Problem 4 - Advanced Engineering Mathematics

Question

Find (a) \(3 \mathbf{a}\), (b) \(\mathbf{a}+\mathbf{b}\), (c) \(\mathbf{a}-\mathbf{b}\), (d) \(\|\mathbf{a}+\mathbf{b}\|\), and (e) \(\|\mathbf{a}-\mathbf{b}\|\). $$ \mathbf{a}=\frac{1}{6} \mathbf{i}-\frac{1}{6} \mathbf{j}, \mathbf{b}=\frac{1}{2} \mathbf{i}+\frac{5}{6} \mathbf{j} $$

Step-by-Step Solution

Answer

(a) \(3\mathbf{a}=\frac{1}{2}\mathbf{i}-\frac{1}{2}\mathbf{j}\), (b) \(\mathbf{a}+\mathbf{b}=\frac{2}{3}\mathbf{i}+\frac{2}{3}\mathbf{j}\), (c) \(\mathbf{a}-\mathbf{b}=-\frac{1}{3}\mathbf{i}-1\mathbf{j}\), (d) \(\|\mathbf{a}+\mathbf{b}\|=\frac{2\sqrt{2}}{3}\), (e) \(\|\mathbf{a}-\mathbf{b}\|=\frac{\sqrt{10}}{3}\).

Step 1: Calculate 3a

Given \( \mathbf{a} = \frac{1}{6} \mathbf{i} - \frac{1}{6} \mathbf{j} \), we need to multiply each component of \( \mathbf{a} \) by 3.Calculation:\[3 \mathbf{a} = 3 \left( \frac{1}{6} \mathbf{i} - \frac{1}{6} \mathbf{j} \right) = \frac{3}{6} \mathbf{i} - \frac{3}{6} \mathbf{j} = \frac{1}{2} \mathbf{i} - \frac{1}{2} \mathbf{j}\]

Step 2: Find a+b

For vector addition, add the i-components and j-components of vectors \( \mathbf{a} \) and \( \mathbf{b} \) separately.\( \mathbf{a} = \frac{1}{6} \mathbf{i} - \frac{1}{6} \mathbf{j} \) and \( \mathbf{b} = \frac{1}{2} \mathbf{i} + \frac{5}{6} \mathbf{j} \).\[\mathbf{a} + \mathbf{b} = \left( \frac{1}{6} + \frac{1}{2} \right)\mathbf{i} + \left( -\frac{1}{6} + \frac{5}{6} \right)\mathbf{j} = \frac{2}{3} \mathbf{i} + \frac{2}{3} \mathbf{j}\]

Step 3: Find a-b

For vector subtraction, subtract the i-components and j-components of vectors \( \mathbf{b} \) from \( \mathbf{a} \) respectively.\( \mathbf{a} = \frac{1}{6} \mathbf{i} - \frac{1}{6} \mathbf{j} \) and \( \mathbf{b} = \frac{1}{2} \mathbf{i} + \frac{5}{6} \mathbf{j} \).\[\mathbf{a} - \mathbf{b} = \left( \frac{1}{6} - \frac{1}{2} \right)\mathbf{i} + \left( -\frac{1}{6} - \frac{5}{6} \right)\mathbf{j} = -\frac{1}{3} \mathbf{i} - 1 \mathbf{j}\]

Step 4: Calculate magnitude of a+b

Formulate the magnitude of the vector \( \mathbf{c} = \mathbf{a} + \mathbf{b} \), which is \( \frac{2}{3} \mathbf{i} + \frac{2}{3} \mathbf{j} \).Magnitude formula: \( \|\mathbf{c}\| = \sqrt{(c_i)^2 + (c_j)^2} \).\[\|\mathbf{a} + \mathbf{b}\| = \sqrt{\left( \frac{2}{3} \right)^2 + \left( \frac{2}{3} \right)^2} = \sqrt{\frac{4}{9} + \frac{4}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}\]

Step 5: Calculate magnitude of a-b

Formulate the magnitude of the vector \( \mathbf{d} = \mathbf{a} - \mathbf{b} \), which is \( -\frac{1}{3} \mathbf{i} - 1 \mathbf{j} \).Magnitude formula: \( \|\mathbf{d}\| = \sqrt{(d_i)^2 + (d_j)^2} \).\[\|\mathbf{a} - \mathbf{b}\| = \sqrt{\left( -\frac{1}{3} \right)^2 + ( -1 )^2} = \sqrt{\frac{1}{9} + 1} = \sqrt{\frac{10}{9}} = \frac{\sqrt{10}}{3}\]

