Problem 25 - Calculus for Biology and Medicine

Question

Find the angle between \(\mathbf{x}=[0,-1,3]^{\prime}\) and \(\mathbf{y}=[-3,1,1]^{\prime}\).

Step-by-Step Solution

Answer

The angle between the vectors is \( \theta = \cos^{-1}\left( \frac{2}{\sqrt{110}} \right) \).

Step 1: Understand the Dot Product Formula

The formula to find the angle \( \theta \) between two vectors, \( \mathbf{x} \) and \( \mathbf{y} \), is given by \( \cos \theta = \frac{\mathbf{x} \cdot \mathbf{y}}{\|\mathbf{x}\|\|\mathbf{y}\|} \). So, first, we need to find the dot product \( \mathbf{x} \cdot \mathbf{y} \) and the magnitudes \( \|\mathbf{x}\| \) and \( \|\mathbf{y}\| \).

Step 2: Calculate the Dot Product

The dot product is calculated as follows: \( \mathbf{x} \cdot \mathbf{y} = 0\times(-3) + (-1)\times1 + 3\times1 = 0 - 1 + 3 = 2 \). So, \( \mathbf{x} \cdot \mathbf{y} = 2 \).

Step 3: Calculate the Magnitude of \( \mathbf{x} \)

The magnitude of \( \mathbf{x} \) is \( \|\mathbf{x}\| = \sqrt{0^2 + (-1)^2 + 3^2} = \sqrt{0 + 1 + 9} = \sqrt{10} \).

Step 4: Calculate the Magnitude of \( \mathbf{y} \)

The magnitude of \( \mathbf{y} \) is \( \|\mathbf{y}\| = \sqrt{(-3)^2 + 1^2 + 1^2} = \sqrt{9 + 1 + 1} = \sqrt{11} \).

Step 5: Calculate \( \cos \theta \)

Substitute the values into the cosine formula: \( \cos \theta = \frac{2}{\sqrt{10} \cdot \sqrt{11}} = \frac{2}{\sqrt{110}} \).

Step 6: Find the Angle \( \theta \)

To find \( \theta \), compute the inverse cosine: \( \theta = \cos^{-1}\left( \frac{2}{\sqrt{110}} \right) \). Use a calculator to find \( \theta \).

Key Concepts

Dot Product

In vector calculus, one of the most vital operations is the dot product, a method to multiply two vectors. The dot product between two vectors, \( \mathbf{a} \) and \( \mathbf{b} \), is calculated as the sum of the products of their corresponding components. For example, if \( \mathbf{a} = [a_1, a_2, a_3] \) and \( \mathbf{b} = [b_1, b_2, b_3] \), then the dot product \( \mathbf{a} \cdot \mathbf{b} \) is given by:

• \( a_1b_1 + a_2b_2 + a_3b_3 \)

This operation results in a scalar, not a vector.
You'll find the dot product useful for calculating the angle between two vectors and determining if they are orthogonal. If the dot product is zero, the vectors are perpendicular.
In the exercise, the dot product of \( \mathbf{x} = [0, -1, 3]^{\prime} \) and \( \mathbf{y} = [-3, 1, 1]^{\prime} \) was computed as 2.

Vector Magnitude

A vector's magnitude, also known as its 'length' or 'norm,' is an essential property in understanding the vector's size. Calculating the magnitude involves taking the square root of the sum of the squares of a vector's components.
For a vector \( \mathbf{v} = [v_1, v_2, v_3] \), its magnitude \( \|\mathbf{v}\| \) is given by:

• \( \sqrt{v_1^2 + v_2^2 + v_3^2} \)

The magnitude gives you a sense of the vector's overall size.
In the context of our exercise, the magnitudes of \( \mathbf{x} \) and \( \mathbf{y} \) were calculated as \( \sqrt{10} \) and \( \sqrt{11} \) respectively. Knowing these magnitudes allows us to understand the distances or lengths represented by the vectors.

Angle Between Vectors

Finding the angle between two vectors can reveal the relationship between their directions. The angle \( \theta \) between vectors \( \mathbf{a} \) and \( \mathbf{b} \) can be determined using the formula:

• \( \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \)

This equation leverages the components you've calculated: dot product and magnitudes.
After finding \( \cos \theta \), use an inverse cosine function to find \( \theta \). This angle can help in various geometric interpretations and applications, relating the two vectors' orientation.
In our exercise, using \( \cos \theta = \frac{2}{\sqrt{110}} \), we can derive the angle to understand how vectors \( \mathbf{x} \) and \( \mathbf{y} \) align to one another in space.