Problem 23 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The angle between \( \mathbf{x} \) and \( \mathbf{y} \) is approximately 36.87°.

Step 1: Understand the Vector Dot Product

The angle \( \theta \) between two vectors \( \mathbf{x} \) and \( \mathbf{y} \) can be found using the dot product formula: \[ \cos(\theta) = \frac{\mathbf{x} \cdot \mathbf{y}}{||\mathbf{x}|| \cdot ||\mathbf{y}||} \]Here, \( \mathbf{x} \cdot \mathbf{y} \) is the dot product of the vectors, and \( ||\mathbf{x}|| \) and \( ||\mathbf{y}|| \) are the magnitudes of the vectors \( \mathbf{x} \) and \( \mathbf{y} \), respectively.

Step 2: Calculate the Dot Product

To find the dot product \( \mathbf{x} \cdot \mathbf{y} \), use the formula:\[ \mathbf{x} \cdot \mathbf{y} = x_1y_1 + x_2y_2 = 3 \times 3 + 1 \times (-1) = 9 - 1 = 8 \]

Step 3: Calculate the Magnitudes of the Vectors

The magnitude of a vector \( \mathbf{a} = [a_1, a_2] \) is given by:\[ ||\mathbf{a}|| = \sqrt{a_1^2 + a_2^2} \]For \( \mathbf{x} = [3, 1] \):\[ ||\mathbf{x}|| = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \]For \( \mathbf{y} = [3, -1] \):\[ ||\mathbf{y}|| = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10} \]

Step 4: Use the cosine formula

Substitute the calculated values into the cosine formula:\[ \cos(\theta) = \frac{8}{\sqrt{10} \cdot \sqrt{10}} = \frac{8}{10} = 0.8 \]

Step 5: Find the Angle Using the Inverse Cosine

Take the inverse cosine to find \( \theta \):\[ \theta = \cos^{-1}(0.8) \]Using a calculator or cosine inverse tables, we find:\[ \theta \approx 36.87^\circ \]

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a way to multiply two vectors, resulting in a scalar quantity. It provides information about the alignment of two vectors. When two vectors are aligned perfectly in the same direction, their dot product is maximized. If they are perpendicular, the dot product is zero.

For vectors \( \mathbf{x} = [x_1, x_2] \) and \( \mathbf{y} = [y_1, y_2] \), the dot product is calculated as:

• \( \mathbf{x} \cdot \mathbf{y} = x_1 \cdot y_1 + x_2 \cdot y_2 \)

Thus, for \( \mathbf{x} = [3,1] \) and \( \mathbf{y} = [3,-1] \), the calculation becomes:
\[ 3 \times 3 + 1 \times (-1) = 9 - 1 = 8 \]

This value, 8, is the dot product of our vectors. It indicates the extent to which these vectors are pointing in the same general direction.

Vector Magnitude

The vector magnitude, or length, is a measure of how long a vector is. It's calculated using the coordinates of the vector. The magnitude is like the distance from the origin to the point defined by the vector in space.

For a vector \( \mathbf{a} = [a_1, a_2] \), its magnitude is defined as:

• \( ||\mathbf{a}|| = \sqrt{a_1^2 + a_2^2} \)

For example, to find the magnitude of \( \mathbf{x} = [3, 1] \), compute:
\[ ||\mathbf{x}|| = \sqrt{3^2 + 1^2} = \sqrt{10} \]
Similarly, for \( \mathbf{y} = [3, -1] \):
\[ ||\mathbf{y}|| = \sqrt{3^2 + (-1)^2} = \sqrt{10} \]

Both vectors have magnitudes of \( \sqrt{10} \), which reflects their lengths, independent of direction.

Angle Between Vectors

The angle between two vectors is one of the fundamental concepts in vector calculus. It tells us how much one vector deviates from another. This angle can be determined using the cosine of the angle, derived from the dot product and magnitudes of the vectors.

The formula used to find the cosine of the angle \( \theta \) between two vectors \( \mathbf{x} \) and \( \mathbf{y} \) is:

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

Substituting our values, we get:
\[ \cos(\theta) = \frac{8}{\sqrt{10} \cdot \sqrt{10}} = \frac{8}{10} = 0.8 \]

To find \( \theta \), apply the inverse cosine function:
\[ \theta = \cos^{-1}(0.8) \]
After computing, this gives us:
\[ \theta \approx 36.87^\circ \]
This angle indicates how far apart the directions of the two vectors are in a plane.