Problem 24 - Calculus for Biology and Medicine

Question

Find the angle between \(\mathbf{x}=[-1,2]^{\prime}\) and \(\mathbf{y}=[-2,-4]^{\prime}\).

Step-by-Step Solution

Answer

The angle between the vectors is approximately \( 126.87^\circ \).

Step 1: Understand the Dot Product

The dot product between two vectors \( \mathbf{x} \) and \( \mathbf{y} \) is a scalar quantity representing the product of their magnitudes and the cosine of the angle \( \theta \) between them. The formula for the dot product is: \[ \mathbf{x} \cdot \mathbf{y} = x_1y_1 + x_2y_2 \] where \( \mathbf{x} = [-1, 2]' \) and \( \mathbf{y} = [-2, -4]' \).

Step 2: Calculate the Dot Product

Using the components of \( \mathbf{x} \) and \( \mathbf{y} \), calculate the dot product: \[ \mathbf{x} \cdot \mathbf{y} = (-1)(-2) + (2)(-4) = 2 - 8 = -6 \] So, the dot product is \( -6 \).

Step 3: Calculate the Magnitudes of the Vectors

The magnitude of a vector \( \mathbf{v} = [a, b]' \) is given by: \[ \| \mathbf{v} \| = \sqrt{a^2 + b^2} \] Calculate the magnitudes:\[ \| \mathbf{x} \| = \sqrt{(-1)^2 + (2)^2} = \sqrt{1 + 4} = \sqrt{5} \] \[ \| \mathbf{y} \| = \sqrt{(-2)^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} \]

Step 4: Find the Cosine of the Angle

Using the formula for the dot product, we have: \[ \cos \theta = \frac{\mathbf{x} \cdot \mathbf{y}}{\| \mathbf{x} \| \| \mathbf{y} \|} \] Substitute the calculated values: \[ \cos \theta = \frac{-6}{\sqrt{5} \times \sqrt{20}} = \frac{-6}{\sqrt{100}} = \frac{-6}{10} = -0.6 \]

Step 5: Calculate the Angle

Take the inverse cosine (arccos) of \( -0.6 \) to find \( \theta \): \[ \theta = \cos^{-1}(-0.6) \] Using a calculator, \( \theta \approx 126.87^\circ \) or approximately \( 2.214 \) radians.

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a key operation in vector algebra. When computing the dot product of two vectors, think of it as translating vector information into a single number, which often reveals important relationships between the vectors.
To find the dot product of two vectors \( \mathbf{x} = [x_1, x_2]' \) and \( \mathbf{y} = [y_1, y_2]' \), we use the formula:

• \( \mathbf{x} \cdot \mathbf{y} = x_1y_1 + x_2y_2 \)

For our example vectors \( \mathbf{x} = [-1, 2]' \) and \( \mathbf{y} = [-2, -4]' \), the dot product is calculated as follows:

• \( \mathbf{x} \cdot \mathbf{y} = (-1)(-2) + (2)(-4) = 2 - 8 = -6 \)

Here, the dot product is \(-6\), indicating that the angle between the vectors is obtuse. This is because a negative dot product implies the vectors point in generally opposite directions.

Magnitude of a Vector

The magnitude of a vector represents its length and is essential for understanding the vector's size regardless of its direction.
To calculate the magnitude of a vector \( \mathbf{v} = [a, b]' \), we use the Pythagorean theorem in two-dimensional space:

• \( \| \mathbf{v} \| = \sqrt{a^2 + b^2} \)

Let's compute the magnitudes of \( \mathbf{x} = [-1, 2]' \) and \( \mathbf{y} = [-2, -4]' \):

• \( \| \mathbf{x} \| = \sqrt{(-1)^2 + (2)^2} = \sqrt{1 + 4} = \sqrt{5} \)
• \( \| \mathbf{y} \| = \sqrt{(-2)^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} \)

Understanding the magnitudes helps us standardize the vectors when calculating the angle between them, as we relate length to direction with the help of cosine.

Cosine of Angle Between Vectors

The cosine of the angle between two vectors offers a precise way to describe their orientation with respect to each other. Traditionally, the cosine tells us how much one vector "points in the direction of" another.
To find the cosine of the angle \( \theta \) between vectors \( \mathbf{x} \) and \( \mathbf{y} \), use the following formula:

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

Inserting our earlier results, we have:

• \( \cos \theta = \frac{-6}{\sqrt{5} \times \sqrt{20}} = \frac{-6}{\sqrt{100}} = \frac{-6}{10} = -0.6 \)

The negative result, \(-0.6\), indicates that the angle is greater than 90 degrees, confirming that the vectors are oriented in generally opposite directions. This measure helps us conclude about the spatial relationship between the vectors beyond just their directions.