Problem 42 - Trigonometry

Question

For each pair of vectors, find \(\mathbf{U}+\mathbf{V}, \mathbf{U}-\mathbf{V}\), and \(2 \mathbf{U}-3 \mathbf{V}\). $$\mathbf{U}=\langle 2,0\rangle, \mathbf{V}=\langle 0,-7\rangle$$

Step-by-Step Solution

Answer

\( \mathbf{U} + \mathbf{V} = \langle 2, -7 \rangle \), \( \mathbf{U} - \mathbf{V} = \langle 2, 7 \rangle \), and \( 2\mathbf{U} - 3\mathbf{V} = \langle 4, 21 \rangle \).

Step 1: Calculate \( \mathbf{U} + \mathbf{V} \)

To perform vector addition \( \mathbf{U} + \mathbf{V} \), add the corresponding components of \( \mathbf{U} \) and \( \mathbf{V} \).\[\mathbf{U} + \mathbf{V} = \langle 2, 0 \rangle + \langle 0, -7 \rangle = \langle 2+0, 0+(-7) \rangle = \langle 2, -7 \rangle.\]

Step 2: Calculate \( \mathbf{U} - \mathbf{V} \)

To perform vector subtraction \( \mathbf{U} - \mathbf{V} \), subtract the corresponding components of \( \mathbf{V} \) from \( \mathbf{U} \).\[\mathbf{U} - \mathbf{V} = \langle 2, 0 \rangle - \langle 0, -7 \rangle = \langle 2-0, 0-(-7) \rangle = \langle 2, 7 \rangle.\]

Step 3: Calculate \( 2\mathbf{U} - 3\mathbf{V} \)

First, find \( 2\mathbf{U}\) by multiplying each component of \( \mathbf{U} \) by 2 and \( 3\mathbf{V} \) by multiplying each component of \( \mathbf{V} \) by 3.\[2\mathbf{U} = 2 \times \langle 2, 0 \rangle = \langle 4, 0 \rangledivandtripleV3\mathbf{V} = 3 \times \langle 0, -7 \rangle = \langle 0, -21 \rangle\]Now subtract the vector \( 3\mathbf{V} \) from \( 2\mathbf{U} \).\[2\mathbf{U} - 3\mathbf{V} = \langle 4, 0 \rangle - \langle 0, -21 \rangle = \langle 4-0, 0-(-21) \rangle = \langle 4, 21 \rangle.\]

Key Concepts

Vector Addition

Vector addition is a fundamental operation in vector arithmetic. It is used to combine two vectors to form a new vector. When adding two vectors like \( \mathbf{U} = \langle 2, 0 \rangle \) and \( \mathbf{V} = \langle 0, -7 \rangle \), we simply add their corresponding components:

• Add the first components of each vector: \( 2 + 0 = 2 \).
• Add the second components of each vector: \( 0 + (-7) = -7 \).

Thus, the resulting vector from the addition of \( \mathbf{U} \) and \( \mathbf{V} \) is \( \langle 2, -7 \rangle \).
This operation is visually interpreted as placing the tail of vector \( \mathbf{V} \) at the head of vector \( \mathbf{U} \), resulting in a new vector pointing directly from the start of \( \mathbf{U} \) to the end of \( \mathbf{V} \). This process demonstrates how multiple effects can combine simultaneously, an essential concept in physics and engineering.

Vector Subtraction

Vector subtraction is the opposite of vector addition. When performing vector subtraction, we find the difference between two vectors. To subtract \( \mathbf{V} = \langle 0, -7 \rangle \) from \( \mathbf{U} = \langle 2, 0 \rangle \):

• Subtract the first components: \( 2 - 0 = 2 \).
• Subtract the second components: \( 0 - (-7) = 7 \).

So, \( \mathbf{U} - \mathbf{V} = \langle 2, 7 \rangle \).
In graphical terms, this subtraction can be visualized as adding \( -\mathbf{V} \) to \( \mathbf{U} \), essentially reversing the direction of \( \mathbf{V} \) and then applying vector addition. This method is often used to determine relative positions and directions, such as in determining velocity differences in physics.

Scalar Multiplication

Scalar multiplication involves multiplying a vector by a scalar, which stretches or shrinks the vector while maintaining its direction. Let's consider multiplying vector \( \mathbf{U} = \langle 2, 0 \rangle \) by the scalar 2:

• Multiply each component by the scalar: \( 2 \times 2 = 4 \) and \( 2 \times 0 = 0 \).

Thus, \( 2\mathbf{U} = \langle 4, 0 \rangle \).
Now, for \( 3\mathbf{V} \) where \( \mathbf{V} = \langle 0, -7 \rangle \):

• Multiply each component by 3: \( 3 \times 0 = 0 \) and \( 3 \times -7 = -21 \).

Thus, \( 3\mathbf{V} = \langle 0, -21 \rangle \).
Scalar multiplication is particularly useful in physics for scaling forces or operations in multiple dimensions. Finally, we subtract \( 3\mathbf{V} \) from \( 2\mathbf{U} \) to get \( 2\mathbf{U} - 3\mathbf{V} = \langle 4, 21 \rangle \). This operation illustrates how alterations in magnitude can significantly change vector direction and length, a key aspect in vector transformations.