Problem 50 - Algebra and Trigonometry

Question

Find the unit vector in the same direction as \(\mathbf{V}\). \(\mathbf{v}=-3 \mathbf{j}\)

Step-by-Step Solution

Answer

The unit vector is \(-\textbf{j}\).

Step 1: Identify the given vector

The given vector is \(\textbf{v} = -3 \textbf{j}\).

Step 2: Determine the magnitude of the vector

The magnitude of a vector \(\textbf{v} = a \mathbf{i} + b\textbf{j}\) is given by \(|\textbf{v}| = \sqrt{a^2 + b^2}\). In this case, \(\textbf{v} = -3 \textbf{j}\), so we have: \(|\textbf{v}| = \sqrt{0^2 + (-3)^2} = \sqrt{9} = 3\).

Step 3: Divide each component by the magnitude

To find the unit vector \(\textbf{u}\) in the same direction as \(\textbf{v}\), divide each component of \(\textbf{v}\) by the magnitude \(|\textbf{v}|\). In this case, \(\textbf{v} = -3 \textbf{j}\) and \(|\textbf{v}| = 3\): \(\textbf{u} = \frac{\textbf{v}}{|\textbf{v}|} = \frac{-3 \textbf{j}}{3} = -\textbf{j}\).

Key Concepts

Vector Magnitude

A vector's magnitude represents its length or size. To find the magnitude of a vector in the form \(\textbf{v} = a \mathbf{i} + b \textbf{j}\), we use the formula: \[ \|\textbf{v}\| = \sqrt{a^2 + b^2} \]. This formula is derived from the Pythagorean Theorem and helps us understand how 'long' the vector is.
For the vector given in the exercise, \(\textbf{v} = -3 \textbf{j}\), the calculation is simple because it only has a y-component. Plugging in the values, we get:
\[ \|\textbf{v}\| = \sqrt{0^2 + (-3)^2} = \sqrt{9} = 3 \].
This gives us the magnitude of \(\textbf{v}\), which is 3.
Remember: The magnitude is always a non-negative number.

Component Form

Vectors are often broken down into their component form to make operations like addition or finding magnitude easier. A vector in two-dimensional space can be expressed as \(\textbf{v} = a \textbf{i} + b \textbf{j}\), where 'a' represents the x-component and 'b' represents the y-component.

In the exercise, the vector \(\textbf{v} = -3 \textbf{j}\) is already in component form. The x-component is 0, and the y-component is -3. By understanding the components, we can easily apply various calculations, like finding the magnitude or performing vector addition.

Component form simplifies the representation and calculation of vectors, especially in higher dimensions where vectors can have more components.

Vector Normalization

Vector normalization involves transforming a vector into a unit vector. A unit vector has a magnitude of 1 and retains the same direction as the original vector. The unit vector \(\textbf{u}\) is found by dividing each component of the vector \(\textbf{v}\) by its magnitude \(|\textbf{v}|\).
Using the provided exercise, the given vector is \(\textbf{v} = -3 \textbf{j}\) with a magnitude of 3. We find the unit vector as follows:

\[ \textbf{u} = \frac{\textbf{v}}{|\textbf{v}|} = \frac{-3 \textbf{j}}{3} = -\textbf{j} \].
This results in the unit vector \(\textbf{u}\) being \(-\textbf{j}\), which has a magnitude of 1.
Normalization is important in various applications, from physics to computer graphics, where having a standard length simplifies calculations and analyses.