Problem 87 - Algebra and Trigonometry Real Mathematics, Real People

Question

Prove the following. $$\|\mathbf{u}-\mathbf{v}\|^{2}=\|\mathbf{u}\|^{2}+\|\mathbf{v}\|^{2}-2 \mathbf{u} \cdot \mathbf{v}$$

Step-by-Step Solution

Answer

The proof has been carried out by breaking down the left-hand side of the equation using the commutative property of scalar product and the identity \(\|\mathbf{u}\|^{2} = \mathbf{u} \cdot \mathbf{u}\)

Step 1 Expand the Norm

Expand the left-hand side. The norm of a vector squared is the scalar product of a vector with itself. So, \(\|\mathbf{u}-\mathbf{v}\|^{2}=(\mathbf{u}-\mathbf{v}) \cdot (\mathbf{u}-\mathbf{v})\).

Step 2 Distribute the Scalar Product

Distribute the scalar product: (\(\mathbf{u}-\mathbf{v}) \cdot (\mathbf{u}-\mathbf{v}) = \mathbf{u} \cdot \mathbf{u} - \mathbf{u} \cdot \mathbf{v} - \mathbf{v} \cdot \mathbf{u} + \mathbf{v} \cdot \mathbf{v}\). But the scalar product is commutative, meaning \(\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}\). So, the expression simplifies to \(\mathbf{u} \cdot \mathbf{u} - 2\mathbf{u} \cdot \mathbf{v} + \mathbf{v} \cdot \mathbf{v}\).

Step 3 Conclude with the Norm Identity

Notice that \(\mathbf{u} \cdot \mathbf{u} = \|\mathbf{u}\|^{2}\) and \(\mathbf{v} \cdot \mathbf{v} = \|\mathbf{v}\|^{2}\). Therefore, the expression can be rewritten as \(\|\mathbf{u}\|^{2} - 2\mathbf{u} \cdot \mathbf{v} + \|\mathbf{v}\|^{2}\), which is the right-hand side of the original equation. Thus, the given formula has been proved.

Key Concepts

Scalar Product in Vectors

The scalar product, also known as the dot product, is an essential operation in vector mathematics. It combines two vectors into a single number, called a scalar. The scalar product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is denoted by \( \mathbf{a} \cdot \mathbf{b} \), and is calculated as follows:

• Each corresponding component of the vectors is multiplied together.
• The resulting products are summed up to produce the final scalar.

Mathematically, for vectors in three dimensions \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \), the scalar product is:\[\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3\]This operation is widely used because it provides valuable information about the vectors, such as their angle or their orthogonality. For instance, if the result is zero, the vectors are perpendicular to each other.

Vector Norm and Dot Product

Vector norms measure the size, or length, of a vector. Specifically, the norm of a vector \( \mathbf{u} \), denoted as \( \|\mathbf{u}\| \), is the square root of the scalar product of the vector with itself. It is calculated as:\[\|\mathbf{u}\| = \sqrt{\mathbf{u} \cdot \mathbf{u}}\]The square of the norm, \( \|\mathbf{u}\|^2 \), simplifies to the scalar product \( \mathbf{u} \cdot \mathbf{u} \), providing a convenient way to express vector-related equations in terms of their dot products. This is useful in various applications like physics and engineering, where delimiting precise magnitudes or distances is crucial.
In the context of the given exercise, the vector norm helps express the distance between two vectors \( \mathbf{u} \) and \( \mathbf{v} \), encapsulated by the expression:\[\|\mathbf{u} - \mathbf{v}\|^2 = (\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} - \mathbf{v})\]Breaking this down using the distributive property of the dot product reveals familiar components such as \( \|\mathbf{u}\|^2 \) and \( \|\mathbf{v}\|^2 \).

Commutative Property of Dot Product

An important property of the dot product is its commutative nature. This means that the order in which you multiply the vectors does not matter. For vectors \( \mathbf{u} \) and \( \mathbf{v} \), the dot product satisfies:\[\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}\]This symmetry is a powerful trait, simplifying the manipulation of expressions involving vectors. In the given exercise, when distributing the dot product within the expression \( (\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} - \mathbf{v}) \), commutativity allows replacement of terms like \( \mathbf{u} \cdot \mathbf{v} \) with \( \mathbf{v} \cdot \mathbf{u} \).
This step is crucial for simplifying the expression to show the equivalence \( \|\mathbf{u}\|^2 - 2\mathbf{u} \cdot \mathbf{v} + \|\mathbf{v}\|^2 \). Understanding commutativity is key in vector algebra to rearrange terms easily, illustrate equivalences, and spot implicit symmetries in vector operations.