Problem 17 - Algebra and Trigonometry

Question

Find the vectors \(\mathbf{u}+\mathbf{v}, \mathbf{u}-\mathbf{v},\) and \(3 \mathbf{u}-\frac{1}{2} \mathbf{v}\) $$ \mathbf{u}=\mathbf{i}+\mathbf{j}, \mathbf{v}=-\mathbf{j}-2 \mathbf{k} $$

Step-by-Step Solution

Answer

\(\mathbf{u}+\mathbf{v}=\mathbf{i}-2\mathbf{k}, \mathbf{u}-\mathbf{v}=\mathbf{i}+2\mathbf{j}+2\mathbf{k}, 3\mathbf{u}-\frac{1}{2}\mathbf{v}=3\mathbf{i}+\frac{7}{2}\mathbf{j}+\mathbf{k}\).

Step 1: Calculate \(\mathbf{u}+\mathbf{v}\)

First, add the corresponding components of \(\mathbf{u} = \mathbf{i} + \mathbf{j}\) and \(\mathbf{v} = -\mathbf{j} - 2\mathbf{k}\). This involves adding each vector component wise:1. Find the \(\mathbf{i}\) component: \(1 + 0 = 1\)2. Find the \(\mathbf{j}\) component: \(1 + (-1) = 0\)3. Find the \(\mathbf{k}\) component: \(0 + (-2) = -2\)Thus, \(\mathbf{u} + \mathbf{v} = \mathbf{i} - 2\mathbf{k}\).

Step 2: Calculate \(\mathbf{u}-\mathbf{v}\)

Next, subtract the components of \(\mathbf{v}\) from the components of \(\mathbf{u}\):1. \(\mathbf{i}\) component: \(1 - 0 = 1\)2. \(\mathbf{j}\) component: \(1 - (-1) = 2\)3. \(\mathbf{k}\) component: \(0 - (-2) = 2\)Thus, \(\mathbf{u} - \mathbf{v} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\).

Step 3: Calculate \(3\mathbf{u}-\frac{1}{2}\mathbf{v}\)

First, calculate \(3\mathbf{u}\) and \(\frac{1}{2}\mathbf{v}\):- \(3\mathbf{u}\): 1. \(\mathbf{i}\) component: \(3 \times 1 = 3\) 2. \(\mathbf{j}\) component: \(3 \times 1 = 3\) 3. \(\mathbf{k}\) component: \(3 \times 0 = 0\) So, \(3\mathbf{u} = 3\mathbf{i} + 3\mathbf{j}\).- \(\frac{1}{2}\mathbf{v}\): 1. \(\mathbf{i}\) component: \(\frac{1}{2} \times 0 = 0\) 2. \(\mathbf{j}\) component: \(\frac{1}{2} \times (-1) = -\frac{1}{2}\) 3. \(\mathbf{k}\) component: \(\frac{1}{2} \times (-2) = -1\) So, \(\frac{1}{2}\mathbf{v} = -\frac{1}{2}\mathbf{j} - \mathbf{k}\).Now subtracting \(\frac{1}{2}\mathbf{v}\) from \(3\mathbf{u}\):1. \(\mathbf{i}\) component: \(3 - 0 = 3\)2. \(\mathbf{j}\) component: \(3 - (-\frac{1}{2}) = 3 + \frac{1}{2} = \frac{7}{2}\)3. \(\mathbf{k}\) component: \(0 - (-1) = 1\)Thus, \(3\mathbf{u} - \frac{1}{2}\mathbf{v} = 3\mathbf{i} + \frac{7}{2}\mathbf{j} + \mathbf{k}\).

Key Concepts

Vector Components

To better understand vectors, it's essential to break them down into their components. A vector in three-dimensional space is expressed using the unit vectors: \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \), which represent the x, y, and z axes respectively. Each vector can be viewed as the sum of its component vectors along these three directions.
For example, consider the vectors \( \mathbf{u} = \mathbf{i} + \mathbf{j} \) and \( \mathbf{v} = -\mathbf{j} - 2\mathbf{k} \).
These vectors can be deconstructed into:

• The \( \mathbf{i} \) component (x-axis): which are 1 for \( \mathbf{u} \) and 0 for \( \mathbf{v} \).
• The \( \mathbf{j} \) component (y-axis): which are 1 for \( \mathbf{u} \) and -1 for \( \mathbf{v} \).
• The \( \mathbf{k} \) component (z-axis): which are 0 for \( \mathbf{u} \) and -2 for \( \mathbf{v} \).

Understanding these components is key for any further vector operations like scalar multiplication or subtraction.

Scalar Multiplication of Vectors

Scalar multiplication involves multiplying a vector by a scalar (a real number), which stretches or shrinks all components of the vector by that factor.
Let's take the vector \( \mathbf{u} = \mathbf{i} + \mathbf{j} \) and perform scalar multiplication with the scalar 3, denoted as \( 3\mathbf{u} \):

• The \( \mathbf{i} \) component: \( 3 \times 1 = 3 \)
• The \( \mathbf{j} \) component: \( 3 \times 1 = 3 \)
• The \( \mathbf{k} \) component: \( 3 \times 0 = 0 \)

So, \( 3\mathbf{u} = 3\mathbf{i} + 3\mathbf{j} \).
Another example is \( \frac{1}{2}\mathbf{v} \), scaling the vector \( \mathbf{v} = -\mathbf{j} - 2\mathbf{k} \) by \( \frac{1}{2} \):

• The \( \mathbf{j} \) component: \( \frac{1}{2} \times -1 = -\frac{1}{2} \)
• The \( \mathbf{k} \) component: \( \frac{1}{2} \times -2 = -1 \)

This results in \( \frac{1}{2}\mathbf{v} = -\frac{1}{2}\mathbf{j} - \mathbf{k} \).
Scalar multiplication is useful for adjusting vector lengths without changing their direction, unless the scalar is negative, which will also invert the vector's direction.

Subtraction of Vectors

Subtracting vectors involves taking the difference between the corresponding components. This is fundamentally similar to addition, but with the subtraction operation instead.
Let's subtract \( \mathbf{v} \) from \( \mathbf{u} \), where \( \mathbf{u} = \mathbf{i} + \mathbf{j} \) and \( \mathbf{v} = -\mathbf{j} - 2\mathbf{k} \):

• \( \mathbf{i} \) component: \( 1 - 0 = 1 \)
• \( \mathbf{j} \) component: \( 1 - (-1) = 2 \)
• \( \mathbf{k} \) component: \( 0 - (-2) = 2 \)

Therefore, \( \mathbf{u} - \mathbf{v} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} \).
The subtraction operation essentially transitions between two positions in space, moving in the direction from one vector to the other. This can be especially useful in physics and engineering when calculating relative positions or velocities.