Key Concepts

Vector Addition

Vector addition is a fundamental operation where two or more vectors are combined to produce a resultant vector. It is performed by adding corresponding components of the vectors. For example, if you have two vectors \( \mathbf{a} = a_i \mathbf{i} + a_j \mathbf{j} \) and \( \mathbf{b} = b_i \mathbf{i} + b_j \mathbf{j} \), the resultant vector \( \mathbf{c} = \mathbf{a} + \mathbf{b} \) would be calculated as follows:
\[ \mathbf{c} = (a_i + b_i) \mathbf{i} + (a_j + b_j) \mathbf{j} \]
This means you simply sum the \(i\)-components of both vectors, and then the \(j\)-components. Using simple arithmetic, this operation is quite straightforward:

• For \(\mathbf{a} = \frac{1}{6} \mathbf{i} - \frac{1}{6} \mathbf{j}\) and \(\mathbf{b} = \frac{1}{2} \mathbf{i} + \frac{5}{6} \mathbf{j}\), just plug in these values.
• First, add the \(i\)-components: \(\frac{1}{6} + \frac{1}{2} = \frac{2}{3}\).
• Then add the \(j\)-components: \(-\frac{1}{6} + \frac{5}{6} = \frac{2}{3}\).

The addition of the vectors gives us \(\mathbf{a} + \mathbf{b} = \frac{2}{3} \mathbf{i} + \frac{2}{3} \mathbf{j}\).
Vector addition is used in many fields such as physics, engineering, and computer graphics to calculate things like net forces, velocities, and transitions.

Vector Subtraction

Vector subtraction is similar to vector addition but involves taking the difference between corresponding components of vectors. If you have vectors \( \mathbf{a} = a_i \mathbf{i} + a_j \mathbf{j} \) and \( \mathbf{b} = b_i \mathbf{i} + b_j \mathbf{j} \), you find \( \mathbf{c} = \mathbf{a} - \mathbf{b} \) as follows:
\[ \mathbf{c} = (a_i - b_i) \mathbf{i} + (a_j - b_j) \mathbf{j} \]
This concept implies:

• Taking the \(i\)-component of \( \mathbf{a} \) and subtracting the \(i\)-component of \( \mathbf{b} \).
• Following the same process for the \(j\)-components.

For example, given \( \mathbf{a} = \frac{1}{6} \mathbf{i} - \frac{1}{6} \mathbf{j} \) and \( \mathbf{b} = \frac{1}{2} \mathbf{i} + \frac{5}{6} \mathbf{j} \):

• Subtract \(i\)-components: \( \frac{1}{6} - \frac{1}{2} = -\frac{1}{3} \).
• Subtract \(j\)-components: \( -\frac{1}{6} - \frac{5}{6} = -1 \).

The resultant vector is \( \mathbf{a} - \mathbf{b} = -\frac{1}{3} \mathbf{i} - 1 \mathbf{j} \).
Vector subtraction is pivotal in fields like physics to calculate relative velocities and displacements.

Magnitude of a Vector

Finding the magnitude of a vector is an important concept because it tells you the length or size of the vector. It's calculated using the Pythagorean theorem. For a vector \( \mathbf{v} = v_i \mathbf{i} + v_j \mathbf{j} \), its magnitude \( \|\mathbf{v}\| \) is given by:
\[ \|\mathbf{v}\| = \sqrt{(v_i)^2 + (v_j)^2} \]
This formula helps us determine the absolute extent of a vector, essentially treating its components like the sides of a right triangle. Let's look at the calculated magnitudes for vectors \( \mathbf{a} + \mathbf{b} \) and \( \mathbf{a} - \mathbf{b} \):

• For \( \mathbf{a} + \mathbf{b} = \frac{2}{3} \mathbf{i} + \frac{2}{3} \mathbf{j} \):
  • The magnitude is \( \|\mathbf{a} + \mathbf{b}\| = \sqrt{\left(\frac{2}{3}\right)^2 + \left(\frac{2}{3}\right)^2} = \frac{2\sqrt{2}}{3} \).
• For \( \mathbf{a} - \mathbf{b} = -\frac{1}{3} \mathbf{i} - 1 \mathbf{j} \):
  • The magnitude becomes \( \|\mathbf{a} - \mathbf{b}\| = \sqrt{\left(-\frac{1}{3}\right)^2 + (-1)^2} = \frac{\sqrt{10}}{3} \).

With magnitude, you are effectively measuring the 'distance' the vector covers in space, which aids in scenarios such as calculating speeds or forces exerted.