Problem 4 - Precalculus: Functions and Graphs

Question

Find (a) the dot product of the two vectors and (b) the angle between the two vectors. $$8 \mathbf{i}-3 \mathbf{j}, \quad 2 \mathbf{i}-7 \mathbf{j}$$

Step-by-Step Solution

Answer

(a) Dot product is 37; (b) Use \( \cos(\theta) = \frac{37}{\sqrt{3870}} \) to find \( \theta \).

Step 1: Write the Vectors in Component Form

The first vector is \( 8\mathbf{i} - 3\mathbf{j} \) and the second vector is \( 2\mathbf{i} - 7\mathbf{j} \). In component form, these vectors are \( \vec{A} = (8, -3) \) and \( \vec{B} = (2, -7) \).

Step 2: Calculate the Dot Product

The dot product of two vectors \( \vec{A} = (a_1,a_2) \) and \( \vec{B} = (b_1,b_2) \) is calculated using the formula:\[\vec{A} \cdot \vec{B} = a_1 \cdot b_1 + a_2 \cdot b_2\]For the given vectors:\[\vec{A} \cdot \vec{B} = 8 \times 2 + (-3) \times (-7) = 16 + 21 = 37\]

Step 3: Calculate the Magnitude of Each Vector

The magnitude of a vector \( \vec{A} = (a_1, a_2) \) is given by:\[|\vec{A}| = \sqrt{a_1^2 + a_2^2}\]For \( \vec{A} = (8, -3) \):\[|\vec{A}| = \sqrt{8^2 + (-3)^2} = \sqrt{64 + 9} = \sqrt{73}\]For \( \vec{B} = (2, -7) \):\[|\vec{B}| = \sqrt{2^2 + (-7)^2} = \sqrt{4 + 49} = \sqrt{53}\]

Step 4: Calculate the Angle Between the Vectors

The angle \( \theta \) between two vectors \( \vec{A} \) and \( \vec{B} \) can be found using the formula:\[\cos(\theta) = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| \cdot |\vec{B}|}\]Substituting the known values:\[\cos(\theta) = \frac{37}{\sqrt{73} \times \sqrt{53}}\]First, calculate the product of the magnitudes:\[\sqrt{73} \times \sqrt{53} = \sqrt{3870}\]Then:\[\cos(\theta) = \frac{37}{\sqrt{3870}}\]Use a calculator to find \( \theta \).

Key Concepts

Angle Between Vectors

When working with vectors, understanding the angle between them can provide valuable insights into their relative directions. The angle between two vectors is calculated using their dot product and magnitudes. The formula to find the cosine of the angle \( \theta \) between two vectors \( \vec{A} \) and \( \vec{B} \) is:\[\cos(\theta) = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| \cdot |\vec{B}|}\]This formula shows that the cosine of the angle is the dot product of the vectors divided by the product of their magnitudes.

• The dot product \(\vec{A} \cdot \vec{B}\) reflects how much one vector extends in the direction of the other.
• The magnitude \(|\vec{A}|\) and \(|\vec{B}|\) represent the lengths of the vectors.

In our exercise, the cosine of the angle is found using:\[\cos(\theta) = \frac{37}{\sqrt{73} \times \sqrt{53}}\]Calculating \(\theta\) involves using a calculator to determine the precise angle from the cosine value. This angle shows how the two vectors are oriented with respect to each other. An angle close to 90 degrees implies the vectors are perpendicular, while near 0 degrees suggests parallelism.

Vector Magnitude

Magnitude is a fundamental property of vectors. It measures the length or size of the vector and is calculated using the Pythagorean theorem. For a vector \( \vec{A} = (a_1, a_2) \), its magnitude \(|\vec{A}|\) is given by:\[|\vec{A}| = \sqrt{a_1^2 + a_2^2}\]Magnitude helps in understanding the scale of the vector's direction and is crucial for various applications like finding the angle between vectors.In our example:

• For \( \vec{A} = (8, -3) \), the magnitude is calculated as \(|\vec{A}| = \sqrt{8^2 + (-3)^2} = \sqrt{73}\).
• For \( \vec{B} = (2, -7) \), the magnitude is \(|\vec{B}| = \sqrt{2^2 + (-7)^2} = \sqrt{53}\).

These magnitudes are then used in the formula for the angle between vectors as well as providing an understanding of the vector's strength. A larger magnitude indicates a longer vector.

Component Form of Vectors

Vectors in physics and mathematics are frequently represented in their component form, which breaks down the vector into its x (horizontal) and y (vertical) components. This form makes it easier to perform arithmetic operations like addition, subtraction, and dot products.For example, given a vector \( 8\mathbf{i} - 3\mathbf{j} \), its component form is \( \vec{A} = (8, -3) \). The same applies to another vector \( 2\mathbf{i} - 7\mathbf{j} \) with component form \( \vec{B} = (2, -7) \).

• \( \mathbf{i} \) represents the unit vector in the x-direction.
• \( \mathbf{j} \) represents the unit vector in the y-direction.

The component form not only aids in visualizing vectors individually but also in calculations like the dot product. For instance, to find the dot product:\[\vec{A} \cdot \vec{B} = a_1 \cdot b_1 + a_2 \cdot b_2\]This approach simplifies understanding vectors as being split into orthogonal (perpendicular) components, significantly aiding their manipulation in algebraic operations.