Problem 39 - Trigonometry

Question

For each pair of vectors, find \(\mathbf{U}+\mathbf{V}, \mathbf{U}-\mathbf{V}\), and \(2 \mathbf{U}-3 \mathbf{V}\). $$\mathbf{U}=\langle 4,4\rangle, \mathbf{V}=\langle 4,-4\rangle$$

Step-by-Step Solution

Answer

\(\mathbf{U} + \mathbf{V} = \langle 8, 0 \rangle\), \(\mathbf{U} - \mathbf{V} = \langle 0, 8 \rangle\), \(2 \mathbf{U} - 3 \mathbf{V} = \langle -4, 20 \rangle\).

Step 1: Add the Vectors

To find \(\mathbf{U}+\mathbf{V}\), we add the corresponding components of the vectors \(\mathbf{U}\) and \(\mathbf{V}\). For the given vectors, \(\mathbf{U} = \langle 4, 4 \rangle\) and \(\mathbf{V} = \langle 4, -4 \rangle\). Therefore, \(\mathbf{U}+\mathbf{V} = \langle 4 + 4, 4 + (-4) \rangle = \langle 8, 0 \rangle\).

Step 2: Subtract the Vectors

To find \(\mathbf{U}-\mathbf{V}\), we subtract the components of \(\mathbf{V}\) from the components of \(\mathbf{U}\). So, \(\mathbf{U}-\mathbf{V} = \langle 4 - 4, 4 - (-4) \rangle = \langle 0, 8 \rangle\).

Step 3: Compute the Combination of Vectors

For \(2 \mathbf{U} - 3 \mathbf{V}\), first scale \(\mathbf{U}\) and \(\mathbf{V}\): \(2 \mathbf{U} = 2 \times \langle 4, 4 \rangle = \langle 8, 8 \rangle\) and \(3 \mathbf{V} = 3 \times \langle 4, -4 \rangle = \langle 12, -12 \rangle\). Then subtract these scaled vectors: \(2 \mathbf{U} - 3 \mathbf{V} = \langle 8, 8 \rangle - \langle 12, -12 \rangle = \langle 8 - 12, 8 - (-12) \rangle = \langle -4, 20 \rangle\).

Key Concepts

Understanding Vector Addition

Vector addition is an essential operation in vector mathematics. It involves adding two vectors together to get a new vector. This is done by adding each of their corresponding components.
For example, consider the vectors \( \mathbf{U} = \langle 4, 4 \rangle \) and \( \mathbf{V} = \langle 4, -4 \rangle \). When adding these vectors, you combine each component:

• For the x-components: \(4 + 4 = 8\)
• For the y-components: \(4 + (-4) = 0\)

This results in the new vector \( \mathbf{U} + \mathbf{V} = \langle 8, 0 \rangle \).
Think of vector addition as finding a resultant vector when you travel along one vector and then another; the final destination is the result of your vector addition.

Exploring Vector Subtraction

Vector subtraction mirrors addition but with a twist—here you subtract one vector from another. This is similar to adding the negative of a vector.
Consider \( \mathbf{U} = \langle 4, 4 \rangle \) and \( \mathbf{V} = \langle 4, -4 \rangle \). The vector subtraction operation \( \mathbf{U} - \mathbf{V} \) involves:

• Subtracting the x-components: \(4 - 4 = 0\)
• Subtracting the y-components: \(4 - (-4) = 4 + 4 = 8\)

This results in the vector \( \mathbf{U} - \mathbf{V} = \langle 0, 8 \rangle \).
Visualize this as moving along one vector and then moving backwards along the other. The result gives you the net effect of the subtraction.

Diving into Scalar Multiplication

Scalar multiplication involves multiplying a vector by a scalar (a single number), which scales the vector's magnitude without changing its direction (unless the scalar is negative, which also reverses the direction).
For

• \( 2 \mathbf{U} \): You double each component of \( \mathbf{U} = \langle 4, 4 \rangle \), resulting in \( \langle 8, 8 \rangle \).
• \( 3 \mathbf{V} \): Multiply each component of \( \mathbf{V} = \langle 4, -4 \rangle \) by 3, giving \( \langle 12, -12 \rangle \).

Once scaled, vector operations can be performed for combinations such as \( 2 \mathbf{U} - 3 \mathbf{V} \):

• Subtract the x-components: \(8 - 12 = -4\)
• Subtract the y-components: \(8 - (-12) = 8 + 12 = 20\)

Thus, the result is \( \langle -4, 20 \rangle \).
Scalar multiplication plays a crucial role in vector mathematics, enabling the modification of vector length and interactions in vector equations.