Problem 12 - Advanced Engineering Mathematics

Question

Find (a) \(4 \mathrm{a}-2 \mathrm{~b}\) and \((\mathrm{b})-3 \mathrm{a}-5 \mathrm{~b}\). \(\mathbf{a}=\langle 2,0\rangle, \mathbf{b}=\langle 0,-3\rangle\)

Step-by-Step Solution

Answer

(a) \( \langle 8, 6 \rangle \), (b) \( \langle -6, 15 \rangle \)

Step 1: Given Vectors

We are given the vectors \( \mathbf{a} = \langle 2, 0 \rangle \) and \( \mathbf{b} = \langle 0, -3 \rangle \). We need to perform the given operations on these vectors.

Step 2: Calculate 4a

Multiply the vector \( \mathbf{a} = \langle 2, 0 \rangle \) by 4. This gives us:\[ 4 \mathbf{a} = 4 \times \langle 2, 0 \rangle = \langle 4 \times 2, 4 \times 0 \rangle = \langle 8, 0 \rangle \]

Step 3: Calculate 2b

Multiply the vector \( \mathbf{b} = \langle 0, -3 \rangle \) by 2. This gives us:\[ 2 \mathbf{b} = 2 \times \langle 0, -3 \rangle = \langle 2 \times 0, 2 \times -3 \rangle = \langle 0, -6 \rangle \]

Step 4: Subtract to find 4a - 2b

Subtract the vector \( 2 \mathbf{b} \) from \( 4 \mathbf{a} \):\[ 4 \mathbf{a} - 2 \mathbf{b} = \langle 8, 0 \rangle - \langle 0, -6 \rangle = \langle 8 - 0, 0 - (-6) \rangle = \langle 8, 6 \rangle \]

Step 5: Calculate -3a

Multiply the vector \( \mathbf{a} = \langle 2, 0 \rangle \) by -3. This gives us:\[ -3 \mathbf{a} = -3 \times \langle 2, 0 \rangle = \langle -3 \times 2, -3 \times 0 \rangle = \langle -6, 0 \rangle \]

Step 6: Calculate -5b

Multiply the vector \( \mathbf{b} = \langle 0, -3 \rangle \) by -5. This gives us:\[ -5 \mathbf{b} = -5 \times \langle 0, -3 \rangle = \langle -5 \times 0, -5 \times -3 \rangle = \langle 0, 15 \rangle \]

Step 7: Add to find -3a - 5b

Add the vector \( -5 \mathbf{b} \) to \( -3 \mathbf{a} \):\[ -3 \mathbf{a} - 5 \mathbf{b} = \langle -6, 0 \rangle + \langle 0, 15 \rangle = \langle -6 + 0, 0 + 15 \rangle = \langle -6, 15 \rangle \]

Key Concepts

Scalar Multiplication

Scalar multiplication is a simple yet powerful operation in vector arithmetic. When we perform scalar multiplication, we multiply each component of a vector by a scalar (a real number).

This operation changes the magnitude of the vector but keeps its direction the same, unless the scalar is negative, which also reverses the direction. Let's break down how to perform scalar multiplication using an example:

• Consider vector \( \mathbf{a} = \langle 2, 0 \rangle \). If we want to multiply this vector by 4, we apply the scalar multiplication to each component: \( 4 \cdot \langle 2, 0 \rangle = \langle 4 \times 2, 4 \times 0 \rangle = \langle 8, 0 \rangle \).
  
• Notice how each component of the vector \( \mathbf{a} \) is multiplied by 4 to achieve this result.
  
• As another example, multiplying \( \mathbf{a} = \langle 2, 0 \rangle \) by -3 gives \( -3 \cdot \langle 2, 0 \rangle = \langle -6, 0 \rangle \). Here, the direction is reversed because of the negative scalar.

By mastering scalar multiplication, you'll have a solid foundation for performing more complex vector operations.

Vector Addition

Vector addition involves combining two or more vectors to form a new vector. This is done by adding corresponding components of each vector. Vector addition is both commutative and associative, which makes it straightforward in terms of computation.

Here is how you handle vector addition step-by-step:

• Imagine you are given two vectors \( \mathbf{u} = \langle 8, 0 \rangle \) and \( \mathbf{v} = \langle 0, 15 \rangle \). You can add these vectors by adding their corresponding components: \( \mathbf{u} + \mathbf{v} = \langle 8 + 0, 0 + 15 \rangle = \langle 8, 15 \rangle \).
  
• The result \( \langle 8, 15 \rangle \) is a new vector that represents the combination of the original vectors.
  
• This operation is visualized as placing the tail of vector \( \mathbf{v} \) at the head of vector \( \mathbf{u} \), resulting in a diagonal from the tail of \( \mathbf{u} \) to the head of \( \mathbf{v} \).

Understanding vector addition helps in solving problems where you need to combine different directional movements.

Vector Subtraction

Vector subtraction may sound complex, but it's simply the addition of a vector's inverse. When we subtract a vector, we are essentially adding its opposite. Subtraction is used to find the difference between two vectors or to remove a vector's influence.

Here's how vector subtraction works:

• Suppose you have two vectors, \( \mathbf{u} = \langle 8, 6 \rangle \) and another vector \( \mathbf{v} = \langle 0, -6 \rangle \). To subtract \( \mathbf{v} \) from \( \mathbf{u} \), you reverse the direction of \( \mathbf{v} \) by changing its sign and then proceed with vector addition: \( \mathbf{u} - \mathbf{v} = \mathbf{u} + (-\mathbf{v}) = \langle 8, 6 \rangle - \langle 0, -6 \rangle = \langle 8 - 0, 6 - (-6) \rangle = \langle 8, 12 \rangle \).
  
• You can think of this as adding the components of \( \mathbf{u} \) and the additive inverse of \( \mathbf{v} \). This process effectively 'subtracts' \( \mathbf{v} \) from \( \mathbf{u} \).

With vector subtraction, you can address questions that involve changing directions or differences between positions and forces.