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 approximately 80.57 degrees.

Step 1: Understand the Formula

To find the angle \( \theta \) between two vectors \( \mathbf{x} \) and \( \mathbf{y} \), we can use the dot product formula: \( \cos \theta = \frac{\mathbf{x} \cdot \mathbf{y}}{\|\mathbf{x}\| \|\mathbf{y}\|} \).

Step 2: Calculate Dot Product

Calculate the dot product \( \mathbf{x} \cdot \mathbf{y} \) using the formula: \( \mathbf{x} \cdot \mathbf{y} = x_1 \cdot y_1 + x_2 \cdot y_2 + x_3 \cdot y_3 \). Substitute the given values: \( 0\cdot(-3) + (-1)\cdot1 + 3\cdot1 = 0 - 1 + 3 = 2 \).

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

The magnitude \( \|\mathbf{x}\| \) is calculated as \( \sqrt{x_1^2 + x_2^2 + x_3^2} \). For \( \mathbf{x} = [0, -1, 3] \), \( \|\mathbf{x}\| = \sqrt{0^2 + (-1)^2 + 3^2} = \sqrt{0 + 1 + 9} = \sqrt{10} \).

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

The magnitude \( \|\mathbf{y}\| \) is calculated as \( \sqrt{y_1^2 + y_2^2 + y_3^2} \). For \( \mathbf{y} = [-3, 1, 1] \), \( \|\mathbf{y}\| = \sqrt{(-3)^2 + 1^2 + 1^2} = \sqrt{9 + 1 + 1} = \sqrt{11} \).

Step 5: Compute Cosine of Angle

Substitute the dot product and magnitudes into the formula: \( \cos \theta = \frac{\mathbf{x} \cdot \mathbf{y}}{\|\mathbf{x}\| \|\mathbf{y}\|} = \frac{2}{\sqrt{10} \cdot \sqrt{11}} = \frac{2}{\sqrt{110}} \).

Step 6: Calculate Angle \( \theta \)

To find \( \theta \), calculate \( \theta = \cos^{-1}\left(\frac{2}{\sqrt{110}}\right) \). Using a calculator, this gives \( \theta \approx 80.57^\circ \).

Key Concepts

Dot Product

Imagine two vectors as arrows pointing in different directions. The dot product is like comparing how much they align in terms of their direction. It tells us about the cosine of the angle between these vectors.
To calculate the dot product of two vectors \(\mathbf{x}\) and \(\mathbf{y}\), you multiply corresponding components and add the results:

• \( x_1 \times y_1 \)
• \( x_2 \times y_2 \)
• \( x_3 \times y_3 \)

So if \(\mathbf{x} = [0, -1, 3]\) and \(\mathbf{y} = [-3, 1, 1]\), you calculate as follows:

• \(0 \times (-3) = 0\)
• \((-1) \times 1 = -1\)
• \(3 \times 1 = 3\)

Add these up to get the dot product: \(0 - 1 + 3 = 2\). This result tells us there's some alignment between the vectors in terms of direction.

Magnitude of Vectors

The magnitude of a vector is like its length, the distance from the starting point to the tip of the arrow. Think of it as how "strong" the vector is in terms of size.
To find the magnitude of a vector \( \mathbf{x} = [x_1, x_2, x_3] \), you use the formula:

• \( \|\mathbf{x}\| = \sqrt{x_1^2 + x_2^2 + x_3^2} \)

It's a straightforward calculation like finding the hypotenuse of a 3D triangle.
For \( \mathbf{x} = [0, -1, 3] \), you compute:

• \( \sqrt{0^2 + (-1)^2 + 3^2} = \sqrt{0 + 1 + 9} = \sqrt{10} \)

And for \( \mathbf{y} = [-3, 1, 1] \), the magnitude is:

• \( \sqrt{(-3)^2 + 1^2 + 1^2} = \sqrt{9 + 1 + 1} = \sqrt{11} \)

These magnitudes help us determine not only the size but also how two vectors interact with each other.

Cosine Formula

The cosine formula relates the dot product and magnitudes to the angle between two vectors in a way that forms the heart of trigonometry for vectors.
The formula is:

• \( \cos \theta = \frac{\mathbf{x} \cdot \mathbf{y}}{\|\mathbf{x}\| \|\mathbf{y}\|} \)

This equation helps us find the angle \(\theta\) between two vectors. It effectively bridges their directional and spatial relationships.
In our example, with a dot product of 2 and magnitudes \( \sqrt{10} \) and \( \sqrt{11} \), we use the cosine formula:

• \( \cos \theta = \frac{2}{\sqrt{10} \times \sqrt{11}} = \frac{2}{\sqrt{110}} \)

Finally, to find the actual angle, we find \( \theta = \cos^{-1}\left(\frac{2}{\sqrt{110}}\right) \), which tells us the vectors form an angle of about \(80.57^\circ\).
The cosine formula isn't just about numbers—it's a key concept in understanding the spatial relationship of vectors.