Problem 14 - Algebra and Trigonometry

Question

Find the magnitude of the given vector. $$ \langle 1,-6,2 \sqrt{2}\rangle $$

Step-by-Step Solution

Answer

The magnitude of the vector is \( 3\sqrt{5} \).

Step 1: Understand the Vector

The given vector is \( \langle 1, -6, 2\sqrt{2} \rangle \). This vector has components in three dimensions: \( x = 1 \), \( y = -6 \), and \( z = 2\sqrt{2} \). To find the magnitude of a vector in three-dimensional space, we use the formula \( \| \mathbf{v} \| = \sqrt{x^2 + y^2 + z^2} \).

Step 2: Square Each Component

Square each component of the vector: \[ x^2 = 1^2 = 1, \]\[ y^2 = (-6)^2 = 36, \]\[ z^2 = (2\sqrt{2})^2 = 8. \]

Step 3: Sum the Squares

Add the squared values together: \[ 1 + 36 + 8 = 45. \]

Step 4: Take the Square Root

Find the square root of the sum of the squares to determine the magnitude of the vector: \[ \| \mathbf{v} \| = \sqrt{45}. \]

Step 5: Simplify the Square Root

Simplify the square root \( \sqrt{45} \) by recognizing that 45 can be factored as 9 times 5, so \( \sqrt{45} = \sqrt{9 \times 5} = \sqrt{9} \times \sqrt{5} = 3\sqrt{5} \).

Key Concepts

Three-Dimensional Vectors

When we talk about three-dimensional vectors, we refer to vectors that have components in three directions: usually denoted by the x, y, and z axes. Imagine this as describing motion or position in a 3D space, much like a grid, where:

• x: represents movement along the horizontal axis.
• y: represents movement along the vertical axis.
• z: represents movement perpendicular to both x and y axes, often described as depth.

Each component of a vector contributes to the overall direction and magnitude, which is like the vector’s size or length in the space. In the given vector, \( \langle 1, -6, 2\sqrt{2} \rangle \), the components are \( x = 1 \), \( y = -6 \), and \( z = 2\sqrt{2} \). If you were to visualize this vector, it would point to a specific location in a 3D environment, characterized by these three measurements from a common origin point, usually where the value of each component is zero.

Magnitude Formula

To determine how "long" or how much "size" a vector has, we use the magnitude formula. Magnitude measures the distance from the start of the vector (the origin) to its endpoint in space. For a three-dimensional vector with components \( x \), \( y \), and \( z \), the magnitude \( \| \mathbf{v} \| \) is found using:

• The formula: \( \| \mathbf{v} \| = \sqrt{x^2 + y^2 + z^2} \).

This formula comes from the Pythagorean theorem, extended into three dimensions. Each component's square—\( x^2 \), \( y^2 \), \( z^2 \)—is summed up to indicate the squared "length" in each direction. Finally, the square root of this sum provides us the actual magnitude by converting those squared components back into real-world distances, creating a single value that represents the vector's full extent in 3D space.

Square Root Simplification

Often in algebra and vector calculations, you encounter square roots needing simplification for ease of interpretation. This is the case with our vector's magnitude calculation, where we obtained \( \sqrt{45} \).

• First, identify perfect squares within the number under the square root. Here, 45 can be rewritten as \( 9 \times 5 \).
• Simplification process: Break it down: \( \sqrt{45} = \sqrt{9 \times 5} \). Since 9 is a perfect square (\( 9 = 3^2 \)), it can be extracted from the square root: \( \sqrt{9} = 3 \).
• This gives: \( \sqrt{45} = 3\sqrt{5} \).

This simplification helps in presenting the magnitude in a form that is often considered neater or more recognizable in mathematical contexts. It makes further calculations and visualizations more straightforward, especially if combined with other expressions or when requiring an exact value.