Problem 73 - Precalculus with Calculus Previews

Question

In Problems \(65-76, \mathbf{u}=\langle 1,-3,2\rangle, \mathbf{v}=\langle-1,1,1\rangle\), and \(\mathbf{w}=\langle 2,6,9\rangle .\) Find the indicated vector or scalar. $$ \frac{1}{2} \mathbf{u} \cdot \mathbf{v} $$

Step-by-Step Solution

Answer

The result is -1.

Step 1: Calculate Half of Vector u

Start by finding \( \frac{1}{2} \mathbf{u} \). Since \( \mathbf{u} = \langle 1, -3, 2 \rangle \), we calculate each component by multiplying by \( \frac{1}{2} \):\[\frac{1}{2} \mathbf{u} = \left\langle \frac{1}{2} \times 1, \frac{1}{2} \times -3, \frac{1}{2} \times 2 \right\rangle = \langle \frac{1}{2}, -\frac{3}{2}, 1 \rangle.\]

Step 2: Set Up the Dot Product Formula

The dot product of two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \) is given by:\[\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3.\]

Step 3: Calculate the Dot Product

Use the components from \( \frac{1}{2} \mathbf{u} \) and \( \mathbf{v} \):\[\frac{1}{2} \mathbf{u} = \langle \frac{1}{2}, -\frac{3}{2}, 1 \rangle\quad \text{and}\quad \mathbf{v} = \langle -1, 1, 1 \rangle.\]Now calculate:\[\frac{1}{2} \mathbf{u} \cdot \mathbf{v} = \left( \frac{1}{2} \times -1 \right) + \left( -\frac{3}{2} \times 1 \right) + \left( 1 \times 1 \right).\]

Step 4: Simplify the Calculation

Perform each multiplication and then sum the results:\[\frac{1}{2} \cdot -1 = -\frac{1}{2},\quad -\frac{3}{2} \cdot 1 = -\frac{3}{2},\quad 1 \cdot 1 = 1.\]Now add these results:\[-\frac{1}{2} - \frac{3}{2} + 1 = -2 + 1 = -1.\]

Key Concepts

Dot Product

The dot product is a fundamental concept in vector calculus and is used to determine the interaction between two vectors. It is essentially a form of multiplying two vectors but results in a scalar value, not another vector. The dot product is useful in numerous applications, such as finding angles between vectors or determining orthogonality.
To compute the dot product, you multiply each corresponding component of the vectors and then sum those products. For example, if you have vectors

• \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \)
• \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \)

The dot product is calculated as: \[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]The result is a single number, a scalar.
In the exercise above, the dot product helps find the interaction between half of vector \( \mathbf{u} \) and vector \( \mathbf{v} \). By following the steps, you unravel the interactions of dimensions within the space the vectors describe. You multiply each corresponding component from both vectors and sum them to get \(-1\). This scalar represents a specific kind of alignment or interaction between the vectors.

Vector Arithmetic

Vector arithmetic is the cornerstone of vector calculus, and it manipulates vectors using operations like addition, subtraction, and multiplication. These operations enable us to perform various calculations and measurements, which are essential in fields such as physics, computer graphics, and engineering.
Specifically, the arithmetic operations allow vectors to be combined and compared, and they include:

• Addition: Combining two vectors to form a new vector by adding their corresponding components.
• Subtraction: Finding the difference between two vectors by subtracting corresponding components.
• Scalar Multiplication: Scaling a vector by multiplying its components by a scalar, which will be further detailed in another section.

These operations provide flexibility in calculations, such as finding resultants or analyzing motions. In the solution provided, you saw arithmetic used in simplifying expressions involving vectors and scalars to eventually compute the dot product. Understanding vector arithmetic allows you to manipulate vector components to discover meaningful scalar quantities or new vector results.

Scalar Multiplication

Scalar multiplication involves multiplying a vector by a scalar value, effectively scaling the vector. This operation alters the magnitude of the vector without changing its direction (except possibly reversing it if the scalar is negative).
This can be visualized as stretching or shrinking the vector's length proportional to the scalar. Consider vector \( \mathbf{u} = \langle 1, -3, 2 \rangle \) and a scalar \( \frac{1}{2} \). Scalar multiplication is performed as:

• \( \frac{1}{2} \times 1 = \frac{1}{2} \)
• \( \frac{1}{2} \times -3 = -\frac{3}{2} \)
• \( \frac{1}{2} \times 2 = 1 \)

The scaled vector becomes \( \langle \frac{1}{2}, -\frac{3}{2}, 1 \rangle \). This operation is critical in various applications, such as in computer graphics, where it helps in scaling object dimensions efficiently. In solving the given problem, scalar multiplication was employed as the initial step to obtain a new vector that could be used in the subsequent dot product operation, demonstrating its importance in simplifying and solving vector-related queries.