Proving the Vector Identity ( mathbf{u}mathbf{v}^2 - mathbf{u} - mathbf{v}^2 4mathbf{u} cdot mathbf{v} )

Introduction to Vector Identities and Dot Product

Vector identities play a crucial role in linear algebra and have widespread applications in various fields such as physics, engineering, and computer science. One such identity is proving the vector identity

uv22?u?v224u?v

while making use of the properties of the dot product and norms. This article will guide you through the proof of this identity, offering insights and practical steps.

Defining the Dot Product and Norm

Before diving into the proof, it is essential to understand the basic concepts of the dot product and the norm. The dot product, denoted by

u?v∑i1nu?i?v?i, computes the scalar product of two vectors. The norm of a vector is defined as

‖u‖u?u

representing its length or magnitude.

Proof of the Vector Identity

Step 1: Expanding (mathbf{u}mathbf{v}^2)

Let's begin by expanding the term

uv22uv?uvu?u?2u?v?v?v h3. Step 2: Expanding (mathbf{u} - mathbf{v}^2)

Next, we expand the term

u-v22u-v?u-vu?u-2u?v-v?v

Subtracting the two expansions, we get

uv22-(u-v22)(u22?2u?v?v22)-(u22-2u?v-v22)

After simplifying, we find that the

u22?v22-u224u?v

Conclusion

Thus, we have shown that the vector identity (mathbf{u}mathbf{v}^2 - mathbf{u} - mathbf{v}^2 4mathbf{u} cdot mathbf{v}) holds true, proving the initial equation. This proof is not only a valuable mathematical exercise but also an application of fundamental concepts in linear algebra.

Additional Insights

It is interesting to note that if you were to add the vectors instead, you would get the parallelogram identity:

u?v (u-v)22u?2v in the plane, it expresses the relationship that the sum of the squares of the diagonals of a parallelogram equals the sum of the squares of the sides. This identity has applications in geometry and mechanics, where understanding vector algebra is crucial.