Problem 35 - Precalculus Mathematics for Calculus

Question

Two vectors \(u\) and \(v\) are given. Find the angle (expressed in degrees) between \(\mathbf{u}\) and \(v\) $$\mathbf{u}=\mathbf{j}+\mathbf{k}, \quad \mathbf{v}=\mathbf{i}+2 \mathbf{j}-3 \mathbf{k}$$

Step-by-Step Solution

Answer

The angle between the vectors is approximately 98.13 degrees.

Step 1: Write the Vectors in Component Form

The vectors given are: \(\mathbf{u} = \mathbf{j} + \mathbf{k}\) and \(\mathbf{v} = \mathbf{i} + 2\mathbf{j} - 3\mathbf{k}\). First, express these in component form. \(\mathbf{u} = (0, 1, 1)\) and \(\mathbf{v} = (1, 2, -3)\).

Step 2: Use the Dot Product Formula

The formula for the dot product \(\mathbf{u} \cdot \mathbf{v}\) is given by \(a_1 \cdot b_1 + a_2 \cdot b_2 + a_3 \cdot b_3\). Substitute the vector components: \((0) \cdot (1) + (1) \cdot (2) + (1) \cdot (-3)\).

Step 3: Calculate the Dot Product

Calculate the dot product from the formula: \(0 + 2 - 3 = -1\).

Step 4: Calculate the Magnitude of the Vectors

Find the magnitude of each vector using \(\|\mathbf{u}\| = \sqrt{a^2 + b^2 + c^2}\). For \(\mathbf{u}: \|\mathbf{u}\| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}\). For \(\mathbf{v}: \|\mathbf{v}\| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{14}\).

Step 5: Use the Cosine Formula to Find the Angle

The formula for the angle \(\theta\) between two vectors is \(\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|}\). Substitute the values: \(\cos \theta = \frac{-1}{\sqrt{2} \times \sqrt{14}} = \frac{-1}{\sqrt{28}} = \frac{-1}{2\sqrt{7}} = \frac{-\sqrt{7}}{14}\).

Step 6: Find the Angle in Degrees

To find \(\theta\), calculate \(\theta = \cos^{-1}\left(\frac{-\sqrt{7}}{14}\right)\). Use a calculator to find \(\theta \approx 98.13\) degrees.

Key Concepts

Dot Product

The dot product, also known as the scalar product, is one of the fundamental operations you can perform on vectors. When you calculate the dot product of two vectors, you essentially multiply their corresponding components and sum up those products. This mathematical operation gives you a scalar value, hence the name scalar product.

To compute the dot product of vectors \( \mathbf{u} \) and \( \mathbf{v} \), you use the formula:

• \( \mathbf{u} = (a_1, b_1, c_1) \)
• \( \mathbf{v} = (a_2, b_2, c_2) \)
• \( \mathbf{u} \cdot \mathbf{v} = a_1 \cdot a_2 + b_1 \cdot b_2 + c_1 \cdot c_2 \)

The result of a dot product can tell you a lot about the relationship between the two vectors. For instance, if the dot product is zero, it means the vectors are perpendicular. In this exercise, the calculation \( 0 + 2 - 3 = -1 \) gives us the dot product of the given vectors.

Vector Magnitude

The magnitude of a vector represents its length in space. You can think of it as the distance from the vector's tail to its tip in the coordinate system. To find the magnitude, use the formula derived from the Pythagorean theorem.

For a vector \( \mathbf{u} = (a, b, c) \), the magnitude \( \| \mathbf{u} \| \) is computed as:

• \( \| \mathbf{u} \| = \sqrt{a^2 + b^2 + c^2} \)

This formula sums the squares of the vector's components and then takes the square root of the result.
For example, the vector \( \mathbf{u} = \mathbf{j} + \mathbf{k} \) has a magnitude of \( \sqrt{2} \), and vector \( \mathbf{v} = \mathbf{i} + 2 \mathbf{j} - 3 \mathbf{k} \) has a magnitude of \( \sqrt{14} \). Understanding the concept of magnitude is crucial when you want to determine the scale or size of the vector in question.

Angle Between Vectors

The angle between two vectors gives you an idea of their direction relative to each other. To find the angle, you make use of the cosine formula, which links the dot product to the magnitudes of the vectors.

Given two vectors, \( \mathbf{u} \) and \( \mathbf{v} \), the formula to find the angle \( \theta \) is:

• \( \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{u} \| \cdot \| \mathbf{v} \|} \)

This expression states that the cosine of the angle is equal to the dot product of the vectors divided by the product of their magnitudes. Once you have \( \cos \theta \), use the inverse cosine function (\( \cos^{-1} \)) to find \( \theta \).
In the given example, the formula yields \( \theta = \cos^{-1}\left(\frac{-\sqrt{7}}{14}\right) \), allowing you to calculate the angle between vectors \( \mathbf{u} \) and \( \mathbf{v} \) which is approximately \( 98.13 \) degrees. Understanding the angle helps in determining how parallel or perpendicular the vectors are.