Problem 2 - Precalculus: Functions and Graphs

Question

Find (a) the dot product of the two vectors and (b) the angle between the two vectors. $$(4,-7), \quad(-2,3)$$

Step-by-Step Solution

Answer

The dot product is -29. The angle between the vectors is \(\cos^{-1}\left(\frac{-29}{\sqrt{845}}\right)\).

Step 1: Understanding Dot Product Formula

The dot product of two vectors \(\mathbf{a} = (a_1, a_2)\) and \(\mathbf{b} = (b_1, b_2)\) is calculated as: \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2\). We will use this formula to find the dot product of the given vectors \((4, -7)\) and \((-2, 3)\).

Step 2: Calculate Dot Product

Using the formula, substitute \(a_1 = 4\), \(a_2 = -7\), \(b_1 = -2\), and \(b_2 = 3\): \(\mathbf{a} \cdot \mathbf{b} = 4 \cdot (-2) + (-7) \cdot 3\). Calculate each term: \(4 \times -2 = -8\) and \(-7 \times 3 = -21\), then sum them: \(-8 + (-21) = -29\). Thus, the dot product is \(-29\).

Step 3: Vector Magnitudes

To find the angle between the vectors, we need their magnitudes. The magnitude of \(\mathbf{a} = (a_1, a_2)\) is \(\|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2}\). For \(\mathbf{a} = (4, -7)\), \(\|\mathbf{a}\| = \sqrt{4^2 + (-7)^2} = \sqrt{16 + 49} = \sqrt{65}\). Similarly, the magnitude of \(\mathbf{b} = (-2, 3)\) is \(\|\mathbf{b}\| = \sqrt{(-2)^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}\).

Step 4: Cosine of the Angle

The cosine of the angle \(\theta\) between vectors \(\mathbf{a}\) and \(\mathbf{b}\) is given by: \(\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \times \|\mathbf{b}\|}\). Substituting the values: \(\cos \theta = \frac{-29}{\sqrt{65} \times \sqrt{13}}\). Calculate the denominator: \(\sqrt{65} \times \sqrt{13} = \sqrt{845}\). Therefore, \(\cos \theta = \frac{-29}{\sqrt{845}}\).

Step 5: Calculate the Angle

Finally, to find the angle \(\theta\), take the inverse cosine: \(\theta = \cos^{-1}\left(\frac{-29}{\sqrt{845}}\right)\). Use a calculator to find \(\theta\) in degrees or radians as needed. This is the angle between the two vectors

Key Concepts

Dot Product

The dot product is an important concept in vector mathematics. It is the product of two vectors that results in a scalar value. When calculating the dot product, we follow a simple formula. If we have two vectors \( \mathbf{a} = (a_1, a_2) \) and \( \mathbf{b} = (b_1, b_2) \), the dot product is computed as:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \)

This neat formula tells us to multiply corresponding components from each vector and then sum these products.
It's significant because it provides insight into the relationship between the two vectors.
A positive dot product indicates that the vectors are pointing in a similar direction, while a negative value suggests they are pointing in opposite directions.
Using our specific vectors \((4, -7)\) and \((-2, 3)\), the dot product is:

• \( 4 \times -2 + (-7) \times 3 = -8 + (-21) = -29 \)

Thus, the dot product is \(-29\), showing that these vectors point somewhat in opposite directions.

Vector Magnitude

Vector magnitude, often referred to as the 'length' of a vector, is another fundamental idea in vector mathematics. It provides a measure of how long the vector is in the space it exists in. The formula to find a vector's magnitude \( \|\mathbf{a}\| \) for a vector \( \mathbf{a} = (a_1, a_2) \) is:

• \( \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2} \)

This involves squaring each component of the vector, adding those squares together, and then taking the square root of that sum.
It's crucial for various calculations, including determining the angle between vectors.
In our problem, let's compute the magnitudes:

• For \( \mathbf{a} = (4, -7) \), \( \|\mathbf{a}\| = \sqrt{4^2 + (-7)^2} = \sqrt{16 + 49} = \sqrt{65} \)
• For \( \mathbf{b} = (-2, 3) \), \( \|\mathbf{b}\| = \sqrt{(-2)^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \)

Understanding vector magnitude further helps when moving to calculate the angle between vectors.

Angle Between Vectors

Knowing the angle between two vectors is useful in determining how they relate spatially. The cosine of this angle, \( \theta \), can be found using the dot product and magnitudes of the vectors. The formula used is:

• \( \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \times \|\mathbf{b}\|} \)

It is all about employing previously calculated values: the dot product and the magnitudes. This relation yields the cosine of the angle, which we can then manipulate to find \( \theta \) by taking the arccosine.
Performing these calculations:

• \( \cos \theta = \frac{-29}{\sqrt{65} \times \sqrt{13}} \)
• \( \sqrt{65} \times \sqrt{13} = \sqrt{845} \)
• Therefore, \( \cos \theta = \frac{-29}{\sqrt{845}} \)

Finally, to find the exact angle \( \theta \), use a calculator to compute \( \theta = \cos^{-1}\left(\frac{-29}{\sqrt{845}}\right) \). This operation will give you the angle either in degrees or radians, depending on your requirement.