Problem 12 - Calculus for Biology and Medicine

Question

Normalize \([2,0,-4]\).

Step-by-Step Solution

Answer

The normalized vector is \(\left[\frac{\sqrt{5}}{5}, 0, \frac{-2\sqrt{5}}{5}\right]\).

Step 1: Find the Magnitude

To normalize a vector, we must first find its magnitude (or length). The magnitude of a vector \([x, y, z]\) is given by the formula \(\sqrt{x^2 + y^2 + z^2}\). For the vector \([2, 0, -4]\), calculate:\[\sqrt{2^2 + 0^2 + (-4)^2} = \sqrt{4 + 0 + 16} = \sqrt{20} = 2\sqrt{5}\].

Step 2: Divide Each Component by the Magnitude

Next, divide each component of the vector by the magnitude to obtain the normalized vector. The magnitude we found is \(2\sqrt{5}\), so we divide each component by this value. The normalized vector \(\mathbf{v}\) becomes:\[\mathbf{v} = \left[\frac{2}{2\sqrt{5}}, \frac{0}{2\sqrt{5}}, \frac{-4}{2\sqrt{5}}\right] = \left[\frac{1}{\sqrt{5}}, 0, \frac{-2}{\sqrt{5}}\right]\].

Step 3: Rationalize the Denominator

Though it's not necessary for normalization, it's often helpful to present the normalized vector with rationalized denominators. To rationalize, multiply the numerator and denominator of each component by \(\sqrt{5}\):\[\left[\frac{1}{\sqrt{5}}, 0, \frac{-2}{\sqrt{5}}\right] = \left[\frac{1\cdot\sqrt{5}}{\sqrt{5}\cdot\sqrt{5}}, 0, \frac{-2\cdot\sqrt{5}}{\sqrt{5}\cdot\sqrt{5}}\right] = \left[\frac{\sqrt{5}}{5}, 0, \frac{-2\sqrt{5}}{5}\right]\].

Key Concepts

magnitude calculation

Understanding how to calculate the magnitude of a vector is crucial for normalization. The magnitude is essentially the "length" of the vector in space. For any vector \([x, y, z]\), the magnitude is found using the formula \(\sqrt{x^2 + y^2 + z^2}\). This formula comes from the Pythagorean theorem, extending it into three dimensions. When we calculate the magnitude for the vector \([2, 0, -4]\), we substitute the values into the formula:

• Calculate \((2)^2\), \((0)^2\), and \((-4)^2\).
• Add these values: \((4 + 0 + 16) = 20\).
• Finally, take the square root: \(\sqrt{20} = 2\sqrt{5}\).

Finding the magnitude helps us understand how much we need to "shrink" or "stretch" each component to get a unit vector.

rationalizing denominators

After calculating the magnitude and adjusting each vector component, you might end up with a fraction like \(\frac{1}{\sqrt{5}}\). While this is mathematically correct, it's common practice to present the fraction in a form without a square root in the denominator. This is called rationalizing the denominator.To rationalize a denominator, you multiply both the numerator and the denominator by \(\sqrt{5}\). Here's how it works step by step:

• Start with the fraction \(\frac{1}{\sqrt{5}}\).
• Multiply both the top and bottom by \(\sqrt{5}\) to get \((\frac{1 \cdot \sqrt{5}}{\sqrt{5} \cdot \sqrt{5}})\).
• The denominator becomes \((\sqrt{5} \cdot \sqrt{5} = 5)\).
• So, the fraction now is \(\frac{\sqrt{5}}{5}\).

This step is not mandatory for vector calculations but can make expressions neater and easier to interpret.

vector components

Vectors are composed of several components, typically represented in coordinate space. In our exercise, the vector \([2, 0, -4]\) has three components: \([x, y, z] = [2, 0, -4]\).Each component contributes to the vector's position and direction.

• The first component \((x = 2)\) measures horizontal movement along the x-axis.
• The second component \((y = 0)\) indicates no movement along the y-axis.
• The third component \((z = -4)\) tells us about vertical displacement along the z-axis.

When normalizing a vector, we adjust each component based on the magnitude, ensuring the resulting vector has a magnitude of one.For normalization, we divide each component by the calculated magnitude (e.g., \([\frac{2}{2\sqrt{5}}, \frac{0}{2\sqrt{5}}, \frac{-4}{2\sqrt{5}}]\)), resulting in the unit vector with the same direction: \([\frac{1}{\sqrt{5}}, 0, \frac{-2}{\sqrt{5}}]\). This ensures that the vector maintains its direction but has a standardized length.