Problem 49 - Trigonometry

Question

For each pair of vectors, find \(\mathbf{U}+\mathbf{V}, \mathbf{U}-\mathbf{V}\), and \(3 \mathbf{U}+2 \mathbf{V}\). $$\mathbf{U}=2 \mathbf{i}+5 \mathbf{j}, \mathbf{V}=5 \mathbf{i}+2 \mathbf{j}$$

Step-by-Step Solution

Answer

\( \mathbf{U} + \mathbf{V} = 7\mathbf{i} + 7\mathbf{j} \), \( \mathbf{U} - \mathbf{V} = -3\mathbf{i} + 3\mathbf{j} \), \( 3\mathbf{U} + 2\mathbf{V} = 16\mathbf{i} + 19\mathbf{j} \).

Step 1: Add the Vectors

To find \( \mathbf{U} + \mathbf{V} \), we simply add the corresponding components of vectors \( \mathbf{U} \) and \( \mathbf{V} \). \( \mathbf{U} = 2\mathbf{i} + 5\mathbf{j} \), and \( \mathbf{V} = 5\mathbf{i} + 2\mathbf{j} \).\[ \mathbf{U} + \mathbf{V} = (2+5)\mathbf{i} + (5+2)\mathbf{j} = 7\mathbf{i} + 7\mathbf{j} \]

Step 2: Subtract the Vectors

For \( \mathbf{U} - \mathbf{V} \), subtract the corresponding components of vector \( \mathbf{V} \) from \( \mathbf{U} \).\[ \mathbf{U} - \mathbf{V} = (2-5)\mathbf{i} + (5-2)\mathbf{j} = -3\mathbf{i} + 3\mathbf{j} \]

Step 3: Multiply and Add the Vectors

To find \( 3\mathbf{U} + 2\mathbf{V} \), first multiply each vector by the scalar, then add the results.\[ 3\mathbf{U} = 3(2\mathbf{i} + 5\mathbf{j}) = 6\mathbf{i} + 15\mathbf{j} \]\[ 2\mathbf{V} = 2(5\mathbf{i} + 2\mathbf{j}) = 10\mathbf{i} + 4\mathbf{j} \]\[ 3\mathbf{U} + 2\mathbf{V} = (6 + 10)\mathbf{i} + (15 + 4)\mathbf{j} = 16\mathbf{i} + 19\mathbf{j} \]

Key Concepts

Vector Subtraction

Vector subtraction is much like vector addition but involves taking the components of one vector away from another. If you have two vectors, say \( \mathbf{U} = a\mathbf{i} + b\mathbf{j} \) and \( \mathbf{V} = c\mathbf{i} + d\mathbf{j} \), the subtraction \( \mathbf{U} - \mathbf{V} \) is calculated by subtracting the corresponding components:
\[ \mathbf{U} - \mathbf{V} = (a-c)\mathbf{i} + (b-d)\mathbf{j} \]

• First, subtract the \( \mathbf{i} \) components of \( \mathbf{V} \) from \( \mathbf{U} \).
• Next, subtract the \( \mathbf{j} \) components.

Using the exercise's example vectors, \( \mathbf{U} = 2\mathbf{i} + 5\mathbf{j} \) and \( \mathbf{V} = 5\mathbf{i} + 2\mathbf{j} \), we compute:

• \( \mathbf{U} - \mathbf{V} = (2-5)\mathbf{i} + (5-2)\mathbf{j} = -3\mathbf{i} + 3\mathbf{j} \).

This simple process helps visually understand the direction and magnitude that result when one vector is removed from another.

Scalar Multiplication

Scalar multiplication involves multiplying a vector by a scalar (a real number). This process scales the vector's magnitude without changing its direction, unless the scalar is negative—which will reverse the vector's direction.
For a vector \( \mathbf{U} = a\mathbf{i} + b\mathbf{j} \) and a scalar \( k \), the resulting vector is given by:
\[ k\mathbf{U} = k(a\mathbf{i} + b\mathbf{j}) = (ka)\mathbf{i} + (kb)\mathbf{j} \]

• Multiply each component of the vector by the scalar.
• The length of the vector changes depending on the magnitude of the scalar.

In our exercise, we scaled each vector before adding them:

• \( 3\mathbf{U} = 3(2\mathbf{i} + 5\mathbf{j}) = 6\mathbf{i} + 15\mathbf{j} \)
• \( 2\mathbf{V} = 2(5\mathbf{i} + 2\mathbf{j}) = 10\mathbf{i} + 4\mathbf{j} \)

This manipulation is crucial in vector operations, especially when working with forces or other physical quantities.

Component Form of Vectors

Vectors in component form are expressed as a sum of their horizontal (\( \mathbf{i} \)) and vertical (\( \mathbf{j} \)) components. This form is extremely handy in calculations because it allows us to work with vectors mathematically, just as we do with other number pairs.
Consider vector \( \mathbf{U} = a\mathbf{i} + b\mathbf{j} \). This vector has:

• A horizontal component \( a \)
• A vertical component \( b \)

This separation simplifies addition, subtraction, and scalar multiplication:

• Just operate on corresponding components independently.

In the exercise, \( \mathbf{U} = 2\mathbf{i} + 5\mathbf{j} \) and \( \mathbf{V} = 5\mathbf{i} + 2\mathbf{j} \), each vector is fully characterized by its components.
The component form breaks down complex vector operations into simple arithmetic, enabling easy computation and visualization in physics, engineering, and computer graphics.