Problem 18 - Precalculus Mathematics for Calculus

Question

Find the vectors \(\mathbf{u}+\mathbf{v}, \mathbf{u}-\mathbf{v},\) and \(3 \mathbf{u}-\frac{1}{2} \mathbf{v}\) $$\mathbf{u}=(a, 2 b, 3 c), \mathbf{v}=\langle- 4 a, b,-2 c\rangle$$

Step-by-Step Solution

Answer

\( \mathbf{u} + \mathbf{v} = (-3a, 3b, c) \), \( \mathbf{u} - \mathbf{v} = (5a, b, 5c) \), and \( 3\mathbf{u} - \frac{1}{2}\mathbf{v} = (5a, \frac{11}{2}b, 10c) \).

Step 1: Add Vectors

To find \( \mathbf{u} + \mathbf{v} \), add corresponding components of the vectors \( \mathbf{u} \) and \( \mathbf{v} \).\[ \mathbf{u} + \mathbf{v} = (a, 2b, 3c) + (-4a, b, -2c) \]This results in:\[ (a + (-4a), 2b + b, 3c + (-2c)) = (-3a, 3b, c) \]

Step 2: Subtract Vectors

To find \( \mathbf{u} - \mathbf{v} \), subtract the components of \( \mathbf{v} \) from \( \mathbf{u} \).\[ \mathbf{u} - \mathbf{v} = (a, 2b, 3c) - (-4a, b, -2c) \]This results in:\[ (a - (-4a), 2b - b, 3c - (-2c)) = (5a, b, 5c) \]

Step 3: Compute Linear Combination

For the expression \( 3 \mathbf{u} - \frac{1}{2} \mathbf{v} \), compute \( 3 \mathbf{u} \) and \( \frac{1}{2} \mathbf{v} \), then subtract the latter from the former.Start with multiplying \( \mathbf{u} \) by 3:\[ 3 \mathbf{u} = 3(a, 2b, 3c) = (3a, 6b, 9c) \]Multiply \( \mathbf{v} \) by \( \frac{1}{2} \):\[ \frac{1}{2} \mathbf{v} = \frac{1}{2}(-4a, b, -2c) = (-2a, \frac{1}{2}b, -c) \]Subtract \( \frac{1}{2} \mathbf{v} \) from \( 3 \mathbf{u} \):\[ 3 \mathbf{u} - \frac{1}{2} \mathbf{v} = (3a, 6b, 9c) - (-2a, \frac{1}{2}b, -c) \]This results in:\[ (3a + 2a, 6b - \frac{1}{2}b, 9c + c) = (5a, \frac{11}{2}b, 10c) \]

Key Concepts

Addition of Vectors

Adding vectors is a central operation in vector mathematics, which combines two or more vectors to produce a single resultant vector. Imagine you have two vectors, denoted as \(\mathbf{u}\) and \(\mathbf{v}\). Each vector can be represented as a list of components. For instance, let's use the vectors given in the problem: \(\mathbf{u} = (a, 2b, 3c)\) and \(\mathbf{v} = \langle-4a, b, -2c\rangle\).

To add these vectors, you simply add their corresponding components:

• Add the first components: \(a + (-4a)\)
• Add the second components: \(2b + b\)
• Add the third components: \(3c + (-2c)\)

This straightforward process gives us:
\[\mathbf{u} + \mathbf{v} = (-3a, 3b, c) \]Vectors follow the commutative property, meaning that \(\mathbf{u} + \mathbf{v}\) is the same as \(\mathbf{v} + \mathbf{u}\). This property is useful in many areas of physics and engineering, where vector quantities like force and velocity are often combined.

Subtraction of Vectors

Subtracting vectors is very similar to adding them, except instead of combining, you are finding the difference between two vectors. This operation can be visualized as finding a vector that connects the tail of the one you are subtracting from to the head of the vector being subtracted.

With the example vectors \(\mathbf{u} = (a, 2b, 3c)\) and \(\mathbf{v} = \langle-4a, b, -2c\rangle\), vector subtraction is performed by subtracting components of \(\mathbf{v}\) from \(\mathbf{u}\):

• Subtract the first components: \(a - (-4a) = a + 4a\)
• Subtract the second components: \(2b - b\)
• Subtract the third components: \(3c - (-2c) = 3c + 2c\)

This process yields:\[\mathbf{u} - \mathbf{v} = (5a, b, 5c)\]Vector subtraction is not commutative. That means \(\mathbf{u} - \mathbf{v}\) is different from \(\mathbf{v} - \mathbf{u}\). This operation is particularly important in calculating changes in position or velocity, known as displacement vector.

Scalar Multiplication of Vectors

Scalar multiplication of vectors involves multiplying each component of the vector by a scalar, which is a constant value. This operation scales the vector, changing its magnitude while maintaining its direction unless the scalar is negative, which then reverses the direction.

Consider the vector \(\mathbf{u} = (a, 2b, 3c)\) from the problem. If we multiply \(\mathbf{u}\) by a scalar, for example 3, we multiply each component by 3:

• First component: \(3 \times a = 3a\)
• Second component: \(3 \times 2b = 6b\)
• Third component: \(3 \times 3c = 9c\)

Thus, we obtain:\[3\mathbf{u} = (3a, 6b, 9c)\]
Now consider the vector \(\mathbf{v} = \langle-4a, b, -2c\rangle\) and a scalar of \(\frac{1}{2}\). The multiplication proceeds similarly:

• First component: \(\frac{1}{2} \times (-4a) = -2a\)
• Second component: \(\frac{1}{2} \times b = \frac{1}{2}b\)
• Third component: \(\frac{1}{2} \times (-2c) = -c\)

So,\[\frac{1}{2}\mathbf{v} = (-2a, \frac{1}{2}b, -c)\]This concept is pivotal in various applications, like deriving a vector projection or changing units in physics problems.