Problem 30 - Calculus for Biology and Medicine

Question

Let \(\mathbf{x}=[2,0,1]^{\prime}\). Find \(\mathbf{y}\) so that \(\mathbf{x}\) and \(\mathbf{y}\) are perpendicular.

Step-by-Step Solution

Answer

The vector \( \mathbf{y} = [1, 0, -2]' \) is perpendicular to \( \mathbf{x} = [2, 0, 1]' \).

Step 1: Understanding Vectors Perpendicularity

Two vectors \(\mathbf{x}\) and \(\mathbf{y}\) are perpendicular if their dot product is zero. For vectors \(\mathbf{x}=[2,0,1]^{\prime}\) and \(\mathbf{y}=[y_1,y_2,y_3]^{\prime}\), the dot product is given by: \(2y_1 + 0\cdot y_2 + 1\cdot y_3\). Therefore, we need to find \(y_1, y_2, y_3\) such that \(2y_1 + y_3 = 0\).

Step 2: Setting Up the Equation

Given the condition for perpendicularity, write the equation \(2y_1 + y_3 = 0\). This equation represents the relationship between \(y_1\) and \(y_3\) necessary for the vectors to be perpendicular. \(y_2\) can be any value since it doesn't affect the dot product with \(\mathbf{x}\).

Step 3: Solving for One of the Variables

From the equation \(2y_1 + y_3 = 0\), express \(y_3\) in terms of \(y_1\): \(y_3 = -2y_1\). This gives us a family of solutions for different values of \(y_1\).

Step 4: Choosing Specific Values

Set \(y_1 = 1\) to find a specific solution. Substitute into the equation \(y_3 = -2y_1\) to get \(y_3 = -2\). Choose any arbitrary value for \(y_2\), such as \(y_2 = 0\), to keep it simple.

Step 5: Constructing a Solution Vector

Based on the chosen values, construct the vector \(\mathbf{y} = [1, 0, -2]^{\prime}\) as a valid solution that meets the perpendicularity condition.

Key Concepts

Vector Perpendicularity

Vectors are geometric entities that have both magnitude and direction. Two vectors are said to be perpendicular if the angle between them is 90 degrees.
When this condition is met,

• They are orthogonal.
• Their interaction results in a dot product of zero.

Understanding this geometric property helps in visualizing how vectors relate in space. This can be useful when extending concepts in physics, engineering, and computer graphics. Vectors perpendicular to each other do not share any similarity in direction. Thus, they form part of the foundation for many geometric representations.
To find vectors that are perpendicular to a given vector, like \(\mathbf{x}=[2,0,1]'\), it is necessary to manipulate components so that their dot product either cancels out or equals zero.

Dot Product

The dot product is a crucial operation in vector calculus because it measures how much two vectors are aligned.
To calculate the dot product of two vectors \(\mathbf{a}=[a_1, a_2, a_3]'\) and \(\mathbf{b}=[b_1, b_2, b_3]'\), you do the following:

• Multiply their respective components: \(a_1\cdot b_1, a_2\cdot b_2, a_3\cdot b_3\).
• Add the results: \(a_1\cdot b_1 + a_2\cdot b_2 + a_3\cdot b_3\).

This operation essentially combines the directional components of each vector, quantifying their interaction.
If the result of a dot product is zero, the vectors are perpendicular. This is because the cosine of 90 degrees (the angle between perpendicular lines) is zero, making the dot product an excellent tool to determine vector perpendicularity without direct angle measurement.
The dot product is also used to find the magnitude projection of one vector onto another.

Linear Equations

A linear equation forms a straight line when graphed in a coordinate plane. These equations demonstrate how different components interact to form solutions that can be visualized in simple geometric forms.
The equation for vector perpendicularity \(2y_1 + y_3 = 0\) is a linear equation.
In this context:

• \(y_1\) and \(y_3\) are variables that must relate linearly to satisfy the condition.
• The vector \(y_2\) can vary independently since its coefficient in the dot product equation is zero.

Solving this linear equation provides a family of solutions due to its single-dependency on two variables. When you choose a fixed value for \(y_1\), you can easily find \(y_3\) by rearranging and solving the equation.
This approach is fundamental in linear algebra and is often applied to understand structures, predict outcomes, and solve real-world problems that follow linear patterns.