PROF. SHILPA PATRA
Department of Mathematics Narajole Raj College
1. Triple Product
In vector algebra, 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 scalar-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 dot product of one of the vectors with the cross product of the other two.
Definition 2.1. For a given set of three vectorsa,b, andc, the scalar(a×b)·c is called a scalar triple product ofa,b, andc.
Remark: a·bis a scalar, and so (a·b)×chas no meaning.
Note: Given any three vectorsa,b, andc, 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:
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Geometrically, the scalar triple product (a×b)·c is the (signed) volume of the parallelepiped 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 vectora×bis perpendicular to the plane parallel to bothaandb.
Therefore,|(a×b)·c| is|a×b||c||cos(θ)|, whereθ is the angle betweena×b andc. 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. Ifa=a1i+a2j+a3k,b=b1i+b2j+b3k, andc=c1i+c2j+c3k, then
(a×b)·c=
a1 a2 a3 b1 b2 b3
c1 c2 c3
.
Proof. By definition, we have (a×b)·c =
i j k a1 a2 a3
b1 b2 b3
·c
= [(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, andc,
(a×b)·c=a·(b×c).
Proof.
a·(b×c) = (b×c)·a =
b1 b2 b3
c1 c2 c3
a1 a2 a3
=−
a1 a2 a3 c1 c2 c3 b1 b2 b3
by R1↔R3
=
a1 a2 a3 b1 b2 b3 c1 c2 c3
by R2↔R3
= (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 vectorsa,b, andc, 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].
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,cbe any three non-zero vectors. Then,
(a×b)·c= 0 ⇐⇒ a×b=0or(canda×bare perpendicular)
⇐⇒ (aandbare parallel)or(a, b, andcare coplanar)
⇐⇒ a, b, andcare coplanar.
Example 2.6. If2i−j+ 3k,3i+ 2j+k,i+mj+ 4kare coplanar, find the value of m .
Proof. Since the given three vectors are coplanar, we have
2 −1 3
3 2 1
1 m 4
= 0 =⇒ m=−3
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−−→
ADare 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
[−−→ AB,−→
AC,−−→ AD] =
10 −12 −4
−6 10 −6
−4 2 10
= 0.
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 vectorsa,b,care 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 vectorsa+b,b+c,c+aare coplanar.
Theorem 2.9. If a,b,candp,q,r be any systems of three vectors, and if p= x1a+y1b+z1c,q=x2a+y2b+z2c, andr=x3a+y3b+z3c, then
[p,q,r] =
x1 y1 z1
x2 y2 z2
x3 y3 z3
[a,b,c].
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
(a×b) +
y1 y2
z1 z2
(b×c) +
z1 z2
x1 x2
(c×a).
Hence, we get
[p,q,r] = (p×q)·(x3a+y3b+z3c)
=
x3
y1 y2
z1 z2
+y3
z1 z2
x1 x2
+z3
x1 x2
y1 y2
[a,b,c]
=
x3
y1 z1 y2 z2
+y3
z1 x1 z2 x2
+z3
x1 y1 x2 y2
[a,b,c]
=
x1 y1 z1 x2 y2 z2
x3 y3 z3
[a,b,c].
Example 2.10. Ifa,b,care three vectors, prove that [a+c,a+b,a+b+c] =
−[a,b,c].
Proof. Using theorem 2.9, we get
[a+c,a+b,a+b+c] =
1 0 1 1 1 0 1 1 1
[a,b,c]
= −[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 determinant of the 3×3 matrix that has the three vectors either as its rows or its columns
a·(b×c) = det
a1 a2 a3 b1 b2 b3 c1 c2 c3
= det (a,b,c).
• 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 ofu,v,w isu×(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)×wis perpendicular tou×vwhich is normal to a plane determined byuandv. So, (u×v)×w is coplanar withuandv. By the same argument,u×(v×w) is coplanar withvand 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.
We now obtain a formula for the vector triple product which reflects the fact that u×(v×w) , as it is coplanar with vand w, may be expressed asαv+βw for someα, β∈R.
Theorem 3.2. (The triple product expansion, or Lagrange’s formula):
For all vectorsu,v, andw
u×(v×w) = (u·w)v−(u·v)w.
Proof. Letu= (ux, uy, uz),v= (vx, vy, vz), andw= (wx, wy, wz).
Thexcomponent 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, they andz components ofu×(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 linear combination ofuandv.
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 αwhen a,b andc are unit vectors with bandc perpendicular, and the angles between aandb, and between aandc, are bothπ/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).
With the given conditions, we have|a|=|b|=|c|= 1, c·b= 0, a·b=c·a= 1×1×cos(π3) = 12 and so
α= 0−(1 2)2=1
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