Problem 8 - Advanced Engineering Mathematics

Question

In Problems \(1-12, \mathbf{a}=\langle 2,-3,4\rangle, \mathbf{b}=\langle-1,2,5\rangle\), and \(\mathbf{c}=\langle 3,6,-1\rangle .\) Find the indicated scalar or vector. \((2 b) \cdot(3 c)\)

Step-by-Step Solution

Answer

The result is 24.

Step 1: Multiply Vector b by Scalar 2

To find \(2\mathbf{b}\), multiply each component of vector \(\mathbf{b} = \langle -1, 2, 5 \rangle\) by 2. This gives us \(2\mathbf{b} = \langle 2(-1), 2(2), 2(5)\rangle = \langle -2, 4, 10 \rangle\).

Step 2: Multiply Vector c by Scalar 3

Next, compute \(3\mathbf{c}\) by multiplying each component of vector \(\mathbf{c} = \langle 3, 6, -1 \rangle\) by 3. This results in \(3\mathbf{c} = \langle 3(3), 3(6), 3(-1)\rangle = \langle 9, 18, -3 \rangle\).

Step 3: Compute the Dot Product of (2b) and (3c)

Now, compute the dot product of \(2\mathbf{b}\) and \(3\mathbf{c}\). Use the formula for dot product: \(\langle a_1, a_2, a_3 \rangle \cdot \langle b_1, b_2, b_3 \rangle = a_1b_1 + a_2b_2 + a_3b_3\). Substitute \(2\mathbf{b} = \langle -2, 4, 10 \rangle\) and \(3\mathbf{c} = \langle 9, 18, -3 \rangle\): \(-2 \cdot 9 + 4 \cdot 18 + 10 \cdot (-3) = -18 + 72 - 30 = 24\).

Key Concepts

Scalar Multiplication

Scalar multiplication is a fundamental concept in vector calculus. It involves multiplying a vector by a scalar, which is just a single number. This operation scales the vector, changing its magnitude but not its direction (unless the scalar is negative, which reverses the direction). For example, consider the vector \( \mathbf{b} = \langle -1, 2, 5 \rangle \). To calculate \( 2\mathbf{b} \), as described in the first step of our solution, we multiply each component of \( \mathbf{b} \) by 2:

• The first component: \(-1 \times 2 = -2\)
• The second component: \(2 \times 2 = 4\)
• The third component: \(5 \times 2 = 10\)

This yields the scaled vector \( 2\mathbf{b} = \langle -2, 4, 10 \rangle \).
This operation stretches the vector twice as long, maintaining its original direction.

Dot Product

The dot product, also known as the scalar product, is an operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This operation is crucial in vector calculus because it combines the magnitudes and directions of two vectors to find a specific scalar quantity.

Calculating the dot product involves multiplying corresponding components of the vectors and summing those products. For vectors \( \langle a_1, a_2, a_3 \rangle \) and \( \langle b_1, b_2, b_3 \rangle \), the formula is:

• \( a_1 \times b_1 \)
• \( a_2 \times b_2 \)
• \( a_3 \times b_3 \)

Then sum these products together: \( a_1b_1 + a_2b_2 + a_3b_3 \).
In our exercise, the dot product of \( 2\mathbf{b} = \langle -2, 4, 10 \rangle \) and \( 3\mathbf{c} = \langle 9, 18, -3 \rangle \) was calculated as \(-18 + 72 - 30 = 24\).
This result tells us about the spatial relationship between the vectors.

Vector Operations

Vector operations are essential tools in vector calculus and applications involving physics and engineering. They help us perform various calculations using vectors, which have both magnitude and direction.

Key vector operations include:

• Scalar multiplication: Adjusts the magnitude of a vector.
• Dot product: Combines two vectors into a scalar, reflecting the extent to which they point in the same direction.
• Cross product: (Not in the current problem) Produces a new vector that is orthogonal to the other two.

In our task, we used scalar multiplication and the dot product to manipulate and combine vectors effectively. Such operations are extremely helpful in physics, where they can describe forces, velocities, and other vector quantities.

Step-by-Step Solution

Employing a step-by-step solution approach is a powerful method for solving problems, especially in mathematics and vector calculus. This strategy involves breaking down a complex problem into manageable parts and solving each part systematically.

For this specific exercise, the steps were:

• Multiply vector \( \mathbf{b} = \langle -1, 2, 5 \rangle \) by scalar 2.
• Multiply vector \( \mathbf{c} = \langle 3, 6, -1 \rangle \) by scalar 3.
• Compute the dot product of the resulting vectors \( 2\mathbf{b} \) and \( 3\mathbf{c} \).

By tackling one operation at a time, we simplify the process and minimize errors. This approach not only leads to the correct solution of \( 24 \) but also enhances understanding of each mathematical operation involved.