Of all the concepts under vector algebra, the vector triple product is one that involves the cross-product of three vectors. A fact that may be noticed and used when calculating the vector triple product is that the cross product of a specific vector and the result of the cross product of the other two vectors must be calculated. In applications including mechanics, electromagnetism, and geometry, it is a vector operator which helps in the expression of vector connections and streamlines challenging issues. It is widely used in engineering and physics.
The vector triple product involves three vectors - A, B and C. It comes from taking the cross-product of A with the cross-product of B and C. The vector triple product is expressed as:
A × (B × C)
Vector Triple Product Formula is:
A × ( B × C) = (A⋅ C) B − (A⋅ B) C
The vector triple product follows the BAC-CAB Rule. This rule aids weight calculations because it enables numerous vector operations to be done, such as numerous scalar and vector multiplication.
To prove the BAC-CAB Rule, let:
B × C = D = (Dx , Dy, Dz)
Now, compute A × D using the determinant form of the cross product:
A x D = [ i j k ]
[Ax Ay Az]
[Dx Dy Dz ]
Expanding and substituting D in terms of B and C, the terms rearrange into:
A × (B × C) = (A ⋅ C) B− (A ⋅ B) C
This proves the vector triple product formula.
The vector triple product has unique properties that distinguish it from other vector operations:
± (A × B) × C / ∣ (A × B) × C ∣
A × (B × C) = (A × B) × C
This distinction is crucial in computations.
These properties highlight the versatility and utility of the vector triple product in advanced vector operations.
Example 1: Simplify a Vector Triple Product
Simplify A × ( B × C) for A = i + 2j − k, B = 3i − j + 2k and C= 2i + j − k
Solution:
Using the BAC - CAB Rule:
A ×( B × C) = (A · C) B − (A ⋅ B)C
Step 1: Compute A⋅ C:
A ⋅ C = (1) (2) + (2) (1) + (−1) (−1) = 2 + 2 + 1 = 5
Step 2: Compute A ⋅ B:
A ⋅ B = (1) (3) + (2) (−1) + (−1) (2) = 3−2−2 = −1
Step 3: Substitute values:
A × (B × C) = (5) B − (−1)C
= 5 (3i − j + 2k) + C
= (15i − 5j + 10k) + (2i + j − k)
= 17i − 4j + 9k
Answer: A × (B × C) = 17i − 4j + 9k.
Example 2: If A × B = C, B × C = A and A, B, and C be moduli of the vectors A, B, and C, respectively, then find the values of A & B.
Solution:
Magnitudes of both sides
For first equation
AB=C ABsin=C …….(1)
For the second equation
BC=A BCsin=A …….(2)
Substituting the value of C from Equation 1 into Equation 2
B(ABsin)sin=A
AB2sinsin=A B2sinsin=1
∴ A is perpendicular to both B & C, and C is perpendicular to both A & B.
Therefore, A, B, & C are mutually perpendicular. Then sin=sin=1(as ==90)
B2 = 1 B = 1
From equation 1
A1sin90=C
A = C
The vector triple product is such an important tool in vector algebra, letting them simplify complex vector relationships in a three-dimensional space. Its proof based on the BAC-CAB rule presents it as elegant and practical. Mastering concepts like the example of the vector triple product lets students and professionals resolve their problems in physics, geometry, and engineering problems.
(Session 2025 - 26)