Question

a. Show that \((\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}+\mathbf{v})=|\mathbf{u}|^{2}+2 \mathbf{u} \cdot \mathbf{v}+|\mathbf{v}|^{2}\). b. Show that \((\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}+\mathbf{v})=|\mathbf{u}|^{2}+|\mathbf{v}|^{2}\) if \(\mathbf{u}\) is perpendicular to \(\mathbf{v}\). c. Show that \((\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}-\mathbf{v})=|\mathbf{u}|^{2}-|\mathbf{v}|^{2}\).

Short answer

Question: Prove the given equalities involving vector dot products and magnitudes for the expressions (a) \((\mathbf{u}+\mathbf{v})\cdot(\mathbf{u}+\mathbf{v})\), (b) \((\mathbf{u}+\mathbf{v})\cdot(\mathbf{u}+\mathbf{v})\) when \(\mathbf{u}\) is perpendicular to \(\mathbf{v}\), and (c) \((\mathbf{u}+\mathbf{v})\cdot(\mathbf{u}-\mathbf{v})\).

Step-by-step solution

a. Show that \((\mathbf{u}+\mathbf{v})\cdot(\mathbf{u}+\mathbf{v})=|\mathbf{u}|^{2}+2 \mathbf{u} \cdot\mathbf{v}+|\mathbf{v}|^{2}\).

First, we start by taking the dot product of \((\mathbf{u}+\mathbf{v})\) with itself: \((\mathbf{u}+\mathbf{v})\cdot(\mathbf{u}+\mathbf{v}) = (\mathbf{u}+\mathbf{v})\cdot\mathbf{u} + (\mathbf{u}+\mathbf{v})\cdot\mathbf{v}.\) Next, we distribute the dot product for both terms: \(\mathbf{u}\cdot\mathbf{u} + \mathbf{u}\cdot\mathbf{v} + \mathbf{v}\cdot\mathbf{u} + \mathbf{v}\cdot\mathbf{v}.\) Since the dot product is commutative, \(\mathbf{u}\cdot\mathbf{v} = \mathbf{v}\cdot\mathbf{u}\), and we can rewrite the expression as: \(|\mathbf{u}|^{2} + \mathbf{u}\cdot\mathbf{v} + \mathbf{u}\cdot\mathbf{v} + |\mathbf{v}|^{2} = |\mathbf{u}|^{2} + 2\mathbf{u}\cdot\mathbf{v} + |\mathbf{v}|^{2}.\)

b. Show that \((\mathbf{u}+\mathbf{v})\cdot(\mathbf{u}+\mathbf{v})=|\mathbf{u}|^{2}+|\mathbf{v}|^{2}\) if \(\mathbf{u}\) is perpendicular to \(\mathbf{v}\).

From part (a), we know that \((\mathbf{u}+\mathbf{v})\cdot(\mathbf{u}+\mathbf{v}) = |\mathbf{u}|^{2} + 2\mathbf{u}\cdot\mathbf{v} + |\mathbf{v}|^{2}\). If \(\mathbf{u}\) is perpendicular to \(\mathbf{v}\), their dot product is \(0\). Therefore, \(\mathbf{u}\cdot\mathbf{v} = 0\). Setting \(\mathbf{u}\cdot\mathbf{v}\) to \(0\) in the equation from part (a), we get: \(|\mathbf{u}|^{2} + 2(0) + |\mathbf{v}|^{2} = |\mathbf{u}|^{2} + |\mathbf{v}|^{2}\).

c. Show that \((\mathbf{u}+\mathbf{v})\cdot(\mathbf{u}-\mathbf{v})=|\mathbf{u}|^{2}-|\mathbf{v}|^{2}\).

We take the dot product of \((\mathbf{u}+\mathbf{v})\) and \((\mathbf{u}-\mathbf{v})\): \((\mathbf{u}+\mathbf{v})\cdot(\mathbf{u}-\mathbf{v}) = (\mathbf{u}+\mathbf{v})\cdot\mathbf{u} - (\mathbf{u}+\mathbf{v})\cdot\mathbf{v}.\) Now, distribute the dot product for both terms: \(\mathbf{u}\cdot\mathbf{u} + \mathbf{u}\cdot\mathbf{v} - \mathbf{v}\cdot\mathbf{u} - \mathbf{v}\cdot\mathbf{v}\). Again, the dot product is commutative, so \(\mathbf{u}\cdot\mathbf{v} = \mathbf{v}\cdot\mathbf{u}\). Therefore, we get: \(|\mathbf{u}|^{2} + \mathbf{u}\cdot\mathbf{v} - \mathbf{u}\cdot\mathbf{v} - |\mathbf{v}|^{2} = |\mathbf{u}|^{2} - |\mathbf{v}|^{2}\).

Key concepts

Vector Magnitude

Understanding vector magnitude is crucial in vector algebra, especially when dealing with concepts like the dot product. The vector magnitude or length of a vector \(\mathbf{u}\) is denoted by \(|\mathbf{u}|\). It represents the distance from the origin to the point defined by the vector in space. The formula to calculate this is \(|\mathbf{u}| = \sqrt{u_1^2 + u_2^2 + ... + u_n^2}\), which means summing the squares of its components and then taking the square root.

This principle is similar to calculating the hypotenuse in the Pythagorean theorem. Knowing the magnitude allows for a deeper understanding of how vectors interact, especially when they are combined or compared.

For example, in finding \((\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v})\), the magnitude of vectors \(\mathbf{u}\) and \(\mathbf{v}\) is useful for simplifying the expression, visually understanding it as a form of vector addition and measures of change.

Perpendicular Vectors

Perpendicular vectors, also termed orthogonal vectors, have a special relationship in vector algebra: their dot product is zero. If two vectors \(\mathbf{u}\) and \(\mathbf{v}\) are perpendicular, then \(\mathbf{u} \cdot \mathbf{v} = 0\).

This is essential in simplifying calculations and proving vector-related equations. For instance, in the problem section (b), the expression \((\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v})\), simplifies significantly when the vectors are perpendicular. This means removing the middle term related to the dot product of these vectors from the equation, as it effectively becomes zero.

• It provides a much simpler expression.
• Illustrates how vector relationships can reduce complex algebraic computations.
• Facilitates understanding geometric concepts purely from algebra.

Recognizing perpendicular vectors also helps in various applications like physics, where they can represent forces in equilibrium.

Algebraic Proofs

Algebraic proofs involve showing step-by-step logical deductions to establish the truth of mathematical statements. When working with vectors, these proofs often require manipulating dot products and magnitudes.

Take the example of proving \((\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v}) = |\mathbf{u}|^{2} + 2 \mathbf{u} \cdot \mathbf{v} + |\mathbf{v}|^{2}\). The process involves:

• Expanding using distribution law: Each component multiplied separately, like \(\mathbf{u}\) with itself and \(\mathbf{v}\).
• Applying commutative property of dot products: Such as \(\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}\).
• Combining terms based on dot product results: Recognizing that dot products of identical vectors yield their magnitude squared.

Algebraic proofs not only confirm the initial equations but also provide deeper insight into the inherent symmetries and properties of the vectors and their operations.