Problem 43 - Algebra and Trigonometry

Question

In Exercises \(39-46,\) find the unit vector that has the same direction as the vector \(\mathbf{v}\) $$\mathbf{v}=3 \mathbf{i}-2 \mathbf{j}$$

Step-by-Step Solution

Answer

The unit vector that has the same direction as the vector \(\mathbf{v}=3\mathbf{i}-2\mathbf{j}\) is \(\mathbf{u} = \frac{3}{\sqrt{13}}\mathbf{i} - \frac{2}{\sqrt{13}}\mathbf{j}\)

Step 1: Calculate the Magnitude of the Vector

The magnitude of a vector \(\mathbf{v} = a\mathbf{i} + b\mathbf{j}\) is given by the formula \(\|\mathbf{v}\| = \sqrt{a^2 + b^2}\). So for the given vector \(\mathbf{v}=3\mathbf{i}-2\mathbf{j}\), its magnitude would be \(\|\mathbf{v}\|= \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}\)

Step 2: Calculate the Unit Vector

After having the magnitude of the vector, the unit vector \(\mathbf{u}\) is then given by dividing the vector by its magnitude. So, the unit vector of mathbf{v} would be \(\mathbf{u} = \mathbf{v}/\|\mathbf{v}\| = (3\mathbf{i} - 2\mathbf{j})/ \sqrt{13}\).

Step 3: Simplify Unit Vector

By simplifying the vector \(\mathbf{u}\), we get \(\mathbf{u} = \frac{3}{\sqrt{13}}\mathbf{i} - \frac{2}{\sqrt{13}}\mathbf{j}\)

Key Concepts

Vector Magnitude

Vector magnitude is essential for understanding the 'size' or 'length' of a vector, irrespective of its direction. It is calculated by taking the root of the sum of the squares of its components. In the context of a 2-dimensional vector like \( \mathbf{v}=a\mathbf{i}+b\mathbf{j} \), the magnitude is found using the formula \( \|\mathbf{v}\| = \sqrt{a^2 + b^2} \).

For the vector given in the exercise, \( \mathbf{v}=3\mathbf{i}-2\mathbf{j} \), we calculate the magnitude by inserting the 'a' and 'b' values into the formula. So we have \( \|\mathbf{v}\| = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \). Understanding this step is crucial because the magnitude is used in subsequent steps to normalize the vector, turning it into a unit vector.

Vector Operations

Vector operations include addition, subtraction, scalar multiplication, and division. When working with unit vectors, an important operation is scalar division, which is used to scale a vector to its unit length.

In our exercise, the given vector \( \mathbf{v}=3\mathbf{i}-2\mathbf{j} \) is divided by its magnitude \( \sqrt{13} \) to achieve the unit vector \( \mathbf{u} \). This means each component of the vector \( \mathbf{v} \) is divided by \( \sqrt{13} \) to reduce its magnitude to 1, resulting in \( \mathbf{u} = (3\mathbf{i} - 2\mathbf{j})/ \sqrt{13} \).

Understanding this operation is key to simplification and to ensuring the direction of the vector remains unchanged while its length is adjusted to 1, adhering to the definition of a unit vector.

Simplifying Vectors

Simplifying vectors is a crucial step in making them more understandable and in preparing them for further operations. After performing the necessary vector operations, as shown in the previous step where we scaled vector \( \mathbf{v} \) down to a unit vector \( \mathbf{u} \) by dividing it by its magnitude, we simplify \( \mathbf{u} \) by ensuring its components are in the simplest form.

Thus, from \( \mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|} \), the simplification process gives us \( \mathbf{u} = \frac{3}{\sqrt{13}}\mathbf{i} - \frac{2}{\sqrt{13}}\mathbf{j} \). It is helpful to have the components as fractions over the magnitude to easily see the proportions of each part. This step, although seemingly trivial, is vital for a clear understanding of the vector's individual components and allows for easier arithmetic for any subsequent vector operations.