Problem 23 - Calculus for Biology and Medicine

Question

Find the angle between \(\mathbf{x}=[1,2]^{\prime}\) and \(\mathbf{y}=[3,-1]^{\prime}\).

Step-by-Step Solution

Answer

The angle between \(\mathbf{x}\) and \(\mathbf{y}\) is approximately \(81.87^{\circ}\).

Step 1: Understand the Dot Product

The dot product of two vectors \( \mathbf{x} = [1, 2]^{\prime} \) and \( \mathbf{y} = [3, -1]^{\prime} \) can be calculated using the formula: \( \mathbf{x} \cdot \mathbf{y} = x_1y_1 + x_2y_2 \).

Step 2: Calculate the Dot Product

Substitute the components of vectors \( \mathbf{x} \) and \( \mathbf{y} \) into the dot product formula: \( \mathbf{x} \cdot \mathbf{y} = 1 \times 3 + 2 \times (-1) = 3 - 2 = 1 \).

Step 3: Calculate the Magnitudes of the Vectors

The magnitude of a vector \( \mathbf{x} = [x_1, x_2]^{\prime} \) is given by the formula \( \|\mathbf{x}\| = \sqrt{x_1^2 + x_2^2} \). For \( \mathbf{x} \), \( \| \mathbf{x} \| = \sqrt{1^2 + 2^2} = \sqrt{5} \). For \( \mathbf{y} \), \( \| \mathbf{y} \| = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10} \).

Step 4: Use the Cosine Formula

The cosine of the angle \( \theta \) between two vectors can be found using the formula: \( \cos \theta = \frac{\mathbf{x} \cdot \mathbf{y}}{\|\mathbf{x}\| \cdot \|\mathbf{y}\|} \). Here, \( \cos \theta = \frac{1}{\sqrt{5} \times \sqrt{10}} = \frac{1}{\sqrt{50}} \).

Step 5: Calculate the Angle

The angle \( \theta \) can be found by taking the inverse cosine (arccos) of the result from Step 4: \( \theta = \cos^{-1} \left( \frac{1}{\sqrt{50}} \right) \). Calculating this gives \( \theta \approx 81.87^{\circ} \).

Key Concepts

Dot Product

The concept of the dot product is fundamental when working with vectors. Imagine it as a way to multiply two vectors to get a scalar—a single number. This operation is not like regular multiplication though. For two vectors, \( \mathbf{x} = [1, 2]^{\prime} \) and \( \mathbf{y} = [3, -1]^{\prime} \), the dot product is calculated as:

• Multiply the components of each vector: \( x_1 \) by \( y_1 \) and \( x_2 \) by \( y_2 \).
• Add up these products to get the final dot product.

Substituting our specific values, we get:\[ \mathbf{x} \cdot \mathbf{y} = 1 \times 3 + 2 \times (-1) = 3 - 2 = 1 \]The dot product tells us about the relationship between the two vectors. If the dot product is positive, the vectors point in similar directions; if it's zero, they're perpendicular; and if it's negative, they point in opposite directions.

Vector Magnitude

To grasp vector magnitude, think of it as the length of the vector in space. It's an extension of the Pythagorean theorem into higher dimensions.For any vector \( \mathbf{x} = [x_1, x_2]^\prime \), its magnitude \( \| \mathbf{x} \| \) is found using the formula:\[ \| \mathbf{x} \| = \sqrt{x_1^2 + x_2^2} \]For our vectors:

• \( \| \mathbf{x} \| = \sqrt{1^2 + 2^2} = \sqrt{5} \)
• \( \| \mathbf{y} \| = \sqrt{3^2 + (-1)^2} = \sqrt{10} \)

Understanding magnitude is essential because it lets us measure the actual distance and direction of vectors. We'll use magnitude in calculating the cosine of the angle between vectors.

Cosine of Angle Between Vectors

The cosine of the angle between vectors offers insight into how similar the directions of two vectors are. Using the dot product and vector magnitudes calculated earlier, we can find this cosine value.The formula is:\[ \cos \theta = \frac{\mathbf{x} \cdot \mathbf{y}}{\|\mathbf{x}\| \cdot \|\mathbf{y}\|} \]Substitute in the values:\[ \cos \theta = \frac{1}{\sqrt{5} \times \sqrt{10}} = \frac{1}{\sqrt{50}} \]Once we have the cosine of the angle, we can find the angle itself by taking the inverse cosine (arccos) of \( \cos \theta \). In this example:\[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{50}}\right) \approx 81.87^{\circ} \]Understanding the cosine of angle between vectors helps to clarify the spatial relationship in terms of similarity and difference in direction.