Problem 34 - Precalculus with Limits

Question

Finding the Angle Between Two Vectors In Exercises \(31-40,\) find the angle \(\theta\) between the vectors. $$\mathbf{u}=2 \mathbf{i}-3 \mathbf{j}$$ $$\mathbf{v}=\mathbf{i}-2 \mathbf{j}$$

Step-by-Step Solution

Answer

The angle \(\theta\) between the vectors \(\mathbf{u}\) and \(\mathbf{v}\) is \(\arccos(\frac{8}{\sqrt{13}*\sqrt{5}})\).

Step 1: Calculate the Dot Product

First, calculate the dot product of the vectors \(\mathbf{u}\) and \(\mathbf{v}\). This is done by multiplying the corresponding components of each vector and then summing up these values. For these vectors, the dot product is \((2*1) + ((-3)*(-2)) = 2 + 6 = 8.\)

Step 2: Calculate the Magnitudes

Next, calculate the magnitudes of the given vectors. The magnitude of a vector \(\mathbf{a} = a\mathbf{i}+b\mathbf{j}\) is calculated as \(\sqrt{a^2 + b^2}\). So for \(\mathbf{u}\), we find \(||\mathbf{u}|| = \sqrt{(2)^2 + (-3)^2} = \sqrt{4+9} = \sqrt{13}\), and for \(\mathbf{v}\), we find \(||\mathbf{v}|| = \sqrt{(1)^2 + (-2)^2} = \sqrt{1+4} = \sqrt{5}\).

Step 3: Calculate \(\cos(\theta)\)

Now that we have the dot product of \(\mathbf{u}\) and \(\mathbf{v}\) and the magnitudes of \(\mathbf{u}\) and \(\mathbf{v}\), we can calculate \(\cos(\theta)\) using the formula \(\cos(\theta) = \frac{\mathbf{u}.\mathbf{v}}{||\mathbf{u}||*||\mathbf{v}||}\). Substituting the previously calculated values, we find \(\cos(\theta) = \frac{8}{\sqrt{13}*\sqrt{5}}\).

Step 4: Calculate \(\theta\)

Finally, to find the angle \(\theta\), take the arccosine of the previously calculated cosine. So \(\theta = \arccos(\frac{8}{\sqrt{13}*\sqrt{5}})\). This is the angle between the vectors, which is the answer.

Key Concepts

Dot Product

The dot product is a fundamental operation in vector algebra. It allows you to determine how two vectors interact with each other to produce a scalar value. This scalar value can tell us how much one vector goes in the direction of another.

To calculate the dot product of two vectors \(\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j}\) and \(\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j}\):

• Multiply the corresponding components of the vectors together.
• Add these products together.

Thus, the formula for the dot product is given by:\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \]The dot product of our vectors \(\mathbf{u}=2\mathbf{i}-3\mathbf{j}\) and \(\mathbf{v}=\mathbf{i}-2\mathbf{j}\) resulted in \((2 \times 1) + ((-3) \times (-2)) = 8\).

This calculation is critical to further find the angle between the two vectors.

Magnitude of a Vector

The magnitude of a vector measures its length within its coordinate space. It represents how far a vector travels from the origin point to its terminal point, regardless of its direction. Calculating the magnitude of a vector like \(\mathbf{a} = a\mathbf{i} + b\mathbf{j}\) involves using the following formula:

• The square root of the sum of the squares of its components.

\[ ||\mathbf{a}|| = \sqrt{a^2 + b^2} \]For our vector \(\mathbf{u}=2\mathbf{i}-3\mathbf{j}\), we calculated:\[ ||\mathbf{u}|| = \sqrt{2^2 + (-3)^2} = \sqrt{13} \]And for \(\mathbf{v}=\mathbf{i}-2\mathbf{j}\):\[ ||\mathbf{v}|| = \sqrt{1^2 + (-2)^2} = \sqrt{5} \]
Understanding the concept of magnitude helps in finding the cosine of the angle between two vectors.

Cosine Rule for Angles

The cosine rule for angles between vectors relates the dot product and magnitudes of two vectors to their cosine of the angle between them. This rule is a clever method in vector geometry that links these components.
When you have the dot product and magnitudes of two vectors, the angle \(\theta\) between the vectors can be found with this formula:\[ \cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \times ||\mathbf{v}||} \]This allows us to directly relate the calculations we did with dot products and magnitudes to the angle of interaction between our vectors. In our problem, it becomes:\[ \cos(\theta) = \frac{8}{\sqrt{13} \times \sqrt{5}} \]

Applying the cosine rule powers our understanding of spatial relationships between directional forces like vectors.

Arccosine Function

The arccosine function is a trigonometric tool that helps in finding the angle when you already know the cosine of that angle. It is the inverse of the cosine function, providing a way to derive angle information from a known cosine value.

In our exercise, we used the arccosine function to determine the actual angle \(\theta\) between the two vectors \(\mathbf{u}\) and \(\mathbf{v}\). After calculating \(\cos(\theta)\), the arccosine function allows us to retrace that back into angle measure:\[ \theta = \arccos\left(\frac{8}{\sqrt{13} \times \sqrt{5}}\right) \]

This final step translates the conceptual understanding of vector interactions into concrete angle measurements, essential for truly grasping vector geometry complexities.