Problem 3 - Precalculus: Functions and Graphs

Question

Find \(a+b, a-b, 4 a+5 b, 4 a-5 b,\) and \(\|a\|\) $$\mathbf{a}=-(7,-2), \quad \mathbf{b}=4(-2,1)$$

Step-by-Step Solution

Answer

\(a+b = (-15,2), a-b = (1,-6), 4a+5b = (-68,12), 4a-5b = (12,-28), \|a\| = \sqrt{53}\)

Step 1: Simplify the vectors

First, let's express vector \( \mathbf{b} \) in a simplified form. Given that \( \mathbf{b} = 4(-2,1) \), distribute the 4 to get: \[ \mathbf{b} = (4 \times -2, 4 \times 1) = (-8, 4) \]

Step 2: Calculate \( a+b \)

Add the corresponding components of vectors \( \mathbf{a} \) and \( \mathbf{b} \). We have:\[ \mathbf{a} + \mathbf{b} = ( -7 + (-8), -2 + 4) = (-15, 2) \]

Step 3: Calculate \( a-b \)

Subtract the corresponding components of vectors \( \mathbf{b} \) from \( \mathbf{a} \). We find:\[ \mathbf{a} - \mathbf{b} = ( -7 - (-8), -2 - 4) = (1, -6) \]

Step 4: Calculate \( 4a+5b \)

We need to compute \( 4\mathbf{a} + 5\mathbf{b} \). Start by finding \( 4\mathbf{a} \) and \( 5\mathbf{b} \):\[ 4\mathbf{a} = 4(-7, -2) = (-28, -8) \]\[ 5\mathbf{b} = 5(-8, 4) = (-40, 20) \]Now calculate:\[ 4\mathbf{a} + 5\mathbf{b} = (-28, -8) + (-40, 20) = (-68, 12) \]

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

Now calculate \( 4\mathbf{a} - 5\mathbf{b} \):\[ 4\mathbf{a} - 5\mathbf{b} = (-28, -8) - (-40, 20) = (-28 + 40, -8 - 20) = (12, -28) \]

Step 6: Calculate \( \|a\| \)

The magnitude of vector \( \mathbf{a} \) is given by:\[ \| \mathbf{a} \| = \sqrt{(-7)^2 + (-2)^2} = \sqrt{49 + 4} = \sqrt{53} \]

Key Concepts

Vector Addition

Vector addition involves adding the corresponding components of two vectors. If you have two vectors, say \( \mathbf{a} = (a_1, a_2) \) and \( \mathbf{b} = (b_1, b_2) \), the resulting vector \( \mathbf{a} + \mathbf{b} \) is

• \( a_1 + b_1 \) in the first component,
• and \( a_2 + b_2 \) in the second component.

This operation is rather intuitive, akin to simply moving along the x and y axes by the specified amounts. In the given exercise with \( \mathbf{a} = (-7, -2) \) and \( \mathbf{b} = (-8, 4) \), we find \( \mathbf{a} + \mathbf{b} = (-15, 2) \).
Here, each element in the vector represents a directional quantity that sums up. Think of it like combining forces from both vectors.

Vector Subtraction

Vector subtraction is similar to vector addition but involves subtracting components instead. To find the vector difference, \( \mathbf{a} - \mathbf{b} \), you will subtract each of vector \( \mathbf{b} \)'s components from the corresponding components of vector \( \mathbf{a} \):

• \( a_1 - b_1 \) is the first component,
• while \( a_2 - b_2 \) is the second component.

This represents the vector that points from the tip of \( \mathbf{b} \) to the tip of \( \mathbf{a} \). For \( \mathbf{a} = (-7, -2) \) and \( \mathbf{b} = (-8, 4) \), the calculation \( \mathbf{a} - \mathbf{b} = (1, -6) \) shows how each component changes according to the vectors’ positions.
This kind of operation can be visualized as finding the path from the endpoint of one vector to the other.

Scalar Multiplication

Scalar multiplication involves multiplying a vector by a real number (called a scalar). This operation changes the length of the vector but not its direction if positive. If negative, it also reverses the direction.
When multiplying a vector \( \mathbf{a} = (a_1, a_2) \) by a scalar \( c \), each component is multiplied by \( c \). Thus:

• The new vector is \( (c \cdot a_1, c \cdot a_2) \).

In our exercise, such operations include \( 4\mathbf{a} \) and \( 5\mathbf{b} \), with calculated results \( 4(-7, -2) = (-28, -8) \) and \( 5(-8, 4) = (-40, 20) \), respectively.
This manipulation is crucial in scaling vectors for applications such as enlargements or reductions in physics and geometry.

Vector Magnitude

The magnitude of a vector quantifies its length and is an essential concept when dealing with vectors, especially because it provides insights into the size of the vector's influence.
The magnitude \( \|\mathbf{a}\| \) of a vector \( \mathbf{a} = (a_1, a_2) \) is calculated using the Pythagorean theorem:

• \( \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2}. \)

This formula shows how to find the length of the hypotenuse in a right triangle formed by the vector components.
For the vector \( \mathbf{a} = (-7, -2) \), the magnitude becomes \( \|\mathbf{a}\| = \sqrt{(-7)^2 + (-2)^2} = \sqrt{53} \).
Understanding this helps in determining distance or force magnitude directions, commonly used in fields like physics and engineering.