Problem 43 - Precalculus with Limits

Question

In Exercises 41-44, graph the vectors and find the degree measure of the angle \(\theta\) between the vectors. \(\mathbf{u} = 5\mathbf{i} + 5\mathbf{j}\) \(\mathbf{v} = -8\mathbf{i} + 8\mathbf{j}\)

Step-by-Step Solution

Answer

The degree measure of the angle between the vectors is 90 degrees.

Step 1: Graph the vectors

Start by graphing the vectors \(\mathbf{u}\) and \(\mathbf{v}\). Based on their component forms, \(\mathbf{u}\) ends at the point (5,5) and \(\mathbf{v}\) ends at the point (-8,8). These are represented on a two-dimensional coordinate plane.

Step 2: Calculate the dot product

Use the dot product formula for vectors: \(\mathbf{u} \cdot \mathbf{v} = ||\mathbf{u}|| \cdot ||\mathbf{v}|| \cdot cos(\theta)\) where \(\theta\) is the angle between the vectors. The left-hand side can be calculated directly using the given vectors: \(5*-8 + 5*8 = 0\).

Step 3: Calculate the magnitudes of the vectors

The magnitudes of \(\mathbf{u}\) and \(\mathbf{v}\) can be calculated using the Pythagorean theorem: ||\(\mathbf{u}\)|| = \(\sqrt{5^2 + 5^2}\) = \(\sqrt{50}\) and ||\(\mathbf{v}\)|| = \(\sqrt{-8^2 + 8^2}\) = \(\sqrt{128}\).

Step 4: Solve for the angle

Substitute values into the formula (since the left-hand side is 0, \(\mathbf{u} \cdot \mathbf{v} = 0\)): 0 = \(\sqrt{50}\) \cdot \(\sqrt{128}\) \cdot cos(\theta). Therefore, cos(\theta) = 0, and \theta is thus either 90 or -90 degrees, but only 90 degrees is valid, since angles between vectors are always taken to be between 0 and 180 degrees.

Key Concepts

Dot Product

The dot product is a method to multiply two vectors and get a scalar (number) as a result. To calculate the dot product of vectors \( \mathbf{u} = \langle u_1, u_2 \rangle \) and \( \mathbf{v} = \langle v_1, v_2 \rangle \), you apply the formula: \[ \mathbf{u} \cdot \mathbf{v} = u_1 \cdot v_1 + u_2 \cdot v_2 \]This operation combines the corresponding components of the vectors, resulting in a single number. In our exercise, the vectors \( \mathbf{u} = 5\mathbf{i} + 5\mathbf{j} \) and \( \mathbf{v} = -8\mathbf{i} + 8\mathbf{j} \) gave:

• \(5 \cdot (-8) + 5 \cdot 8 = -40 + 40 = 0\)

A dot product of zero tells us that the vectors are perpendicular, meaning they form a right angle (90 degrees) with each other. Understanding the dot product aids in finding angles between vectors using the internal relationship they share.

Magnitude of a Vector

Magnitude is essentially the 'length' or 'size' of a vector. It represents how far the vector stretches in space. To find it, use the Pythagorean theorem adapted for vectors. For a vector \( \mathbf{u} = \langle u_1, u_2 \rangle \): \[ ||\mathbf{u}|| = \sqrt{u_1^2 + u_2^2} \]For our example vectors:

• \( ||\mathbf{u}|| = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} \)
• \( ||\mathbf{v}|| = \sqrt{(-8)^2 + 8^2} = \sqrt{64 + 64} = \sqrt{128} \)

The magnitude gives insight into the strength or intensity of the vector. By computing the magnitude of each vector, we could proceed to calculate angles and assess directions, central to the solution process.

Graphing Vectors

Graphing vectors visually demonstrates their direction and magnitude on a coordinate plane. Each vector begins at the origin \((0, 0)\) and extends to the coordinates defined by its components. For a vector \( \mathbf{u} = u_1\mathbf{i} + u_2\mathbf{j} \), the graph ends at \((u_1, u_2)\).Applying this to our vectors:

• Vector \( \mathbf{u} = 5\mathbf{i} + 5\mathbf{j} \) ends at \((5, 5)\).
• Vector \( \mathbf{v} = -8\mathbf{i} + 8\mathbf{j} \) ends at \((-8, 8)\).

By plotting these points on a two-dimensional plane, you can easily identify the vectors' directions and see the angle between them. Graphing helps to make the abstract numerical relationships more concrete, especially in visualizing orthogonal (perpendicular) relationships.