × w is coplanar with u and v. By the same argument, u × (v × w) is coplanar with v and w. For this reason it is vital that we include the parentheses in a vector triple product to indicate which vector product should be performed first. Paper:C4T Unit-4 Topic- Triple product

We now obtain a formula for the vector triple product which reflects the fact that u × (v × w) , as it is coplanar with v and w, may be expressed as αv + βw for some α, β ∈ R. Theorem 3.2. (The triple product expansion, or Lagrange’s formula): For all vectors u, v, and w u × (v × w) = (u · w)v − (u · v)w.

Proof. Let u = (ux, uy, uz), v = (vx, vy, vz), and w = (wx, wy, wz). The x component of u × (v × w) is given by:

(u × (v × w))x = uy(vxwy − vywx) − uz(vzwx − vxwz)

= vx(uywy + uzwz) − wx(uyvy + uzvz)

= vx(uywy + uzwz) − wx(uyvy + uzvz) + (uxvxwx − uxvxwx)

= vx(uxwx + uywy + uzwz) − wx(uxvx + uyvy + uzvz)

= (u · w)vx − (u · v)wx. Similarly, the y and z components of u × (v × w) are given by:

(u × (v × w))y = (u · w)vy − (u · v)wy . (u × (v × w))z = (u · w)vz − (u · v)wz By combining these three components we obtain: u × (v × w) = (u · w)v − (u · v)w.

 Example 3.3. Express (u × v) × w as a of u and v. Proof. Since the cross product is anticommutative, this formula may also be written as: (u × v) × w = −w × (u × v) = −(w · v)u + (w · u)v.  From Lagrange’s formula it follows that the vector triple product satisfies: u × (v × w) + v × (w × u) + w × (u × v) = 0 which is the Jacobi identity for the cross product. Another useful formula follows: (u × v) × w = u × (v × w) − v × (u × w). Example 3.4. Let α = (a × b) · (a × c). Show that α = (c · a)|a|2 − ((c · a))(a · b). Evaluate α when a, b and c are unit vectors with b and c perpendicular, and the angles between a and b, and between a and c, are both π/3. Proof. (a × b) · (a × c) = [a × b, a, c] = [a, c, a × b] = a · (c × (a × b)) = a · ((c · b)a − (c · a)b) = (c · b)(a · a) − (c · a)(a · b) = (c · a)|a|2 − ((c · a))(a · b). Paper:C4T Unit-4 Topic- Triple product

With the given conditions, we have |a| = |b| = |c| = 1, c · b = 0, a · b = c · a = π 1 1 × 1 × cos( 3 ) = 2 and so 1 1 α = 0 − ( )2 = . 2 4