Problem 42 - Precalculus with Limits

Question

Finding the Angle Between Two Vectors In Exercises \(41-44\) , graph the vectors and find the degree measure of the angle \(\theta\) between the vectors. $$\mathbf{u}=6 \mathbf{i}+3 \mathbf{j}$$ $$\mathbf{v}=-4 \mathbf{i}+4 \mathbf{j}$$

Step-by-Step Solution

Answer

Calculations reveal that the angle between the two vectors is determined by \(\theta=\arccos{\frac{-12}{\sqrt{45}*\sqrt{32}}}\). This evaluates to an angle in radians, which can be converted to degrees using the conversion factor \(1 rad = \frac{180}{\pi} deg\).

Step 1: Calculate the Dot Product of the 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 \(|\mathbf{v}|\) represent the magnitudes of the vectors and \(\theta\) is the angle between them. The dot product can also be calculated as \(\mathbf{u} \cdot \mathbf{v} = u_x v_x + u_y v_y\). So, let's compute the dot product: \(\mathbf{u} \cdot \mathbf{v} = (6*-4) + (3*4) =-24+12= -12\)

Step 2: Calculate the Magnitude of Each Vector

We compute the magnitude of each vector using the formula \(|A|=\sqrt{a_x^2 + a_y^2}\). The magnitude of vector \(\mathbf{u}\) is \(|\mathbf{u}|=\sqrt{6^2 + 3^2}=\sqrt{36+9}=\sqrt{45}\) and the magnitude of vector \(\mathbf{v}\) is \(|\mathbf{v}|= \sqrt{(-4)^2 + 4^2}=\sqrt{16+16}=\sqrt{32}\)

Step 3: Compute the Angle \(\theta\)

We can find \(\theta\) with either arcsin or arccos. Here we will use arccos. Rearranging the dot product formula gives us \(\cos(\theta)=\frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|}\), then \(\theta=\arccos{(\frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|})}\). Substituting the previous results into this formula gives us \(\theta=\arccos{(\frac{-12}{\sqrt{45}*\sqrt{32}})}\). In order to find the degree measure of the angle, recall that \(1 rad = \frac{180}{\pi} deg\). Applying this conversion factor gives us \(\theta_{deg}=\theta_{rad}* \frac{180}{\pi}\)

Key Concepts

Dot Product

The dot product is a fundamental operation in vector algebra that combines two vectors into a single scalar quantity. It is deeply connected to the angle between vectors.
In mathematical terms, the dot product of two vectors \( \mathbf{u} = (u_x, u_y) \) and \( \mathbf{v} = (v_x, v_y) \) is calculated as:

• \( \mathbf{u} \cdot \mathbf{v} = u_x v_x + u_y v_y \)

This definition is quite handy as it translates geometric concepts into algebraic expressions. Moreover, the dot product allows the calculation of the angle \( \theta \) between the vectors using another elegant formula:

• \( \mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos(\theta) \)

Thus, the dot product is not just a measure of vector alignment but also a gateway to deeper vector relationships.

Magnitude of Vector

The magnitude of a vector provides a measure of its length in space, giving us an idea of how "long" or "big" a vector is.
To calculate the magnitude of a vector \( \mathbf{A} = (a_x, a_y) \), the Pythagorean theorem from geometry is used, expressed as:

• \( |\mathbf{A}| = \sqrt{a_x^2 + a_y^2} \)

This formula essentially finds the hypotenuse of a right triangle with sides \( a_x \) and \( a_y \).
In the context of finding angles between vectors, knowing each vector's magnitude is crucial. It acts as a scaling factor when calculating how much of one vector is pointing in the direction of another through the dot product formula.
Thus, vector magnitude plays a vital role in both visualizing and solving vector-related problems.

Trigonometric Functions

Trigonometric functions, like cosine and arccosine, are key tools in the study of angles formed by vectors.
When dealing with the angle between two vectors \( \mathbf{u} \) and \( \mathbf{v} \), the dot product formula uses the cosine of the angle \( \theta \) to tie everything together:

• \( \mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos(\theta) \)
• Rearranging gives \( \cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|} \)

The trigonometric function \( \cos(\theta) \) helps us encapsulate how much two vectors are aligned with each other. The arccosine function, \( \arccos \), then helps us find the angle itself from the cosine value:

• \( \theta = \arccos(\cos(\theta)) \)

Once \( \theta \) is found, it can be converted from radians to degrees if needed, especially when interpretation in geometric terms is required.
This conversion is critical as it bridges the gap between abstract mathematical calculations and real-world, intuitive understanding of angles.