Problem 35 - Precalculus Mathematics for Calculus

Question

Find \(2 u,-3 v, u+v,\) and \(3 u-4 v\) for the given vectors \(\mathbf{u}\) and \(\mathbf{v}\) $$\mathbf{u}=2 \mathbf{i}, \quad \mathbf{v}=3 \mathbf{i}-2 \mathbf{j}$$

Step-by-Step Solution

Answer

\(2\mathbf{u} = 4\mathbf{i}, -3\mathbf{v} = -9\mathbf{i} + 6\mathbf{j}, \mathbf{u} + \mathbf{v} = 5\mathbf{i} - 2\mathbf{j}, 3\mathbf{u} - 4\mathbf{v} = -6\mathbf{i} + 8\mathbf{j}.\)

Step 1: Calculate \(2\mathbf{u}\)

First, multiply the vector \(\mathbf{u} = 2\mathbf{i}\) by 2. This means multiplying each component of the vector by 2, resulting in \(2\mathbf{u} = 2 \times 2\mathbf{i} = 4\mathbf{i}\).

Step 2: Calculate \(-3\mathbf{v}\)

Next, multiply the vector \(\mathbf{v} = 3\mathbf{i} - 2\mathbf{j}\) by -3. This involves multiplying each component by -3, resulting in \(-3\mathbf{v}= -3 \times 3\mathbf{i} + (-3) \times (-2)\mathbf{j} = -9\mathbf{i} + 6\mathbf{j}\).

Step 3: Calculate \(\mathbf{u} + \mathbf{v}\)

Add vectors \(\mathbf{u} = 2\mathbf{i}\) and \(\mathbf{v} = 3\mathbf{i} - 2\mathbf{j}\) component-wise. This gives: \[\mathbf{u} + \mathbf{v} = (2\mathbf{i} + 3\mathbf{i}) + (0\mathbf{j} - 2\mathbf{j}) = 5\mathbf{i} - 2\mathbf{j}.\]

Step 4: Calculate \(3\mathbf{u} - 4\mathbf{v}\)

First, calculate \(3\mathbf{u}\) by multiplying \(\mathbf{u} = 2\mathbf{i}\) by 3: \[3\mathbf{u} = 3 \times 2\mathbf{i} = 6\mathbf{i}.\]Then calculate \(4\mathbf{v}\) by multiplying \(\mathbf{v} = 3\mathbf{i} - 2\mathbf{j}\) by 4: \[4\mathbf{v} = 4 \times 3\mathbf{i} + 4 \times (-2)\mathbf{j} = 12\mathbf{i} - 8\mathbf{j}.\]Now subtract \(4\mathbf{v}\) from \(3\mathbf{u}\): \[3\mathbf{u} - 4\mathbf{v} = 6\mathbf{i} - 12\mathbf{i} + 0\mathbf{j} + 8\mathbf{j} = -6\mathbf{i} + 8\mathbf{j}.\]

Key Concepts

Scalar Multiplication

Scalar multiplication is a basic operation in vector mathematics. It involves multiplying a numerical value (the scalar) with a vector. This operation scales the vector by the scalar's magnitude.

Consider a vector \( \mathbf{u} = 2\mathbf{i} \). If we apply scalar multiplication with a factor of 2, we perform the following calculation:

• Multiply each component of \( \mathbf{u} \) by 2: \( 2 \times 2\mathbf{i} = 4\mathbf{i} \).

This operation does not affect the direction of the vector, only its magnitude. Imagine stretching or shrinking the vector without changing its path.

Likewise, when multiplying \( \mathbf{v} = 3\mathbf{i} - 2\mathbf{j} \) by \(-3\), we scale down the vector and reverse its direction due to the negative sign:

• Calculate each component’s product: \(-3 \times 3\mathbf{i} = -9\mathbf{i} \) and \((-3) \times (-2)\mathbf{j} = 6\mathbf{j} \).

Scalar multiplication is fundamental in changing the size of vectors in various applications like physics and computer graphics.

Vector Addition

Vector addition combines two or more vectors to produce a resultant vector. This involves adding corresponding components of the vectors together. Consider two vectors \( \mathbf{u} \) and \( \mathbf{v} \).

For \( \mathbf{u} = 2\mathbf{i} \) and \( \mathbf{v} = 3\mathbf{i} - 2\mathbf{j} \), the addition is performed as follows:

• Add the \( \mathbf{i} \) components: \( \mathbf{u} + \mathbf{v} = (2\mathbf{i} + 3\mathbf{i}) = 5\mathbf{i} \).
• Add the \( \mathbf{j} \) components: \( \mathbf{j} + (-2\mathbf{j}) = -2\mathbf{j} \).

So, \( \mathbf{u} + \mathbf{v} = 5\mathbf{i} - 2\mathbf{j} \). This operation can be visualized as placing the vectors tail-to-head and finding the vector from the start of the first to the end of the second vector. In practical applications, vector addition helps in calculating the resultant force or velocity of an object.

Component-Wise Operations

Component-wise operations are crucial when dealing with vectors as they allow you to handle each dimension independently. Let's delve into the specifics of this approach when performing vector operations such as addition and subtraction.

For example, when adding vectors \( \mathbf{u} = 2\mathbf{i} \) and \( \mathbf{v} = 3\mathbf{i} - 2\mathbf{j} \), we consider each axis separately:

• Add \( \mathbf{i} \) components: \( 2 \mathbf{i} + 3 \mathbf{i} = 5 \mathbf{i} \).
• Add \( \mathbf{j} \) components: \( 0\mathbf{j} + (-2\mathbf{j}) = -2\mathbf{j} \).

Similarly, for subtraction such as \( 3\mathbf{u} - 4\mathbf{v} \), you also deal with each component:

• After finding \( 3\mathbf{u} = 6\mathbf{i} \) and \( 4\mathbf{v} = 12\mathbf{i} - 8\mathbf{j} \), subtract the components:
• \( \mathbf{i} \) components: \( 6\mathbf{i} - 12\mathbf{i} = -6\mathbf{i} \).
• \( \mathbf{j} \) components: \( 0\mathbf{j} - (-8\mathbf{j}) = 8\mathbf{j} \).

Component-wise operations provide clarity when managing complex problems involving multiple vectors, allowing easy calculations on each dimension.