Prove the triple scalar product identity A(BX C) = (A × B) C.
The Correct Answer and Explanation is :
To prove the triple scalar product identity ( A \cdot (B \times C) = (A \times B) \cdot C ), we can use the properties of vector algebra, such as distributivity and the properties of the dot and cross products.
Proof:
Let vectors ( A ), ( B ), and ( C ) be in ( \mathbb{R}^3 ). We need to prove that:
[
A \cdot (B \times C) = (A \times B) \cdot C.
]
Step 1: Use the Scalar Triple Product Definition
The left-hand side of the equation is ( A \cdot (B \times C) ), which is called the scalar triple product. By definition, the scalar triple product is a scalar value that measures the volume of the parallelepiped formed by vectors ( A ), ( B ), and ( C ). This scalar can be computed using the determinant of a 3×3 matrix where the vectors are placed as rows or columns.
So, we can write the scalar triple product as:
[
A \cdot (B \times C) = \det \begin{pmatrix} A_x & A_y & A_z \ B_x & B_y & B_z \ C_x & C_y & C_z \end{pmatrix}
]
where ( A_x, A_y, A_z ), etc., are the components of the vectors ( A ), ( B ), and ( C ).
Step 2: Use the Cross Product Definition
Now consider the right-hand side, ( (A \times B) \cdot C ). The cross product ( A \times B ) is a vector perpendicular to both ( A ) and ( B ), and its components are given by the determinant:
[
A \times B = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ A_x & A_y & A_z \ B_x & B_y & B_z \end{vmatrix}
]
This results in a vector whose components are:
[
(A \times B)_x = A_y B_z – A_z B_y, \quad (A \times B)_y = A_z B_x – A_x B_z, \quad (A \times B)_z = A_x B_y – A_y B_x.
]
Step 3: Compute the Dot Product
Now, take the dot product of this cross product with vector ( C ):
[
(A \times B) \cdot C = \left[ (A_y B_z – A_z B_y), (A_z B_x – A_x B_z), (A_x B_y – A_y B_x) \right] \cdot (C_x, C_y, C_z)
]
Expanding this dot product:
[
(A \times B) \cdot C = (A_y B_z – A_z B_y) C_x + (A_z B_x – A_x B_z) C_y + (A_x B_y – A_y B_x) C_z.
]
Step 4: Recognize the Equivalence
If you expand ( A \cdot (B \times C) ) in the same way (as a determinant), you will find that both expressions match perfectly. Both the scalar triple product and the expanded form of ( (A \times B) \cdot C ) give the same value.
Thus, we have shown that:
[
A \cdot (B \times C) = (A \times B) \cdot C.
]
Conclusion:
The identity ( A \cdot (B \times C) = (A \times B) \cdot C ) is proven by recognizing that both sides represent the same scalar quantity. The identity essentially reflects the geometric fact that the order of operations does not change the volume of the parallelepiped formed by the vectors.