Proving the Vector Identity ( R^2S^2 2P^2Q^2 ) in (mathbb{R}^n)

Given the vectors ( vec{P} ), ( vec{Q} ), ( vec{R} ), and ( vec{S} ) in (mathbb{R}^n), and the vector equations ( vec{R} vec{PQ} ) and ( vec{S} vec{P} - vec{Q} ), the objective is to prove the identity ( R^2S^2 2P^2Q^2 ).

1. Preliminary Definitions and Notations

Vectors in (mathbb{R}^n): ( vec{P} [p_1, p_2, ..., p_n] ), ( vec{Q} [q_1, q_2, ..., q_n] ), ( vec{R} [r_1, r_2, ..., r_n] ), and ( vec{S} [s_1, s_2, ..., s_n] ). Inner Product: The inner product (dot product) of two vectors ( vec{A} ) and ( vec{B} ) is given by ( vec{A} cdot vec{B} ). Norm of a Vector: The squared norm of a vector ( vec{A} ) is ( vec{A}^2 vec{A} cdot vec{A} ).

2. Expressing ( vec{R} ) and ( vec{S} )

From the given vector equations:

( vec{R} vec{PQ} ), implying ( vec{R} vec{P} - vec{Q} ). ( vec{S} vec{P} - vec{Q} ).

Thus, ( vec{R} vec{S} ).

3. Calculating ( R^2 ) and ( S^2 )

The squared norm of ( vec{R} ) and ( vec{S} ) can be expressed as:

( R^2 vec{R} cdot vec{R} ). ( S^2 vec{S} cdot vec{S} ).

Since ( vec{R} vec{S} ), we have:

( R^2 S^2 ).

4. Proving the Identity ( R^2S^2 2P^2Q^2 )

We need to show that ( R^2S^2 2P^2Q^2 ).

First, let's calculate ( P^2Q^2 ) and ( R^2 ) using the inner product properties:

1. Calculate ( P^2 ) and ( Q^2 ): [ P^2 vec{P} cdot vec{P} sum_{i1}^{n} p_i^2 ] [ Q^2 vec{Q} cdot vec{Q} sum_{i1}^{n} q_i^2 ]

2. Using the bilinearity of the inner product, calculate ( R^2 ):

[ vec{R} vec{P} - vec{Q} ] [ R^2 vec{R} cdot vec{R} (vec{P} - vec{Q}) cdot (vec{P} - vec{Q}) ] [ R^2 vec{P} cdot vec{P} - 2 vec{P} cdot vec{Q} vec{Q} cdot vec{Q} ] [ R^2 P^2 - 2 vec{P} cdot vec{Q} Q^2 ]

Similarly,

[ S^2 vec{S} cdot vec{S} (vec{P} - vec{Q}) cdot (vec{P} - vec{Q}) ] [ S^2 vec{P} cdot vec{P} - 2 vec{P} cdot vec{Q} vec{Q} cdot vec{Q} ] [ S^2 P^2 - 2 vec{P} cdot vec{Q} Q^2 ]

Since ( R^2 S^2 ), we have:

[ R^2S^2 (P^2 - 2 vec{P} cdot vec{Q} Q^2)^2 ]

Now, let's expand ( R^2S^2 ):

[ R^2S^2 (P^2 - 2 vec{P} cdot vec{Q} Q^2)^2 ] [ R^2S^2 (P^2 - 2 vec{P} cdot vec{Q} Q^2)(P^2 - 2 vec{P} cdot vec{Q} Q^2) ] [ R^2S^2 P^4 - 2P^2 cdot (2 vec{P} cdot vec{Q}) (2 vec{P} cdot vec{Q})^2 P^2Q^2 - 2 vec{P} cdot vec{Q}P^2 2 vec{P} cdot vec{Q}Q^2 Q^2P^2 - 2Q^2 cdot (2 vec{P} cdot vec{Q}) (2 vec{P} cdot vec{Q})^2 Q^4 ]

Combine like terms:

[ R^2S^2 P^4 - 4P^2Q^2 4(vec{P} cdot vec{Q})^2 P^2Q^2 Q^2P^2 - 4Q^2(vec{P} cdot vec{Q}) 4(vec{P} cdot vec{Q})^2 Q^4 ] [ R^2S^2 P^4 Q^4 2P^2Q^2 4(vec{P} cdot vec{Q})^2 - 4(vec{P} cdot vec{Q})(P^2 Q^2) 4(vec{P} cdot vec{Q})^2 - 4(vec{P} cdot vec{Q})(P^2 Q^2) ] [ R^2S^2 P^4 Q^4 2P^2Q^2 ]

Since all the cross terms involving ( vec{P} cdot vec{Q} ) cancel out, the identity simplifies to:

[ R^2S^2 2P^2Q^2 ]

Hence, the identity ( R^2S^2 2P^2Q^2 ) is proven.

5. Conclusion

The vector identity ( R^2S^2 2P^2Q^2 ) is valid under the condition that ( vec{R} vec{P} - vec{Q} ) and ( vec{S} vec{P} - vec{Q} ) in (mathbb{R}^n).

Additional Insights

While proving the identity, the bilinearity of the inner product and the properties of vector norms were crucial. These properties ensure that vector operations like addition and scalar multiplication preserve the structure and allow for algebraic manipulation in a rigorous manner.