Problem 1 - Differential Equations and Linear Algebra

Question

Determine the angle between the given vectors \(\mathbf{u}\) and \(\mathbf{v}\) using the standard inner product on \(\mathbb{R}^{n}\). \(\mathbf{u}=(2,3)\) and \(\mathbf{v}=(4,-1).\)

Step-by-Step Solution

Answer

The angle between the given vectors \(\mathbf{u} = (2,3)\) and \(\mathbf{v} = (4,-1)\) is \(\theta = \arccos\left(\frac{5}{\sqrt{13}\sqrt{17}}\right)\).

Step 1: Calculate the inner product

To calculate the inner product between vectors \(\mathbf{u}=(2,3)\) and \(\mathbf{v}=(4,-1)\), we simply multiply their corresponding components and sum the results: \[\langle\mathbf{u},\mathbf{v}\rangle = (2)(4) + (3)(-1) = 8 - 3 = 5\]

Step 2: Calculate the magnitudes

Now we calculate the magnitude of each vector: \[|\mathbf{u}| = \sqrt{(2)^2 + (3)^2} = \sqrt{4 + 9} = \sqrt{13}\] \[|\mathbf{v}| = \sqrt{(4)^2 + (-1)^2} = \sqrt{16 + 1} = \sqrt{17}\]

Step 3: Find the angle using the relation between the inner product and angle

The relation between the inner product and angle between two vectors is given by: \[\langle\mathbf{u},\mathbf{v}\rangle = |\mathbf{u}||\mathbf{v}|\cos{\theta}\] Where \(\theta\) is the angle between the two vectors. Using our computed values, we have: \[5 = \sqrt{13}\sqrt{17}\cos{\theta}\] To find the angle, divide both sides by the magnitudes and solve for \(\cos{\theta}\): \[\cos{\theta} = \frac{5}{\sqrt{13}\sqrt{17}}\] Now, find the angle \(\theta\) by taking the inverse cosine of \(\cos{\theta}\): \[\theta = \arccos\left(\frac{5}{\sqrt{13}\sqrt{17}}\right)\] Thus, the angle between the given vectors \(\mathbf{u}\) and \(\mathbf{v}\) is \(\theta = \arccos\left(\frac{5}{\sqrt{13}\sqrt{17}}\right)\).

Key Concepts

Inner Product

The inner product, also known as the dot product, is a way to multiply two vectors in order to gain a single scalar value. It reveals the extent to which one vector goes in the direction of another. Consider vectors \( \mathbf{u} = (2,3) \) and \( \mathbf{v} = (4,-1) \). The inner product formula involves multiplying their corresponding components and adding the results:

• First component: \( 2 \times 4 = 8 \)
• Second component: \( 3 \times (-1) = -3 \)

Add these results: \( 8 + (-3) = 5 \).
This value, 5, is the inner product of vectors \( \mathbf{u} \) and \( \mathbf{v} \). It signifies how aligned these vectors are, where values closer to zero indicate they are more perpendicular and larger values suggest more alignment. A positive result hints at a generally forward relationship direction between components.

Magnitude of Vectors

The magnitude of a vector represents its length, meaning how long the vector actually is in space. It's a way to gauge the vector’s size. Calculating magnitude is like finding the hypotenuse of a right triangle formed by vector components.For vector \( \mathbf{u} = (2,3) \):

• Square each component: \( 2^2 = 4 \) and \( 3^2 = 9 \)
• Add the squares: \( 4 + 9 = 13 \)
• Take the square root: \( \sqrt{13} \)

Thus, the magnitude of \( \mathbf{u} \) is \( \sqrt{13} \).
Perform the same steps for vector \( \mathbf{v} = (4,-1) \):

• \( 4^2 = 16 \)
• \((-1)^2 = 1 \)
• Add these squares: \( 16 + 1 = 17 \)
• Take the square root: \( \sqrt{17} \)

The magnitude of \( \mathbf{v} \) is \( \sqrt{17} \).
The magnitude helps us understand how "long" each vector is, and this will play a crucial role when finding the angle between them.

Cosine of Angle Between Vectors

To find the angle between two vectors, the cosine of that angle provides valuable information about their spatial relationship. This is derived from the formula involving the inner product and magnitudes of the vectors:\[\langle \mathbf{u}, \mathbf{v} \rangle = |\mathbf{u}| |\mathbf{v}| \cos{\theta}\]This relationship shows how the cosine of the angle \( \theta \) can be extracted, which clarifies the angular situation between the vectors.Using our calculated inner product of 5, and magnitudes of \( \sqrt{13} \) and \( \sqrt{17} \), we structure the equation as follows:\[\cos{\theta} = \frac{5}{\sqrt{13} \cdot \sqrt{17}}\]The value \( \frac{5}{\sqrt{13}\sqrt{17}} \) relates to the cosine of the angle between \( \mathbf{u} \) and \( \mathbf{v} \), providing us a dimensionless number between -1 (vectors pointing in opposite directions) and 1 (vectors perfectly aligned).
Acquiring \( \theta \) itself requires applying the inverse cosine function:\[\theta = \arccos\left(\frac{5}{\sqrt{13} \cdot \sqrt{17}}\right)\]This arc cosine function will yield the actual angle in degrees or radians, thereby fully characterizing the spatial relation between \( \mathbf{u} \) and \( \mathbf{v} \). The procedure highlights the tight interconnection between vector magnitudes, their inner product, and the resulting angle.