Shouldn't the right side be |A|2 |B|2? The left side is a scalar (i.e., a number), so the right side needs to be a scalar as well.JasonHathaway said:Homework Statement
Proof that (A×B) . (B×A) + (A . B)^2= A^2 . B^2
JasonHathaway said:Homework Equations
A×(B×C)=(A . C)B - (A . B)C
The Attempt at a Solution
Assuming K=(A×B)
K . (B×A) + (A . B)^2 = A^2 . B^2
B . (A×K) + (A . B)^2 = A^2 . B^2
B . [A×(A×B)] + (A . B)^2 = A^2 . B^2
B . [(A . B)A - (A . A)B] + (A . B)(A . B) = A^2 . B^2
(A . B)(B . A) - (A . A)(B . B) + (A . B)(A . B) = A^2 . B^2
Strictly speaking, no, but I understand what you're trying to say. The product of a vector with itself (which you write as A2) is normally written as ##A \cdot A## or ##A \times A##, depending on which kind of product you mean. Another argument against A2 is that it can't be extended to, say, A3, because ##A \cdot A \cdot A## isn't defined. (The first dot product produces a scalar, which can't be dotted with a vector.)JasonHathaway said:Isn't the dot product of a vector (A for example) with itself is equal to A^2?
JasonHathaway said:I see that the idea behind this proof is to eliminate the (A . B)(B . A) and (A . B)(A . B)
The formula is (A×B) . (B×A) + (A . B)^2 = A^2 . B^2, where A and B are two vectors.
The formula is derived using the properties of vector multiplication and the distributive property. The proof involves expanding each side of the equation using these properties and then simplifying the expressions to show that they are equal.
This proof is important because it demonstrates the commutative property of the vector product. It also shows that the vector product is not only distributive but also commutative, making it a powerful tool in vector algebra.
Yes, this formula can be used in many real-world applications, such as in physics, engineering, and computer graphics. It can be used to calculate the work done by a force, the torque on a rotating object, or the direction of a magnetic field.
While this formula holds true for two-dimensional and three-dimensional vectors, it may not hold true for higher-dimensional vectors. Additionally, this formula only applies to the scalar product of vectors, not the vector product itself.