Problem 11 - Algebra and Trigonometry

Question

Find the magnitude of the given vector. $$ \langle- 2,1,2\rangle $$

Step-by-Step Solution

Answer

The magnitude of the vector \( \langle -2, 1, 2 \rangle \) is 3.

Step 1: Identify the Vector

The vector given in the problem is \( \langle -2, 1, 2 \rangle \). This vector is in three-dimensional space and has components \( x = -2 \), \( y = 1 \), and \( z = 2 \).

Step 2: Use the Magnitude Formula for a Vector

To find the magnitude of a vector \( \langle a, b, c \rangle \), use the formula \( \sqrt{a^2 + b^2 + c^2} \). For our vector, \( a = -2 \), \( b = 1 \), and \( c = 2 \).

Step 3: Substitute the Components into the Formula

Plug the components \( a = -2 \), \( b = 1 \), and \( c = 2 \) into the formula: \( \sqrt{(-2)^2 + (1)^2 + (2)^2} \).

Step 4: Calculate the Squares of Each Component

Calculate each term inside the square root: \((-2)^2 = 4 \), \(1^2 = 1 \), and \(2^2 = 4 \).

Step 5: Sum the Squared Components

Add the squared values: \(4 + 1 + 4 = 9 \).

Step 6: Take the Square Root to Find the Magnitude

Find the square root of the sum: \( \sqrt{9} = 3 \). This is the magnitude of the vector.

Key Concepts

Understanding Three-Dimensional Vectors

A three-dimensional vector is a mathematical object that has direction and magnitude in three-dimensional space. This space is represented by three axes: the x-axis, y-axis, and z-axis. Each axis corresponds to one dimension. In our example, the vector given is \( \langle -2, 1, 2 \rangle \), which lies in three-dimensional space.

The components \( -2 \), \( 1 \), and \( 2 \) indicate the displacement along the x, y, and z axes respectively. These components tell us how far and in which direction the vector points along each axis. Vectors in three-dimensional space are denoted with pointed brackets and can be represented visually as arrows.

The key aspects of such vectors to remember are:

• It has three components.
• Each component affects its corresponding dimension.
• Its direction and length can be calculated using specific mathematical methods, like magnitude.

Exploring the Magnitude Formula

The magnitude of a vector refers to its length, which is an important feature for describing the vector's size.

When dealing with a three-dimensional vector, we calculate its magnitude using a specific mathematical formula. The formula for finding the magnitude \(|\mathbf{v}|\) of a vector \( \langle a, b, c \rangle \) is given by:

• \( \sqrt{a^2 + b^2 + c^2} \)

This formula involves squaring each of the vector's components, then adding them together, and finally taking the square root of the result.

Here's a breakdown of why this formula works:

• Squaring eliminates negative values ensuring all distances are positive.
• The sum of squares combines all the dimensions.
• The square root converts the sum into the actual length in Euclidean space.

Breaking Down Vector Components

Vector components are the individual parts that make up the vector. Each component corresponds to one dimension in space.

For the vector given in the exercise, \( \langle -2, 1, 2 \rangle \), the components are as follows:

• \(x = -2\) : The component along the x-axis.
• \(y = 1\) : The component along the y-axis.
• \(z = 2\) : The component along the z-axis.

Each component represents a scalar quantity and contributes to the overall direction and magnitude of the vector.

Understanding these components is crucial because:

• They help in visualizing the vector's direction in space.
• Each component shows the extent of movement or influence in that particular dimension.
• They are essential for calculating other vector properties, such as magnitude.

Simplifying Square Root Calculation

The final step in finding a vector's magnitude involves calculating the square root of the summed squares of its components.

Let's walk through this with the vector \( \langle -2, 1, 2 \rangle \):

• First, square each component: \((-2)^2 = 4\), \(1^2 = 1\), and \(2^2 = 4\).
• Next, add them up: \(4 + 1 + 4 = 9\).
• Finally, calculate the square root of this sum: \(\sqrt{9} = 3\).

The square root operation transforms the sum into the vector's actual magnitude, reducing it from a summed square measure to a direct length measurement.

This calculation is a key mathematical process that:

• Converts combined measures into a single, intuitive value.
• Allows comparison of length with other vectors.
• Represents the true physical length irrespective of the direction.