Problem 40 - Precalculus with Limits

Question

Finding the Angle Between Two Vectors In Exercises \(31-40,\) find the angle \(\theta\) between the vectors. $$\begin{aligned} \mathbf{u} &=\cos \left(\frac{\pi}{4}\right) \mathbf{i}+\sin \left(\frac{\pi}{4}\right) \mathbf{j} \\ \mathbf{v} &=\cos \left(\frac{\pi}{2}\right) \mathbf{i}+\sin \left(\frac{\pi}{2}\right) \mathbf{j} \end{aligned}$$

Step-by-Step Solution

Answer

The solution to find the angle \(\theta\) between the vectors \(\mathbf{u}\) and \(\mathbf{v}\) involves three essential steps. First, calculate the dot product of \(\mathbf{u}\) and \(\mathbf{v}\). Next, compute the magnitudes of vectors \(\mathbf{u}\) and \(\mathbf{v}\). Finally, calculate the angle using the formula \(\cos\theta = \frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|}\)

Step 1: Calculate Dot Product of vectors

The dot product of two vectors \(\mathbf{u}\) and \(\mathbf{v}\) is given by \(\mathbf{u}\cdot\mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos\theta\), where \(|\mathbf{u}|\) and \(|\mathbf{v}|\) are magnitudes of \(\mathbf{u}\) and \(\mathbf{v}\), and \(\theta\) is the angle between them. The dot product of \(\mathbf{u}\) and \(\mathbf{v}\) can also be written in terms of their components as \(\mathbf{u}\cdot\mathbf{v} = u_xv_x + u_yv_y\). Substituting the given vectors: \( \mathbf{u} =\cos \left(\frac{\pi}{4}\right) \mathbf{i}+\sin \left(\frac{\pi}{4}\right) \mathbf{j} \), \( \mathbf{v} =\cos \left(\frac{\pi}{2}\right) \mathbf{i}+\sin \left(\frac{\pi}{2}\right) \mathbf{j} \), their dot product becomes: \(\cos\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{2}\right) + \sin\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{2}\right)\)

Step 2: Calculate Magnitude of vectors

The magnitude of a vector is given by \(\sqrt{x^2+y^2}\). Thus, the magnitudes of vectors \(\mathbf{u}\) and \(\mathbf{v}\) are \(|\mathbf{u}| = \sqrt{u_x^2 + u_y^2} = \sqrt{\cos^2\left(\frac{\pi}{4}\right) + \sin^2\left(\frac{\pi}{4}\right)}\) and \(|\mathbf{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{\cos^2\left(\frac{\pi}{2}\right) + \sin^2\left(\frac{\pi}{2}\right)}\) respectively

Step 3: Calculate the Angle

Now, we can evaluate the angle between vectors \(\mathbf{u}\) and \(\mathbf{v}\) using the formula \(\cos\theta = \frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|}\) where \(\mathbf{u}\cdot\mathbf{v}\) is the dot product and \(|\mathbf{u}|\), \(|\mathbf{v}|\) are magnitudes of vectors \(\mathbf{u}\) and \(\mathbf{v}\) respectively. Substitute the values of \(\mathbf{u}\cdot\mathbf{v}\), \(|\mathbf{u}|\) and \(|\mathbf{v}|\) from step 1 and step 2.

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a way to multiply two vectors, yielding a scalar, rather than a vector. To compute the dot product of two vectors \( \mathbf{u} = u_x \mathbf{i} + u_y \mathbf{j} \) and \( \mathbf{v} = v_x \mathbf{i} + v_y \mathbf{j} \), you use the formula: \ \[ \mathbf{u} \cdot \mathbf{v} = u_x v_x + u_y v_y \] \ This formula requires you to multiply the corresponding components of the vectors and then add them up. It's really as straightforward as doing some simple multiplications and an addition.
The significance of the dot product is that it gives a measure of how much of one vector goes in the direction of another. For instance, if the vectors are perpendicular, their dot product will be zero, because they don't align at all.
In the exercise, to find the angle between vectors \( \mathbf{u} \) and \( \mathbf{v} \), you use their dot product as part of the calculation. In this specific case, \( \mathbf{u} \cdot \mathbf{v} = \cos\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{2}\right) + \sin\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{2}\right) \). As one of the cosine components is zero (since \( \cos\left(\frac{\pi}{2}\right) = 0 \)), it simplifies the calculation.

Vector Magnitude

The magnitude of a vector—sometimes called its length—is like measuring how long the vector is, in the context of the geometric space it inhabits. For a vector \( \mathbf{u} = u_x \mathbf{i} + u_y \mathbf{j} \), its magnitude is calculated using the Pythagorean theorem: \ \[ |\mathbf{u}| = \sqrt{u_x^2 + u_y^2} \] \ Think of the vector as a triangle where \( u_x \) and \( u_y \) act as the sides, and the magnitude is the hypotenuse.
Understanding vector magnitudes is critical when talking about vector normalization or when they are a part of physics where distance plays a role.
For the given vectors in the problem: \( |\mathbf{u}| = \sqrt{\cos^2\left(\frac{\pi}{4}\right) + \sin^2\left(\frac{\pi}{4}\right)} \), and this simply results in 1, because the sum of the squares of sine and cosine components is always 1. Likewise, \( |\mathbf{v}| = \sqrt{\cos^2\left(\frac{\pi}{2}\right) + \sin^2\left(\frac{\pi}{2}\right)} \), which also resolves to 1, given by similar trigonometric principles.

Trigonometric Functions

Trigonometric functions, like sine and cosine, are foundational in connecting angles to side lengths in right-angled triangles. They are pivotal in vector calculations, especially in determining directions and projections onto axes.
Cosine, \( \cos\theta \), gives the adjacent side length of a triangle, while sine, \( \sin\theta \), provides the opposite side in respect to the angle \( \theta \). In our vector context, these functions define the direction coefficients in the \( x \) and \( y \) axes respectively.
When finding the angle \( \theta \) between two vectors using their dot product and magnitudes, we use the formula: \ \[ \cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|} \] \ This uses the inverse cosine function, often called arccos, to find the actual angle. This relationship consolidates how trigonometric functions bridge geometry and algebra in vector spaces.
For the exercise, using the computed values, the angle \( \theta \) can directly be determined through \( \cos\theta = \frac{0 + 1}{1 \times 1} = 1 \). Therefore, \( \theta = \cos^{-1}(1) = 0 \text{ radians} \), which is intuitive as \( \mathbf{u} \) and \( \mathbf{v} \) were perfectly aligned with each other at their starting point.