Problem 20 - Precalculus with Calculus Previews

Question

Find \(\mathbf{u}-4 \mathbf{v}\) and \(2 \mathbf{u}+5 \mathbf{v}\). $$ \mathbf{u}=5 \mathbf{i}-10 \mathbf{j}, \mathbf{v}=-10 \mathbf{i} $$

Step-by-Step Solution

Answer

\( \mathbf{u} - 4 \mathbf{v} = 45\mathbf{i} - 10\mathbf{j}, 2 \mathbf{u} + 5 \mathbf{v} = -40\mathbf{i} - 20\mathbf{j} \).

Step 1: Find \( \mathbf{u} - 4 \mathbf{v} \)

First, calculate the expression \( \mathbf{u} - 4 \mathbf{v} \). Substitute the given vectors into the expression: \( \mathbf{u} = 5 \mathbf{i} - 10 \mathbf{j} \) and \( \mathbf{v} = -10 \mathbf{i} \). Therefore, \( 4 \mathbf{v} = 4(-10 \mathbf{i}) = -40 \mathbf{i} \). Now, calculate \( \mathbf{u} - 4 \mathbf{v} = (5 \mathbf{i} - 10 \mathbf{j}) - (-40 \mathbf{i}) = (5 + 40)\mathbf{i} - 10 \mathbf{j} = 45\mathbf{i} - 10\mathbf{j} \).

Step 2: Find \( 2 \mathbf{u} + 5 \mathbf{v} \)

Next, calculate \( 2 \mathbf{u} + 5 \mathbf{v} \). Substitute \( \mathbf{u} = 5 \mathbf{i} - 10 \mathbf{j} \) and \( \mathbf{v} = -10 \mathbf{i} \). Multiply by their respective scalars: \( 2 \mathbf{u} = 2(5 \mathbf{i} - 10 \mathbf{j}) = 10 \mathbf{i} - 20 \mathbf{j} \), and \( 5 \mathbf{v} = 5(-10 \mathbf{i}) = -50 \mathbf{i} \). Now compute \( 2 \mathbf{u} + 5 \mathbf{v} = (10 \mathbf{i} - 20 \mathbf{j}) + (-50 \mathbf{i}) = (10 - 50)\mathbf{i} - 20 \mathbf{j} = -40\mathbf{i} - 20\mathbf{j} \).

Key Concepts

Vector Addition

Vector addition is a fundamental operation in vector mathematics, allowing us to combine two or more vectors. Imagine vectors as arrows pointing from one place to another. Vector addition involves connecting these arrows in a way where the endpoint of the first arrow meets the start point of the second arrow. This process is called the "head-to-tail" method.

To add two vectors, you simply add their corresponding components. Consider the vectors \( \mathbf{u} = 5 \mathbf{i} - 10 \mathbf{j} \) and \( \mathbf{v} = -10 \mathbf{i} \). You would add \( \mathbf{i} \)-components (5 and -10) separately from the \( \mathbf{j} \)-components (-10 and 0), since \( \mathbf{v} \) has no \( \mathbf{j} \) component.

• Calculate the new \( \mathbf{i} \)-component: \( 5 + (-10) = -5 \).
• Calculate the new \( \mathbf{j} \)-component: \( -10 + 0 = -10 \).

The resulting vector from adding \( \mathbf{u} \) and \( \mathbf{v} \) is \( -5 \mathbf{i} - 10 \mathbf{j} \).

Scalar Multiplication

When multiplying a vector by a scalar, you are basically stretching or shrinking its magnitude. Scalar multiplication affects only the size of the vector, not its direction, unless the scalar is negative, which flips the vector direction.

Take the vector \( \mathbf{v} = -10 \mathbf{i} \) and a scalar 4 from the exercise. To perform scalar multiplication, multiply each component of the vector by the scalar.

• For \( 4 \mathbf{v} \): Multiply \( -10 \mathbf{i} \) by 4 to get \(-40 \mathbf{i}\).

Similarly, if you have a more complex vector like \( \mathbf{u} = 5 \mathbf{i} - 10 \mathbf{j} \) and need to multiply it by 2:

• Each component gets multiplied: \( 2(5 \mathbf{i} - 10 \mathbf{j}) = 10 \mathbf{i} - 20 \mathbf{j} \).

Scalar multiplication not only scales vectors proportionally but is a crucial step in combining vectors algebraically, as shown in the exercise solutions.

Vectors in 2D

Vectors in 2D (two dimensions) are representations that have only two components, often labeled as \( \mathbf{i} \) and \( \mathbf{j} \). These components can be visualized on a flat plane, making them easier to manage with basic geometry.

The vector \( \mathbf{u} = 5 \mathbf{i} - 10 \mathbf{j} \) in the exercise is a perfect example of a 2D vector, where \( \mathbf{i} \) represents the horizontal component and \( \mathbf{j} \) the vertical.

• \( \mathbf{i} \) typically denotes movement along the x-axis.
• \( \mathbf{j} \) usually represents movement along the y-axis.

Understanding vectors in 2D helps in visualizing many physical phenomena, such as forces or velocities. The given vectors make it clear how 2D vectors work in operations like addition and scalar multiplication, breaking them down into their constituent parts. This highlights the importance of direction and magnitude, key concepts in vector mathematics.