Problem 2 - Precalculus: Functions and Graphs

Question

Find \(a+b, a-b, 4 a+5 b, 4 a-5 b,\) and \(\|a\|\) $$\mathbf{a}=(-2,6), \quad \mathbf{b}=(2,3)$$

Step-by-Step Solution

Answer

\(a+b = (0, 9); a-b = (-4, 3); 4a+5b = (2, 39); 4a-5b = (-18, 9); \|a\| = 2\sqrt{10}\).

Step 1: Find \(a + b\)

To find \(a + b\), we add the corresponding components of vectors \(\mathbf{a}\) and \(\mathbf{b}\). So, \(a_1 + b_1 = -2 + 2 = 0\) and \(a_2 + b_2 = 6 + 3 = 9\). Therefore, \(a + b = (0, 9)\).

Step 2: Find \(a - b\)

To find \(a - b\), we subtract the components of vector \(\mathbf{b}\) from vector \(\mathbf{a}\). So, \(a_1 - b_1 = -2 - 2 = -4\) and \(a_2 - b_2 = 6 - 3 = 3\). Therefore, \(a - b = (-4, 3)\).

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

First, calculate \(4a\): \(4a_1 = 4(-2) = -8\) and \(4a_2 = 4(6) = 24\), so \(4a = (-8, 24)\). Then, calculate \(5b\): \(5b_1 = 5(2) = 10\) and \(5b_2 = 5(3) = 15\), so \(5b = (10, 15)\). Now, sum \(4a\) and \(5b\): \((-8 + 10, 24 + 15) = (2, 39)\). Therefore, \(4a + 5b = (2, 39)\).

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

Using the previously calculated components, subtract \(5b\) from \(4a\): \((-8 - 10, 24 - 15) = (-18, 9)\). Therefore, \(4a - 5b = (-18, 9)\).

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

To find the magnitude of \(\mathbf{a}\), use the formula \(\|a\| = \sqrt{a_1^2 + a_2^2}\). So, \(a_1^2 = (-2)^2 = 4\) and \(a_2^2 = 6^2 = 36\). Then, \(\|a\| = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}\).

Key Concepts

Vector Subtraction

Subtraction of vectors is quite similar to subtracting numbers. To subtract one vector from another, you simply subtract corresponding components. Let's take a closer look at the exercise problem:
- We are given two vectors, \( \mathbf{a} = (-2,6) \) and \( \mathbf{b} = (2,3) \).- To find \( \mathbf{a} - \mathbf{b} \), you subtract the first component of \( \mathbf{b} \) from the first component of \( \mathbf{a} \), and the second component of \( \mathbf{b} \) from the second component of \( \mathbf{a} \).Here's the step-by-step calculation:

• First component: \( -2 - 2 = -4 \)
• Second component: \( 6 - 3 = 3 \)

Therefore, \( \mathbf{a} - \mathbf{b} = (-4, 3) \).
This simple operation is useful in various applications, such as determining the relative position of objects.

Scalar Multiplication

Scalar multiplication involves multiplying a vector by a scalar (a constant number). This operation changes the magnitude of the vector but not its direction. Let's see how it works in the exercise:
When multiplying a vector by a scalar, you multiply each component of the vector by the scalar:

• For \( 4\mathbf{a} \), multiply each component of \( \mathbf{a} \) by 4: - \( 4(-2) = -8 \) - \( 4(6) = 24 \) - Hence, \( 4\mathbf{a} = (-8, 24) \)
• For \( 5\mathbf{b} \), multiply each component of \( \mathbf{b} \) by 5: - \( 5(2) = 10 \) - \( 5(3) = 15 \) - Hence, \( 5\mathbf{b} = (10, 15) \)

Scalar multiplication is a powerful tool that helps in scaling vectors to desired lengths while maintaining the original direction.

Magnitude of a Vector

Finding the magnitude of a vector is essential in understanding its size or length. The magnitude provides the length of that vector in a coordinate system, and is calculated using the Pythagorean theorem.
The formula for the magnitude of a vector \( \mathbf{a} = (a_1, a_2) \) is:\[\| \mathbf{a} \| = \sqrt{a_1^2 + a_2^2}\]Applying this to the vector \( \mathbf{a} = (-2, 6) \):

• Square the first component: \( (-2)^2 = 4 \)
• Square the second component: \( 6^2 = 36 \)
• Add these squares: \( 4 + 36 = 40 \)
• Take the square root: \( \sqrt{40} = 2\sqrt{10} \)

Thus, the magnitude of \( \mathbf{a} \) is \( 2\sqrt{10} \).
Understanding vector magnitude is crucial in fields like physics and engineering, where it represents quantities like force and velocity.