Problem 11 - Calculus for Biology and Medicine

Question

Normalize \([1,3,-1]^{\prime}\).

Step-by-Step Solution

Answer

The normalized vector is \( \left[ \frac{1}{\sqrt{11}}, \frac{3}{\sqrt{11}}, \frac{-1}{\sqrt{11}} \right]^{\prime} \).

Step 1: Understand the Vector to be Normalized

We have the vector \([1, 3, -1]^{\prime}\). To normalize a vector, we need to scale it so that its magnitude (or length) becomes 1. The input vector here is a three-dimensional column vector.

Step 2: Calculate the Magnitude of the Vector

The magnitude (or norm) of the vector \([1, 3, -1]^{\prime}\) is calculated using the formula \( \sqrt{x_1^2 + x_2^2 + x_3^2} \). For our vector, this becomes \( \sqrt{1^2 + 3^2 + (-1)^2} \), which simplifies to \( \sqrt{1 + 9 + 1} = \sqrt{11} \).

Step 3: Normalize the Vector

To normalize the vector \([1, 3, -1]^{\prime}\), divide each component of the vector by its magnitude, \( \sqrt{11} \). Thus, the normalized vector is \( \left[ \frac{1}{\sqrt{11}}, \frac{3}{\sqrt{11}}, \frac{-1}{\sqrt{11}} \right]^{\prime} \). This ensures that the vector's length is now 1.

Key Concepts

Magnitude of a Vector

Understanding the magnitude of a vector is essential when it comes to normalizing any vector. The magnitude is akin to the vector's "length" in space. It's computed by taking the square root of the sum of the squares of the vector's individual components.

Think of a vector as representing a point or a direction in space. The magnitude measures how far you have to "walk" to reach that point from the origin, in a straight line. For a vector \([x_1, x_2, x_3]^{\prime}\), you calculate this by using the formula:

• \( \sqrt{x_1^2 + x_2^2 + x_3^2} \)

For example, consider the vector \([1, 3, -1]^{\prime}\). Its magnitude is \( \sqrt{1^2 + 3^2 + (-1)^2} = \sqrt{11} \). This means that if you were walking along the vector in a three-dimensional space, you'd travel a length of \( \sqrt{11} \) units. This magnitude serves as a crucial step in finding a normalized vector.

Three-Dimensional Vector

Three-dimensional vectors are pervasive in many fields, from physics to computer graphics. These vectors extend in three different directions, typically represented in terms of x, y, and z axes.

Each component in a three-dimensional vector reflects a distinct directional change:

• \(x\) indicates movement along the horizontal axis.
• \(y\) reflects vertical motion.
• \(z\) represents "depth," moving in or out of the plane.

Our example vector, \([1, 3, -1]^{\prime}\), is a column vector in three-dimensional space. Here, the 1 moves in the x-direction, the 3 in the y-direction, and -1 in the z-direction. This type of vector is crucial in understanding how objects interact in space, as each element conveys vital movement information across the three axes. Vectors like this enable us to visualize geometrical and physical forms with profound simplicity.

Normalizing a Vector

Normalizing a vector is the process of adjusting its length to 1 without changing its direction. This technique is remarkably useful in fields like machine learning, physics, and computer graphics where direction matters, but the length should be uniform.

The steps to normalize a vector include:

• Calculate its magnitude.
• Divide each vector component by this magnitude.

The vector \([1, 3, -1]^{\prime}\) is normalized by dividing each of its entries by its magnitude, \(\sqrt{11}\). This results in the normalized vector: \[ \left[ \frac{1}{\sqrt{11}}, \frac{3}{\sqrt{11}}, \frac{-1}{\sqrt{11}} \right]^{\prime} \]The beauty of a normalized vector lies in its magnitude now being 1, which makes it easy to understand its pure direction. Regardless of the original magnitude, the direction is preserved, and this uniformly scaled vector becomes a unit vector. Normalization is particularly useful when dealing with direction-sensitive calculations, ensuring consistency without altering the vector's trajectory.