Problem 11 - APEX Calculus

Question

Let \(\vec{u}=\langle 1,-2\rangle\) and \(\vec{v}=\langle 1,1\rangle .\) (a) Find \(\vec{u}+\vec{v}, \vec{u}-\vec{v}, 2 \vec{u}-3 \vec{v} .\) (b) Sketch the above vectors on the same axes, along with \(\vec{u}\) and \(\vec{v} .\) (c) Find \(\vec{x}\) where \(\vec{u}+\vec{x}=2 \vec{v}-\vec{x}\)

Step-by-Step Solution

Answer

\(\vec{u} + \vec{v} = \langle 2, -1 \rangle; \vec{u} - \vec{v} = \langle 0, -3 \rangle; 2\vec{u} - 3\vec{v} = \langle -1, -7 \rangle; \vec{x} = \langle \frac{1}{2}, 2 \rangle.\)

Step 1: Finding \( \vec{u} + \vec{v} \)

To find \(\vec{u} + \vec{v}\), you add the corresponding components of the vectors. For \(\vec{u} = \langle 1, -2 \rangle\) and \(\vec{v} = \langle 1, 1 \rangle\), the operation is: \[ \vec{u} + \vec{v} = \langle 1 + 1, -2 + 1 \rangle = \langle 2, -1 \rangle \].

Step 2: Finding \( \vec{u} - \vec{v} \)

To find \(\vec{u} - \vec{v}\), subtract the corresponding components of \(\vec{v}\) from \(\vec{u}\). For \(\vec{u} = \langle 1, -2 \rangle\) and \(\vec{v} = \langle 1, 1 \rangle\), the operation is: \[ \vec{u} - \vec{v} = \langle 1 - 1, -2 - 1 \rangle = \langle 0, -3 \rangle \].

Step 3: Finding \( 2\vec{u} - 3\vec{v} \)

Scale \(\vec{u}\) by 2 and \(\vec{v}\) by 3, then subtract. \(2\vec{u} = \langle 2(1), 2(-2) \rangle = \langle 2, -4 \rangle\) and \(3\vec{v} = \langle 3(1), 3(1) \rangle = \langle 3, 3 \rangle\). Then, \[ 2\vec{u} - 3\vec{v} = \langle 2, -4 \rangle - \langle 3, 3 \rangle = \langle 2 - 3, -4 - 3 \rangle = \langle -1, -7 \rangle \].

Step 4: Sketching the Vectors

To sketch, plot each vector from the origin (0,0). For example, \(\vec{u} = \langle 1, -2 \rangle\) starts at (0,0) and ends at (1,-2). Similarly, sketch \(\vec{v}, \vec{u} + \vec{v}, \vec{u} - \vec{v},\) and \(2\vec{u} - 3\vec{v}\). Each will have its endpoint determined by its components.

Step 5: Solving \( \vec{u} + \vec{x} = 2\vec{v} - \vec{x} \) for \( \vec{x} \)

First, rearrange the equation. \( \vec{u} + \vec{x} = 2\vec{v} - \vec{x} \) becomes \( \vec{u} + 2\vec{x} = 2\vec{v} \). Solve for \(\vec{x}\) by isolating it: \[ 2\vec{x} = 2\vec{v} - \vec{u} \].Replace \(\vec{u}\) and \(\vec{v}\) with their components: \[ 2\vec{x} = 2\langle 1, 1 \rangle - \langle 1, -2 \rangle = \langle 2, 2 \rangle - \langle 1, -2 \rangle = \langle 1, 4 \rangle \].Thus, \( \vec{x} = \langle \frac{1}{2}, 2 \rangle \).

Key Concepts

Vector Addition

When working with vectors, an integral operation is vector addition. This involves summing the corresponding components of two vectors. If you have two vectors, let's say \( \vec{u} = \langle a, b \rangle \) and \( \vec{v} = \langle c, d \rangle \), their sum is found by adding each component separately, which gives \( \vec{u} + \vec{v} = \langle a+c, b+d \rangle \). For our exercise, \( \vec{u} = \langle 1, -2 \rangle \) and \( \vec{v} = \langle 1, 1 \rangle \), thus, \( \vec{u} + \vec{v} = \langle 2, -1 \rangle \).

• The first component: \( 1 + 1 = 2 \)
• The second component: \( -2 + 1 = -1 \)

This results in \( \vec{u} + \vec{v} = \langle 2, -1 \rangle \). Vector addition enables convenient calculation of direction and magnitude when two forces act at a single point in physical applications.

Vector Subtraction

Much like vector addition, vector subtraction involves manipulating each vector's components, but here, you subtract. For two vectors \( \vec{u} = \langle a, b \rangle \) and \( \vec{v} = \langle c, d \rangle \), the subtraction \( \vec{u} - \vec{v} \) results in \( \langle a-c, b-d \rangle \).

• For \( \vec{u} = \langle 1, -2 \rangle \) and \( \vec{v} = \langle 1, 1 \rangle \), we calculate:
• The first component: \( 1 - 1 = 0 \)
• The second component: \( -2 - 1 = -3 \)

This gives us \( \vec{u} - \vec{v} = \langle 0, -3 \rangle \). Subtraction is useful for finding the relative distance between two points or vectors, often used in navigation and physics to determine difference in displacement or other vectors.

Scalar Multiplication

Scalar multiplication involves stretching or compressing a vector by a real number, known as a scalar. This concept is important when we want to change the magnitude of a vector without affecting its direction. When multiplying a vector \( \vec{u} = \langle a, b \rangle \) by a scalar \( k \), the result is \( k\vec{u} = \langle ka, kb \rangle \).

• For example, with \( 2\vec{u} \), if \( \vec{u} = \langle 1, -2 \rangle \), then:
• \( 2\vec{u} = \langle 2(1), 2(-2) \rangle = \langle 2, -4 \rangle \)
• For \(-3\vec{v} \), where \( \vec{v} = \langle 1, 1 \rangle \), it gives:
• \( -3\vec{v} = \langle -3(1), -3(1) \rangle = \langle -3, -3 \rangle \)

When you further combine these operations, like \( 2\vec{u} - 3\vec{v} \), it results in \( \langle -1, -7 \rangle \). Scalar multiplication is critical in physics for vectors representing velocity, force, and other vector quantities scaled by time or another magnitude.

Vector Sketching

Sketching vectors helps visualize their direction, orientation, and magnitude on a coordinate plane. To sketch a vector like \( \vec{u} = \langle a, b \rangle \), start by drawing an arrow from the origin \((0,0)\) to the point \( (a, b) \). This shows the vector's magnitude and direction.

• Take \( \vec{u} = \langle 1, -2 \rangle \): start at \((0,0)\), draw to \((1,-2)\).
• For \( \vec{v} = \langle 1, 1 \rangle \): start at \((0,0)\), draw to \((1,1)\).
• Repeat similarly for other vectors like \( \vec{u} + \vec{v} \), \( \vec{u} - \vec{v} \), and \( 2\vec{u} - 3\vec{v} \).

Visual representation assists in understanding interactions between vectors, making it a valuable tool for solving problems related to forces, movements, and transitions in physics and engineering. By sketching, we can visually confirm vector addition, subtraction, and scalar multiplication results.