Problem 53 - Algebra and Trigonometry

Question

In Exercises 53–56, let. $$\mathbf{u}=-2 \mathbf{i}+3 \mathbf{j}, \mathbf{v}=6 \mathbf{i}-\mathbf{j}, \mathbf{w}=-3 \mathbf{i}$$ Find each specified vector or scalar. $$4 \mathbf{u}-(2 \mathbf{v}-\mathbf{w})$$

Step-by-Step Solution

Answer

The solution is -17 \mathbf{i} + 14 \mathbf{j}.

Step 1: Scalar Multiplication

First, we will perform scalar multiplication operations: \(4 \mathbf{u} = 4*(-2 \mathbf{i}+3 \mathbf{j})= -8 \mathbf{i} + 12 \mathbf{j}\) , \(2 \mathbf{v} = 2*(6 \mathbf{i}-\mathbf{j}) = 12 \mathbf{i} - 2 \mathbf{j}\). So the expression now is \(-8 \mathbf{i} + 12 \mathbf{j} - (12 \mathbf{i} - 2 \mathbf{j}-(-3 \mathbf{i}))\)

Step 2: Simplify the Expression

Now simplify the expression: \(-8 \mathbf{i} + 12 \mathbf{j} - 12 \mathbf{i} + 2 \mathbf{j} +3 \mathbf{i} = (-8-12+3) \mathbf{i} + (12+2) \mathbf{j} = -17 \mathbf{i} + 14 \mathbf{j}\)

Step 3: Final Step

The value of \(4 \mathbf{u}-(2 \mathbf{v}-\mathbf{w})\) is -17 \mathbf{i} + 14 \mathbf{j}

Key Concepts

Scalar Multiplication

Scalar multiplication involves increasing or decreasing the magnitude of a vector by a scalar quantity, without changing its direction. A scalar is a simple quantity that only has magnitude, not direction, such as a single number like 2, 4, or any other real number. To multiply a vector by a scalar, you multiply each component of the vector by that scalar.
Consider the vector \( \mathbf{u} = -2 \mathbf{i} + 3 \mathbf{j} \). If you multiply this vector by the scalar 4, you get the vector \( 4 \mathbf{u} \), where you multiply both the \( \mathbf{i} \)-component and the \( \mathbf{j} \)-component by 4:

• \( 4(-2 \mathbf{i}) = -8 \mathbf{i} \)
• \( 4(3 \mathbf{j}) = 12 \mathbf{j} \)

Thus, \( 4 \mathbf{u} = -8 \mathbf{i} + 12 \mathbf{j} \). This shows the effect of enlarging each part of the vector by a multiple, making the whole vector four times as long, while pointing in the same direction.

Vector Addition

Vector addition involves combining vectors to form a resultant vector. The process is similar to adding two numbers, but instead of adding simple values, we add their respective components.
Suppose we have two vectors: \( \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} \) and \( \mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} \). To add them, you simply combine their \( \mathbf{i} \) components and their \( \mathbf{j} \) components separately:

• The sum of the \( \mathbf{i} \)-components is \( a_1 + b_1 \).
• The sum of the \( \mathbf{j} \)-components is \( a_2 + b_2 \).

As a result, the new vector \( \mathbf{c} \) is \( \mathbf{c} = (a_1 + b_1) \mathbf{i} + (a_2 + b_2) \mathbf{j} \).
In solving an example where you subtract one vector from another, like \( 4 \mathbf{u} - (2 \mathbf{v}) \), you actually add the opposite of the second vector. Remembering that subtracting vector \( \mathbf{b} \) is like adding a negative vector \( -\mathbf{b} \). This way, you remain consistent with addition rules while handling differences in directions.

Simplifying Vectors

Simplifying vectors is the process of combining and reducing vector expressions to their simplest form, typically through vector addition and subtraction following scalar multiplication. This step ensures that all calculations are handled correctly and the expression is as concise as possible.
Start by identifying terms that can be combined. For example, in the expression \(-8 \mathbf{i} + 12 \mathbf{j} - (12 \mathbf{i} - 2 \mathbf{j} - (-3 \mathbf{i}))\), we simplify by distributing and combining like terms:

• Distribute the negative, resulting in: \(-8 \mathbf{i} + 12 \mathbf{j} - 12 \mathbf{i} + 2 \mathbf{j} + 3 \mathbf{i}\)
• Combine the \( \mathbf{i} \)-components and \( \mathbf{j} \)-components:
• \((-8 - 12 + 3) \mathbf{i} = -17 \mathbf{i}\)
• \((12 + 2) \mathbf{j} = 14 \mathbf{j}\)

The overall vector simplifies to \( -17 \mathbf{i} + 14 \mathbf{j} \). This reduction helps in both understanding the final direction and magnitude of the vector, making it clearer for further applications or interpretations.