Problem 32 - Calculus for Biology and Medicine

Question

Let $$ \mathbf{u}=\left[\begin{array}{l} 3 \\ 4 \end{array}\right], \quad \mathbf{v}=\left[\begin{array}{r} 1 \\ -2 \end{array}\right], \quad \text { and } \quad \mathbf{w}=\left[\begin{array}{l} -1 \\ -2 \end{array}\right] $$ Compute \(\mathbf{v}-\frac{1}{2} \mathbf{u}\) and illustrate the result graphically.

Step-by-Step Solution

Answer

The result of \(\mathbf{v} - \frac{1}{2} \mathbf{u}\) is the vector \([-0.5, -4]\).

Step 1: Determine Scalar Multiplication

To solve the expression \(\mathbf{v} - \frac{1}{2} \mathbf{u}\), begin by calculating the product \(\frac{1}{2} \mathbf{u}\). The vector \(\mathbf{u}\) is \(\begin{bmatrix} 3 \ 4 \end{bmatrix}\). To multiply the vector by \(\frac{1}{2}\), apply the scalar multiplication component-wise: \[\frac{1}{2} \mathbf{u} = \frac{1}{2} \begin{bmatrix} 3 \ 4 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} \cdot 3 \ \frac{1}{2} \cdot 4 \end{bmatrix} = \begin{bmatrix} 1.5 \ 2 \end{bmatrix}.\]

Step 2: Subtract Vectors

Using the results from Step 1, subtract \(\frac{1}{2} \mathbf{u}\) from \(\mathbf{v}\). Here, \(\mathbf{v}\) is given by \(\begin{bmatrix} 1 \ -2 \end{bmatrix}\). Perform the subtraction component-wise: \[\mathbf{v} - \frac{1}{2} \mathbf{u} = \begin{bmatrix} 1 \ -2 \end{bmatrix} - \begin{bmatrix} 1.5 \ 2 \end{bmatrix} = \begin{bmatrix} 1 - 1.5 \ -2 - 2 \end{bmatrix} = \begin{bmatrix} -0.5 \ -4 \end{bmatrix}.\]

Step 3: Illustrate Graphically

To illustrate the result, plot the original vectors \(\mathbf{v}\) and \(\mathbf{u}\) on a coordinate plane with the calculated vector \(\mathbf{v} - \frac{1}{2} \mathbf{u}\). \(\mathbf{v}\) starts at the origin and ends at the point \((1, -2)\), and \(\mathbf{u}\) ends at \((3, 4)\). The result \(\begin{bmatrix} -0.5 \ -4 \end{bmatrix}\) also starts at the origin but points to \((-0.5, -4)\). Plotting these illustrates the effect of the scalar multiplication and subsequent subtraction.

Key Concepts

Scalar Multiplication

Scalar multiplication is a fundamental operation in vector algebra where a vector is multiplied by a scalar (a real number). The process involves multiplying each component of the vector by the scalar. This stretches or shrinks the vector, depending on the magnitude of the scalar. If the scalar is positive, the direction remains unchanged; if negative, the direction reverses.

For example, consider the vector \( \mathbf{u} = \begin{bmatrix} 3 \ 4 \end{bmatrix} \). Multiplying \( \mathbf{u} \) by the scalar \( \frac{1}{2} \) results in the vector \( \frac{1}{2} \mathbf{u} = \begin{bmatrix} \frac{3}{2} \ 2 \end{bmatrix} \). This new vector has been shortened by half, maintaining the same direction since the scalar \( \frac{1}{2} \) is positive.

When performing operations like in the exercise, accurately computing scalar multiplication is crucial as it sets the stage for further operations such as addition or subtraction.

Vector Subtraction

Vector subtraction involves taking the difference between two vectors by subtracting corresponding components. This operation can be understood as adding the first vector to the negative of the second.

In the problem, we need to compute \( \mathbf{v} - \frac{1}{2} \mathbf{u} \). Here, \( \mathbf{v} = \begin{bmatrix} 1 \ -2 \end{bmatrix} \) and \( \frac{1}{2} \mathbf{u} = \begin{bmatrix} 1.5 \ 2 \end{bmatrix} \). The subtraction is carried out component-wise:

• The first component: \( 1 - 1.5 = -0.5 \)
• The second component: \( -2 - 2 = -4 \)

Thus, the resultant vector is \( \begin{bmatrix} -0.5 \ -4 \end{bmatrix} \).

This operation essentially finds a new direction and magnitude, showing how the vectors interact in space. It's like moving a point represented by \( \mathbf{v} \) by the negative placement of \( \frac{1}{2} \mathbf{u} \).

Graphical Illustration

Graphical illustration of vectors is a visual representation that helps in understanding their interactions and resultant directions. When graphing vectors, they are often depicted with arrows originating from the origin to their respective points in the plane.

For the vectors in this exercise:

• \( \mathbf{v} \) can be plotted from origin \( (0,0) \) to \( (1,-2) \)
• \( \mathbf{u} \) is drawn from \( (0,0) \) to \( (3,4) \)
• The resulting vector \( \begin{bmatrix} -0.5 \ -4 \end{bmatrix} \) extends from the origin to \( (-0.5, -4) \)

Through plotting, you'll see \( \mathbf{v} - \frac{1}{2} \mathbf{u} \) as a new vector that heads in the combined influence direction of first moving by \( \mathbf{v} \) and then back by \( \frac{1}{2} \mathbf{u} \). This graphical form is pivotal in grasping the geometric implications of algebraic vector operations.