Problem 38 - Precalculus with Limits

Question

Finding the Angle Between Two Vectors In Exercises \(31-40,\) find the angle \(\theta\) between the vectors. $$\mathbf{u}=2 \mathbf{i}-3 \mathbf{j}$$ $$\mathbf{v}=4 \mathbf{i}+3 \mathbf{j}$$

Step-by-Step Solution

Answer

The angle \( \theta \) between vectors \( \mathbf{u} \) and \( \mathbf{v} \) is \( cos^{-1}(-1/\sqrt{65}) \)

Step 1: Calculate the Dot Product

First, calculate the dot product of the vectors \( \mathbf{u} \) and \( \mathbf{v} \). The dot product formula for two vectors \( a\mathbf{i}+b\mathbf{j} \) and \( c\mathbf{i}+d\mathbf{j} \) is \( a*c + b*d \). So, the dot product of \( \mathbf{u} \) and \( \mathbf{v} \) would be \( (2*4) + (-3*3) = 8 - 9 = -1 \).

Step 2: Calculate the Magnitude of each Vector

Next, calculate the magnitude of each vector. The formula for the magnitude of the vector \( a\mathbf{i}+b\mathbf{j} \) is \( \sqrt{a^2 + b^2} \). So, the magnitude of \( \mathbf{u} \) would be \( \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} \) and the magnitude of \( \mathbf{v} \) would be \( \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \).

Step 3: Use the formula to find the angle

By substituting the derived values in the formula of \( cos(\theta) \), we find \( cos(\theta) = \frac{-1}{ \sqrt{13} * 5} \). To find the value of \( \theta \), use the inverse cosine function to calculate \( \theta = cos^{-1}(-1/\sqrt{65}) \).

Key Concepts

Dot Product

If you've come across the term "dot product" in your vector studies, you're on the right track to mastering vector angles. The dot product is a mathematical operation that takes two equal-length sequences of numbers (vectors) and returns a single number. It's also known as the scalar product. This operation is especially useful for finding the angle between two vectors, among other things.

To calculate the dot product of two vectors in two dimensions, say \(\mathbf{u} = a_i \mathbf{i} + a_j \mathbf{j}\) and \(\mathbf{v} = b_i \mathbf{i} + b_j \mathbf{j}\), you multiply the corresponding components and add them up:

• Multiply the \(\mathbf{i}\) components: \(a_i \times b_i\)
• Multiply the \(\mathbf{j}\) components: \(a_j \times b_j\)
• Add the results together: \(a_i \times b_i + a_j \times b_j\)

This gives you the dot product of \(\mathbf{u}\) and \(\mathbf{v}\). It's important to note that the dot product is a scalar, meaning it's just a number, not a vector. This property is particularly helpful because it transforms multi-dimensional problems into manageable arithmetic expressions.

In the provided exercise, the dot product of \(\mathbf{u} = 2\mathbf{i} - 3\mathbf{j}\) and \(\mathbf{v} = 4\mathbf{i} + 3\mathbf{j}\) is calculated as \(8 - 9 = -1\). This result is important for determining the angle between these vectors.

Magnitude of a Vector

To understand vector angles, knowing how to find a vector's magnitude is crucial. The magnitude of a vector is the length or size of the vector. It's the distance from the initial point to the terminal point in vector space. Calculating the magnitude helps in various applications, such as normalizing vectors and working out angles between vectors.

For a two-dimensional vector \(a_i \mathbf{i} + a_j \mathbf{j}\), you can find the magnitude using the formula:

\[\sqrt{a_i^2 + a_j^2}\].

This formula comes from the Pythagorean theorem, because you're essentially calculating the hypotenuse of a right triangle formed by the vector's components.

For the vectors in the exercise, the magnitudes are calculated by:

• For \(\mathbf{u} = 2\mathbf{i} - 3\mathbf{j}\), the magnitude is \(\sqrt{2^2 + (-3)^2} = \sqrt{13}\).
• For \(\mathbf{v} = 4\mathbf{i} + 3\mathbf{j}\), the magnitude is \(\sqrt{4^2 + 3^2} = 5\).

Once you have these magnitudes, you can use them along with the dot product to find the angle between the vectors.

Inverse Cosine

The final step in finding the angle between two vectors involves the inverse cosine function, often called the arccosine. This function is crucial when you have a dot product and magnitudes because it helps convert the cosine value into an actual angle measurement.

Suppose you have calculated the cosine of the angle \(\theta\) between two vectors as:

\[\cos(\theta) = \frac{\text{dot product}}{\text{product of magnitudes}}\].

For practical use, you rearrange this formula to derive the angle \(\theta\) itself by applying the inverse cosine function:
\[\theta = \cos^{-1}\left(\frac{\text{dot product}}{\text{product of magnitudes}}\right)\].

This function will return the angle in radians (often converted to degrees) that's formed between the vectors.

In the given exercise, the process is simplified as:

• First, calculate \(cos(\theta) = \frac{-1}{\sqrt{13} \times 5}\).
• Then, find \(\theta = \cos^{-1}\left(\frac{-1}{\sqrt{65}}\right)\).

This equation provides the exact angle between the two vectors, illustrating how these simple mathematical tools can solve complex geometric problems efficiently.