Problem 15 - Algebra and Trigonometry

Question

Find the vectors \(\mathbf{u}+\mathbf{v}, \mathbf{u}-\mathbf{v},\) and \(3 \mathbf{u}-\frac{1}{2} \mathbf{v}\) $$ \mathbf{u}=\langle 2,-7,3\rangle, \mathbf{v}=\langle 0,4,-1\rangle $$

Step-by-Step Solution

Answer

\(\mathbf{u} + \mathbf{v} = \langle 2, -3, 2 \rangle\); \(\mathbf{u} - \mathbf{v} = \langle 2, -11, 4 \rangle\); \(3\mathbf{u} - \frac{1}{2}\mathbf{v} = \langle 6, -23, \frac{19}{2} \rangle\).

Step 1: Vector Addition

To find \( \mathbf{u} + \mathbf{v} \), add corresponding components of \( \mathbf{u} \) and \( \mathbf{v} \):\\( \begin{aligned} &\langle 2, -7, 3 \rangle + \langle 0, 4, -1 \rangle\ &= \langle 2+0, -7+4, 3+(-1) \rangle \ = \langle 2, -3, 2 \rangle. \end{aligned} \)

Step 2: Vector Subtraction

To find \( \mathbf{u} - \mathbf{v} \), subtract corresponding components of \( \mathbf{v} \) from \( \mathbf{u} \):\\( \begin{aligned} &\langle 2, -7, 3 \rangle - \langle 0, 4, -1 \rangle\ &= \langle 2-0, -7-4, 3-(-1) \rangle \ = \langle 2, -11, 4 \rangle. \end{aligned} \)

Step 3: Scalar Multiplication and Combination

To find \((3\mathbf{u} - \frac{1}{2}\mathbf{v})\), first calculate \(3\mathbf{u}\) and \(-\frac{1}{2}\mathbf{v}\).\Calculate \(3\mathbf{u}\):\\(3 \times \langle 2, -7, 3 \rangle = \langle 3\times2, 3\times(-7), 3\times3 \rangle = \langle 6, -21, 9 \rangle\).\Calculate \(-\frac{1}{2}\mathbf{v}\):\\(-\frac{1}{2} \times \langle 0, 4, -1 \rangle = \langle -\frac{1}{2}\times0, -\frac{1}{2}\times4, -\frac{1}{2}\times(-1) \rangle = \langle 0, -2, \frac{1}{2} \rangle\).\Add the results: \\( \langle 6, -21, 9 \rangle + \langle 0, -2, \frac{1}{2} \rangle = \langle 6+0, -21-2, 9+\frac{1}{2} \rangle = \langle 6, -23, \frac{19}{2} \rangle\).

Key Concepts

Vector Addition

Adding vectors is like combining arrows that represent the vectors. To add two vectors such as \( \mathbf{u} \) and \( \mathbf{v} \), only the corresponding components need to be summed up.
Imagine each component acting independently, combining their individual contributions to provide us with a new vector.

• For our vectors: \( \mathbf{u} = \langle 2, -7, 3 \rangle \) and \( \mathbf{v} = \langle 0, 4, -1 \rangle \)
• Perform addition by summing each corresponding component:
• First component: \( 2 + 0 = 2 \)
• Second component: \( -7 + 4 = -3 \)
• Third component: \( 3 + (-1) = 2 \)

This results in the vector \( \langle 2, -3, 2 \rangle \).
The beauty of vector addition is its simplicity, allowing us clear results with straightforward component-wise calculations.

Vector Subtraction

Subtracting vectors is slightly different but follows a similar component approach to addition. It's like reversing the direction of the vector being subtracted before combining.
With vectors, subtract the components to get the difference between two vectors.

• Given vectors \( \mathbf{u} = \langle 2, -7, 3 \rangle \) and \( \mathbf{v} = \langle 0, 4, -1 \rangle \)
• Subtract the corresponding components of \( \mathbf{v} \) from \( \mathbf{u} \):
• First component: \( 2 - 0 = 2 \)
• Second component: \( -7 - 4 = -11 \)
• Third component: \( 3 - (-1) = 4 \)

This gives us the resulting vector \( \langle 2, -11, 4 \rangle \).
Vector subtraction can be thought of as "adding a negative" and serves to show how one vector differs from another in its magnitude and direction.

Scalar Multiplication

Scalar multiplication involves scaling a vector by multiplying each of its components by a scalar (a constant number). This operation changes the size of the vector, proportional to the scalar value.
When combining scalar multiplication with addition or subtraction, it's crucial to handle each part individually before integrating them back.

• Start with vectors \( \mathbf{u} = \langle 2, -7, 3 \rangle \) and \( \mathbf{v} = \langle 0, 4, -1 \rangle \).
• First, multiply \( \mathbf{u} \) by 3: \( 3 \times \langle 2, -7, 3 \rangle = \langle 6, -21, 9 \rangle \).
• Then, multiply \( \mathbf{v} \) by \(-\frac{1}{2}\): \(-\frac{1}{2} \times \langle 0, 4, -1 \rangle = \langle 0, -2, \frac{1}{2} \rangle \).
• Finally, add these resulting vectors:
  \( \langle 6, -21, 9 \rangle + \langle 0, -2, \frac{1}{2} \rangle = \langle 6, -23, \frac{19}{2} \rangle \).

Scalar multiplication provides flexibility in vector manipulations, allowing fine-tuning in size and direction for various applications.