Problem 38 - Precalculus with Limits

Question

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 the vectors \( \mathbf{u} \) and \( \mathbf{v} \) is approximately 120.5 degrees.

Step 1: Find the Dot Product of Two Vectors

Start by finding the dot product of the vectors \( \mathbf{u} \) and \( \mathbf{v} \). This is done by multiplying their respective \( i \) and \( j \) components and adding them together: (2)* (4) + (-3) * (3) = 8 - 9 = -1.

Step 2: Find the Magnitude of Each Vector

Next, find the magnitude of each vector. This can be done using the formula for the magnitude of a vector which is \( \sqrt{x^2 + y^2} \). For \( \mathbf{u} \) this is \( \sqrt{(2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} \). For \( \mathbf{v} \), this becomes \( \sqrt{(4)^2 +(3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \).

Step 3: Find the Cosine of the Angle Between the Vectors

Now, to find \( \cos{\theta} \), use the formula \( \cos{\theta} = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \cdot ||\mathbf{v}||} \). Substituting the previously calculated values gives \( \cos{\theta} = \frac{-1}{\sqrt{13} \cdot 5} \).

Step 4: Find the angle

Finally, to find the actual angle \( \theta \), use the arccosine (inverse of cosine) of the previously calculated value. In degrees, this is given by \( \theta = \arccos{\left(\frac{-1}{\sqrt{13} \cdot 5}\right)} \) which is approximately 120.5 degrees.

Key Concepts

Dot Product of Vectors

The dot product is a crucial operation when dealing with vectors, especially when you want to find the angle between them. For two vectors \( \mathbf{u} = a\mathbf{i} + b\mathbf{j} \) and \( \mathbf{v} = c\mathbf{i} + d\mathbf{j} \), the dot product is calculated by multiplying corresponding components and summing the results:
\[ \mathbf{u} \cdot \mathbf{v} = ac + bd \]
This operation gives a single scalar value. For example, if you have \( \mathbf{u} = 2\mathbf{i} - 3\mathbf{j} \) and \( \mathbf{v} = 4\mathbf{i} + 3\mathbf{j} \), the dot product is:

• Multiply the \( i \) components: \( 2 \times 4 = 8 \)
• Multiply the \( j \) components: \( -3 \times 3 = -9 \)
• Add them together: \( 8 + (-9) = -1 \)

The dot product, in this case, is \( -1 \). This value is part of the equation to find the cosine of the angle between vectors.

Magnitude of a Vector

The magnitude, or length, of a vector is like finding the length of a straight line. For a vector represented as \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} \), the magnitude is computed by using the Pythagorean theorem:
\[ ||\mathbf{v}|| = \sqrt{a^2 + b^2} \]
This gives you a non-negative scalar that informs you how lengthy the vector is without direction. For instance:

• Magnitude of \( \mathbf{u} = 2\mathbf{i} - 3\mathbf{j} \):
  \[ ||\mathbf{u}|| = \sqrt{(2)^2 + (-3)^2} = \sqrt{13} \]
• Magnitude of \( \mathbf{v} = 4\mathbf{i} + 3\mathbf{j} \):
  \[ ||\mathbf{v}|| = \sqrt{(4)^2 + (3)^2} = 5 \]

These magnitudes are helpful in normalizing vectors or finding angles with the dot product.

Cosine of the Angle Between Vectors

The angle between two vectors is found using a formula relating the dot product and magnitudes of the vectors. The cosine of the angle \( \theta \) is given by:
\[ \cos{\theta} = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \cdot ||\mathbf{v}||} \]
This expression scales the dot product against the sizes of the vectors, offering a value between -1 and 1, representing the cosine of the angle. For our example:

• Dot product \( \mathbf{u} \cdot \mathbf{v} = -1 \)
• Magnitudes: \( ||\mathbf{u}|| = \sqrt{13} \), \( ||\mathbf{v}|| = 5 \)
• Result: \[ \cos{\theta} = \frac{-1}{\sqrt{13} \cdot 5} \]

The actual angle \( \theta \) is then found using the arccosine function, which gives you \( 120.5 \) degrees from this calculation. Understanding this formula is essential as it is a direct link between algebra and geometry in vector space.