Problem 1 - Differential Equations and Linear Algebra

Question

Use the standard inner product in \(\mathbb{R}^{5}\) to determine the angle between the vectors \(\mathbf{v}=(0,-2,1,4,1)\) and \(\mathbf{w}=(-3,1,-1,0,3)\)

Step-by-Step Solution

Answer

The angle between the vectors $\mathbf{v}=(0,-2,1,4,1)$ and $\mathbf{w}=(-3,1,-1,0,3)$ in $\mathbb{R}^{5}$ is $\theta = \frac{\pi}{2}$ or $90^\circ$.

Step 1: Find the inner product of the given vectors

To find the inner product of \(\mathbf{v}\) and \(\mathbf{w}\), we will use the standard formula for inner products of two vectors in \(\mathbb{R}^{5}\): \[\mathbf{v} \cdot \mathbf{w} = v_1 w_1 + v_2 w_2 + v_3 w_3 + v_4 w_4 + v_5 w_5.\] So, for \(\mathbf{v}=(0,-2,1,4,1)\) and \(\mathbf{w}=(-3,1,-1,0,3)\), \[\begin{aligned} \mathbf{v} \cdot \mathbf{w} &= (0)(-3) + (-2)(1) + (1)(-1) + (4)(0) + (1)(3) \\ &= 0 - 2 - 1 + 0 + 3 \\ &= 0.\end{aligned}\]

Step 2: Find norms of the given vectors

Now let's find the norms of the vectors \(\mathbf{v}\) and \(\mathbf{w}\) using the formula for the norm of a vector in \(\mathbb{R}^5\): \[\|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2 + v_4^2 + v_5^2}.\] For vector \(\mathbf{v}=(0,-2,1,4,1)\), we have \[\begin{aligned} \|\mathbf{v}\| &= \sqrt{(0)^2 + (-2)^2 + (1)^2 + (4)^2 + (1)^2} \\ &= \sqrt{0 + 4 + 1 + 16 + 1} \\ &= \sqrt{22}.\end{aligned}\] Similarly, for vector \(\mathbf{w}=(-3,1,-1,0,3)\), we have \[\begin{aligned} \|\mathbf{w}\| &= \sqrt{(-3)^2 + (1)^2 + (-1)^2 + (0)^2 + (3)^2} \\ &= \sqrt{9 + 1 + 1 + 0 + 9} \\ &= \sqrt{20}.\end{aligned}\]

Step 3: Calculate the angle between the vectors

Now we have all the components we need to find the angle \(\theta\): the inner product \(\mathbf{v} \cdot \mathbf{w} = 0\), and the norms \(\|\mathbf{v}\| = \sqrt{22}\), \(\|\mathbf{w}\| = \sqrt{20}\). Using the formula \(\mathbf{v} \cdot \mathbf{w} = \|\mathbf{v}\| \|\mathbf{w}\| \cos(\theta)\), we get: \[\begin{aligned} \cos(\theta) &= \frac{\mathbf{v} \cdot \mathbf{w}}{\|\mathbf{v}\| \|\mathbf{w}\|} \\ &= \frac{0}{\sqrt{22}\sqrt{20}} \\ &= 0. \end{aligned}\] Since \(\cos(\theta) = 0\), this means that the angle between the vectors \(\mathbf{v}\) and \(\mathbf{w}\) is \(\theta = \frac{\pi}{2}\) or \(90^\circ\).

Key Concepts

Inner Product

The inner product, also known as the dot product, is a fundamental operation in vector algebra. It takes two vectors and returns a scalar quantity. This operation is essential for various applications, including finding the angle between vectors, which is precisely what we’re aiming for in this exercise.

In a 5-dimensional vector space \(\mathbb{R}^5\), the inner product of two vectors \(\mathbf{v}\) and \(\mathbf{w}\) is given by the sum of the products of their corresponding components. That is, \[\mathbf{v} \cdot \mathbf{w} = v_1 w_1 + v_2 w_2 + v_3 w_3 + v_4 w_4 + v_5 w_5.\] If the result of this operation is zero, as in our example, it indicates that the vectors are orthogonal to each other, meaning they form a right angle (90 degrees).

Norm of a Vector

The norm of a vector, often referred to as the vector's magnitude or length, is a measure of its size. To find a vector's norm in \(\mathbb{R}^5\), we use the formula \[\|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2 + v_4^2 + v_5^2}.\] It’s the square root of the sum of the squares of its components, which is analogous to the Pythagorean theorem in higher dimensions.

This metric is vital when comparing vectors or determining their scaling, and it plays a critical role in calculating the angle between two vectors. The length of a vector provides context to the inner product and helps normalize the vectors for various operations.

Cosine of Angle

The cosine of the angle between two vectors is derived from the inner product and the norms of the vectors. It’s a critical value that can tell us about the orientation of the vectors with respect to one another. Specifically, the cosine formula for the angle \(\theta\) between two nonzero vectors is \[\cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{w}}{\|\mathbf{v}\| \|\mathbf{w}\|}.\] This ratio effectively measures the relative directionality of the vectors.

When \(\cos(\theta)\) is 0, as demonstrated in our exercise, it means that \(\theta = \frac{\pi}{2}\) radians, or 90 degrees, indicating the vectors are perpendicular. The cosine value ranges from -1 to 1, where 1 indicates parallel vectors in the same direction and -1 indicates parallel vectors in opposite directions.

Vector Space \(\mathbb{R}^5\)

A vector space, such as \(\mathbb{R}^5\), is a mathematical construct that contains all possible 5-dimensional vectors. Each vector in this space has five components, which can represent any five-dimensional quantities. As an abstract concept, vector spaces follow particular axioms, such as vector addition and scalar multiplication.

In the context of our problem, \(\mathbb{R}^5\) is the domain of the vectors we're dealing with. Understanding vector spaces allows mathematicians and scientists to generalize problems in multiple dimensions and apply consistent rules for vector operations. This space underpins many mathematical and physical applications, from computer graphics to quantum mechanics.