Problem 26 - Calculus for Biology and Medicine

Question

Find the angle between \(\mathbf{x}=[1,-3,2]^{\prime}\) and \(\mathbf{y}=[3,1,-4]^{\prime}\).

Step-by-Step Solution

Answer

The angle between the vectors is approximately \( 112.62^\circ \).

Step 1: Calculate the Dot Product

The dot product of two vectors \( \mathbf{x} \) and \( \mathbf{y} \) is given by the formula: \[ \mathbf{x} \cdot \mathbf{y} = x_1y_1 + x_2y_2 + x_3y_3 \]For \( \mathbf{x} = [1, -3, 2]^{\prime} \) and \( \mathbf{y} = [3, 1, -4]^{\prime} \), the dot product is:\[ 1 \times 3 + (-3) \times 1 + 2 \times (-4) = 3 - 3 - 8 = -8 \]

Step 2: Calculate the Magnitude of Each Vector

The magnitude of a vector \( \mathbf{x} = [x_1, x_2, x_3]^{\prime} \) is given by:\[ \| \mathbf{x} \| = \sqrt{x_1^2 + x_2^2 + x_3^2} \]For \( \mathbf{x} = [1, -3, 2]^{\prime} \), the magnitude is:\[ \| \mathbf{x} \| = \sqrt{1^2 + (-3)^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14} \]For \( \mathbf{y} = [3, 1, -4]^{\prime} \), the magnitude is:\[ \| \mathbf{y} \| = \sqrt{3^2 + 1^2 + (-4)^2} = \sqrt{9 + 1 + 16} = \sqrt{26} \]

Step 3: Use the Dot Product Formula to Find the Angle

The formula for the angle \( \theta \) between two vectors using the dot product is:\[ \cos \theta = \frac{\mathbf{x} \cdot \mathbf{y}}{\| \mathbf{x} \| \| \mathbf{y} \|} \]Plug in the values calculated:\[ \cos \theta = \frac{-8}{\sqrt{14} \times \sqrt{26}} \]Calculate \( \sqrt{14} \times \sqrt{26} = \sqrt{364} \). Therefore,\[ \cos \theta = \frac{-8}{\sqrt{364}} = \frac{-8}{2\sqrt{91}} = \frac{-4}{\sqrt{91}} \]

Step 4: Calculate the Angle \( \theta \)

To find the angle \( \theta \), take the inverse cosine (arccos) of \( \cos \theta \):\[ \theta = \cos^{-1} \left( \frac{-4}{\sqrt{91}} \right) \]Using a calculator, compute this value to find:\[ \theta \approx 112.62^\circ \]

Key Concepts

Understanding the Dot Product

The dot product is a fundamental operation in vector calculus that combines two vectors to produce a scalar quantity. It reveals important information about the relationship between the two vectors, such as whether they are orthogonal (perpendicular) or have some directional similarity or opposition.
In mathematical terms, the dot product of two vectors \( \mathbf{x} = [x_1, x_2, x_3]\) and \( \mathbf{y} = [y_1, y_2, y_3]\) is expressed as:

• \( \mathbf{x} \cdot \mathbf{y} = x_1y_1 + x_2y_2 + x_3y_3 \)

This operation combines corresponding components of the vectors through multiplication and then sums the results. This sum reflects how much one vector is aligned with another.
In the example of vectors \( \mathbf{x} = [1, -3, 2]\) and \( \mathbf{y} = [3, 1, -4]\), the dot product calculated is \(-8\), indicating they are not orthogonal and have a certain angle between them. A negative dot product often suggests that the vectors are pointing in broadly opposite directions.

Calculating the Magnitude of a Vector

The magnitude of a vector (also referred to as the length or norm) is a measure of its size or length in space, independent of its direction. It provides a single value that represents the extent of the vector, making it an invaluable tool in vector calculus.
The magnitude for a vector \( \mathbf{x} = [x_1, x_2, x_3]\) is calculated using the formula:

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

In this formula, each component of the vector is squared, the results are summed, and the square root is taken to yield the magnitude.
For the given vectors, the magnitudes are computed as:

• \( \mathbf{x} = [1, -3, 2] \) yields \( \sqrt{14} \)
• \( \mathbf{y} = [3, 1, -4] \) yields \( \sqrt{26} \)

These results quantify the vectors' lengths and are crucial for determining the angle between them when used alongside the dot product.

Using Inverse Cosine to Find the Angle Between Vectors

One of the more intriguing applications of the dot product and vector magnitude is using them together to determine the angle between two vectors. This involves using the inverse cosine function (often denoted as \( \cos^{-1}\) or arccos), which helps to find an angle whose cosine value is known.
The formula used to calculate the angle \( \theta \) between vectors \( \mathbf{x} \) and \( \mathbf{y} \) is:

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

First, the dot product \( \mathbf{x} \cdot \mathbf{y} \) and the magnitudes \( \| \mathbf{x} \| \) and \( \| \mathbf{y} \| \) are needed. With these, the cosine of the angle is determined. Then, by taking \( \cos^{-1} \) of this value, the angle \( \theta \) itself can be found.
In the example, \( \cos \theta = \frac{-4}{\sqrt{91}} \), and using a calculator to find the inverse cosine yields \( \theta \approx 112.62^\circ \). This value indicates that the vectors are neither parallel nor orthogonal but have an obtuse angle, showcasing the importance of understanding and applying vector calculus.