Scalar Triple Product

The scalar triple product, which combines three vectors to create a scalar quantity, is a vital concept in vector algebra. The spatial interactions between vectors, including linear dependency and volume determination, are shown by it. In order to maintain clarity, we will also go over similar ideas that frequently come up in mathematical settings involving vectors, such as the transpose of a matrix and the conjugate transpose of a matrix. 

1.0What is a Scalar Triple Product?

As per the Scalar Triple Product Definition, the scalar triple product involves three vectors A, B, and C, defined as:

[a b c ] = ( a × b) . c

Now, from the above formula, the following conclusions can be obtained:

i) The result is always a scalar quantity.

ii) The cross product of the vectors is computed first, followed by the dot product, which provides the scalar triple product.

iii) The physical importance of the scalar triple product formula is presented as a measure of the volume of the parallelepiped whose 3 coterminous edges are the three vectors a, b and c.

2.0Properties of Scalar Triple Product

• Order Invariance: The scalar triple product is invariant under cyclic permutations:
  A⋅ (B×C) = B⋅ (C×A) = C⋅ (A×B) 
• Geometric Significance: The absolute value obtained via the scalar triple product represents the volume (V) of the parallelepiped spanned by the three vectors.
• Zero Value: If the scalar triple product equals zero, it indicates that the vectors are coplanar.

The volume of a parallelepiped is determined by the triple product. If it is 0, then only when any one of the three vectors has zero magnitude can such a situation occur. The plane that contains a and b is perpendicular to the direction of the cross product of a and b. Therefore, only when vector c is also in the same plane will the dot product of the resultant with c be zero. This is due to the fact that the resultant and C will have an angle of 90° and cos 90°.

Thus, we can quickly determine the volume of a particular parallelepiped by using the scalar triple product.

3.0Geometrical Interpretation of Scalar Triple Product

Given any three vectors, a, b, and c, we now know that the scalar triple product is defined as a · (b × c), which is also equivalent to the determinant of the three vectors' components. Now, let's examine the geometric meaning of the scalar triple product.

As seen in this picture, the coterminous edges serve as representations of the three vectors. The area of the base is obtained by taking the cross-product of vectors a and b. Additionally, the direction of the cross product of vectors is perpendicular to them both. The component of vector c in the direction of the cross product of a and b, in this instance, provides the height because volume is the product of area and height. The component may be found using c cos α.

Therefore, if the coterminous edges of a Parallelepiped are represented by three vectors, a, b, and c, then,

The volume of parallelepiped = ( a × b) c cos α =  ( a × b). c

Where α is the angle between  ( a × b)  and c.

4.0Scalar Triple Product: Solved Problems

Problem 1: Calculate A⋅ (B×C).

If given vectors are A = (5,7,8), B = (9,6,2), C = (6,8,4).  

Solution:

Step 1: Compute B × C

$\mathbf{B}\times \mathbf{C}=\left|\begin{array}{lll}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 9& 6& 2\\ 6& 8& 4\end{array}\right|$

= i (6×4 − 8×2) − j (9×4 − 6×2) + k (9×8−6×6)

= i (8) − j (24) + k (36) = 8i – 24j + 36k

Step 2: Compute A⋅ (B×C)

= A⋅ (8i – 24j + 36k) 

= (5i+7j+8k)(8i – 24j + 36k)

= 5(8) + 7(–24) + 8(36)

= 40 – 168 + 288 = 40

Problem 2: Evaluate the volume (V) of a parallelepiped whose coterminous edges are given as 5i – 3j + 4k, 7i + 5j – 9k, & –6i – 5j + 7k.

Solution:

To find the volume (V) of the parallelepiped with edges given as 5i – 3j + 4k, 7i + 5j – 9k, & –6i – 5j + 7k, we need to evaluate its scalar triple product Using three coterminous edges.

[5i – 3j + 4k, 7i + 5j – 9k, –6i – 5j + 7k]

=$\left|\begin{array}{ccc}i& j& k\\ 5& -3& 4\\ 7& 5& -9\\ -6& -5& 7\end{array}\right|$

= 5(35 - 45) + 3(49 - 54) + 4(-35 - (-30))

= -50 - 15 - 20 = 85

Hence, the volume of the parallelepiped = 85 units

Problem 3: Three vectors, a, b, and c, represent the directions of three roads meeting at a point O in a city. The vectors are given a = 3i+2j+4k, b = i+4j−3k, c = 2i−j+3k. Find the angle between the vectors a, b, and c using the scalar triple product.

Solution: To calculate the angle between the three vectors using the scalar triple product, the following formula can be used: 

$cos\alpha =\frac{|a\cdot (b\times c)|}{|a||b||c|}$

The cross product:

$b\times c=\left|\begin{array}{ccc}i& j& k\\ 1& 4& -3\\ 2& -1& 3\end{array}\right|$

$b\times c=i(12-3)-j(3+6)+k(-1-8)$

$b\times c=9i-9j-9k$

Triple product of

$a\cdot (b\times c)=(3i+2j+4k)\cdot (9i-9j-9k)$

$a\cdot (b\times c)=9\times 3+2(-9)+4(-9)$

$a\cdot (b\times c)=27-18-36=-27$

Magnitudes of all the three vectors 

$|a|=\sqrt{(3^2+2^2+4^2)}=\sqrt{29}$

$|b|=\sqrt{(1^2+4^2+(-3{)}^2)}=\sqrt{26}$

$|c|=\sqrt{(2^2+(-1{)}^2+(3{)}^2)}=\sqrt{14}$

Now, let’s substitute the values in the above formula: 

$cos\alpha =\frac{|-27|}{|\sqrt{29}||\sqrt{26}||\sqrt{14}|}$

$cos\alpha =\frac{|-27|}{|\sqrt{29\times 26\times 14}|}$

$cos\alpha =\frac{27}{\sqrt{10588}}$

$cos\alpha =\frac{27}{102.9}\approx 0.262$

$or\alpha =co{s}^{-1}(0.262)\approx 74.5$

5.0Applications of Scalar Triple Product

1. Volume Calculation: Used to compute the volume of a parallelepiped formed by three vectors.
2. Coplanarity Check: A zero scalar triple product indicates coplanar vectors.
3. Physics: Helps in torque and angular momentum calculations.

By understanding concepts like the transpose of a matrix and conjugate transpose of a matrix, students can approach vector problems with confidence and connect them to broader mathematical frameworks.