Problem 5 - Precalculus: Functions and Graphs

Question

Find \(a+b, a-b, 4 a+5 b, 4 a-5 b,\) and \(\|a\|\) $$\mathbf{a}=\mathbf{i}+2 \mathbf{j}, \quad \mathbf{b}=3 \mathbf{i}-5 \mathbf{j}$$

Step-by-Step Solution

Answer

\(a+b = 4\mathbf{i} - 3\mathbf{j}\), \(a-b = -2\mathbf{i} + 7\mathbf{j}\), \(4a+5b = 19\mathbf{i} - 17\mathbf{j}\), \(4a-5b = -11\mathbf{i} + 33\mathbf{j}\), \(\|a\| = \sqrt{5}\).

Step 1: Find \(a+b\)

To find \( \mathbf{a} + \mathbf{b} \), add the corresponding components from each vector.\[ \mathbf{a} + \mathbf{b} = (1+3)\mathbf{i} + (2-5)\mathbf{j} = 4\mathbf{i} - 3\mathbf{j} \]

Step 2: Find \(a-b\)

Subtract the components of vector \( \mathbf{b} \) from the components of vector \( \mathbf{a} \).\[ \mathbf{a} - \mathbf{b} = (1-3)\mathbf{i} + (2+5)\mathbf{j} = -2\mathbf{i} + 7\mathbf{j} \]

Step 3: Find \(4a+5b\)

First, scale each vector by the respective coefficient and then add.\[ 4\mathbf{a} = 4(1\mathbf{i} + 2\mathbf{j}) = 4\mathbf{i} + 8\mathbf{j} \] \[ 5\mathbf{b} = 5(3\mathbf{i} - 5\mathbf{j}) = 15\mathbf{i} - 25\mathbf{j} \] \[ 4\mathbf{a} + 5\mathbf{b} = (4 + 15)\mathbf{i} + (8 - 25)\mathbf{j} = 19\mathbf{i} - 17\mathbf{j} \]

Step 4: Find \(4a-5b\)

Scale each vector by the respective coefficient and then subtract.\[ 4\mathbf{a} = 4(1\mathbf{i} + 2\mathbf{j}) = 4\mathbf{i} + 8\mathbf{j} \] \[ 5\mathbf{b} = 5(3\mathbf{i} - 5\mathbf{j}) = 15\mathbf{i} - 25\mathbf{j} \] \[ 4\mathbf{a} - 5\mathbf{b} = (4 - 15)\mathbf{i} + (8 + 25)\mathbf{j} = -11\mathbf{i} + 33\mathbf{j} \]

Step 5: Find \(\|a\|\)

The magnitude of vector \( \mathbf{a} = 1\mathbf{i} + 2\mathbf{j} \) is calculated by the formula \( \|\mathbf{a}\| = \sqrt{x^2 + y^2} \).\[ \|\mathbf{a}\| = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \]

Key Concepts

Vector Addition

Vector addition is like finding the total of two different paths taken in a journey. Imagine you have two vectors, \( \mathbf{a} = \mathbf{i} + 2 \mathbf{j} \) and \( \mathbf{b} = 3 \mathbf{i} - 5 \mathbf{j} \). To add these vectors, you simply add the corresponding components. This means you:

• Add the \( \mathbf{i} \)-components: \( 1 + 3 = 4 \)
• Add the \( \mathbf{j} \)-components: \( 2 - 5 = -3 \)

After adding, the resulting vector is \( 4 \mathbf{i} - 3 \mathbf{j} \). Vector addition is straightforward since you just focus on those parallel components.

Vector Subtraction

Vector subtraction determines the difference between two vectors, much like finding the change in position. Using our earlier vectors, \( \mathbf{a} \) and \( \mathbf{b} \), subtracting \( \mathbf{b} \) from \( \mathbf{a} \) involves:

• Subtracting the \( \mathbf{i} \)-components: \( 1 - 3 = -2 \)
• Subtracting the \( \mathbf{j} \)-components: \( 2 + 5 = 7 \)

Thus, the vector \( \mathbf{a} - \mathbf{b} \) becomes \( -2 \mathbf{i} + 7 \mathbf{j} \). The subtraction switches the direction and magnitude, representing the path you need to take to trace back part of your journey.

Scalar Multiplication

Scalar multiplication stretches or shrinks a vector, depending on the scalar value. For instance, if you multiply vector \( \mathbf{a} \) by 4, you multiply each component by 4:

• \( 4(\mathbf{i}) = 4 \mathbf{i} \)
• \( 4(2 \mathbf{j}) = 8 \mathbf{j} \)

Similarly, multiplying vector \( \mathbf{b} \) by 5 results in:

• \( 5(3 \mathbf{i}) = 15 \mathbf{i} \)
• \( 5(-5 \mathbf{j}) = -25 \mathbf{j} \)

Scalar multiplication can change both the length and the direction of a vector, and when combined with addition or subtraction, allows you to further manipulate vectors as in calculating \( 4\mathbf{a} + 5\mathbf{b} \) or \( 4\mathbf{a} - 5\mathbf{b} \). This power of altering vectors helps model various real-world scenarios, such as forces moving differently through space.

Magnitude of a Vector

The magnitude of a vector tells you how long the vector is, giving the "distance" from the origin. For our vector \( \mathbf{a} = \mathbf{i} + 2\mathbf{j} \), the process to find the magnitude is:

• Square each component: \( 1^2 = 1 \) and \( 2^2 = 4 \)
• Add these squares: \( 1 + 4 = 5 \)
• Take the square root: \( \sqrt{5} \)

This results in a magnitude of \( \sqrt{5} \). Understanding magnitudes helps you compare vector sizes independently of directions, important in fields like physics where forces' strengths are crucial.