Problem 38 - Trigonometry

Question

For each vector, find \(\frac{1}{2} \mathbf{V},-\mathbf{V}\), and \(4 \mathbf{V}\). $$V=-\sqrt{3} \mathbf{i}-\mathbf{j}$$

Step-by-Step Solution

Answer

\( \frac{1}{2} \mathbf{V} = -\frac{\sqrt{3}}{2} \mathbf{i} - \frac{1}{2} \mathbf{j} \), \(-\mathbf{V} = \sqrt{3} \mathbf{i} + \mathbf{j} \), \(4 \mathbf{V} = -4\sqrt{3} \mathbf{i} - 4\mathbf{j} \).

Step 1: Understanding the Problem

We are given a vector \( \mathbf{V} = -\sqrt{3} \mathbf{i} - \mathbf{j} \). The task is to find three transformations of this vector: \( \frac{1}{2} \mathbf{V} \), \(-\mathbf{V}\), and \(4 \mathbf{V} \). Each of these operations involves multiplying the vector by a scalar value.

Step 2: Calculating \( \frac{1}{2} \mathbf{V} \)

To find \( \frac{1}{2} \mathbf{V} \), we multiply each component of \( \mathbf{V} \) by \( \frac{1}{2} \). Thus, \( \frac{1}{2} (-\sqrt{3} \mathbf{i} - \mathbf{j}) = \left(-\frac{\sqrt{3}}{2} \right) \mathbf{i} + \left(-\frac{1}{2}\right) \mathbf{j} \).

Step 3: Calculating \( -\mathbf{V} \)

Finding \(-\mathbf{V}\) involves multiplying the vector \( \mathbf{V} \) by \(-1\). Thus, \(-(-\sqrt{3} \mathbf{i} - \mathbf{j}) = \sqrt{3} \mathbf{i} + \mathbf{j} \).

Step 4: Calculating \( 4 \mathbf{V} \)

For \( 4 \mathbf{V} \), we need to multiply the vector by \(4\). This gives: \( 4(-\sqrt{3} \mathbf{i} - \mathbf{j}) = -4\sqrt{3} \mathbf{i} - 4\mathbf{j} \).

Key Concepts

Scalar Multiplication

Scalar multiplication is a foundational concept in understanding vectors. A scalar is simply a number that we use to scale a vector. When we multiply a vector by a scalar, each component of the vector is multiplied by that scalar. This operation stretches or compresses the vector but does not change its direction unless the scalar is negative.

For example, consider the vector \( \mathbf{V} = -\sqrt{3} \mathbf{i} - \mathbf{j} \). If we multiply it by \( \frac{1}{2} \), the resulting vector \( \frac{1}{2} \mathbf{V} \) is \( \left(-\frac{\sqrt{3}}{2}\right) \mathbf{i} + \left(-\frac{1}{2}\right) \mathbf{j} \). Here, each component of \( \mathbf{V} \) is halved, effectively shrinking the vector's length by half.

• If we multiply by a positive scalar greater than one, the vector lengthens.
• If the scalar is a fraction, the vector shortens.
• A negative scalar inverts the direction while scaling the size.

Vector Operations

Vector operations allow us to manipulate and combine vectors in different ways. Basic operations include addition, subtraction, and scalar multiplication. In physics and engineering, these operations help describe forces, velocities, and other vector-based quantities.

• Scalar Multiplication: As described, this operation scales a vector by a scalar. For example, multiplying \( \mathbf{V} \) by \(-1\) yields \(-\mathbf{V} = \sqrt{3} \mathbf{i} + \mathbf{j} \), reversing its direction.
• Vector Addition: Combine two vectors by adding corresponding components. If \( \mathbf{A} = a\mathbf{i} + b\mathbf{j} \) and \( \mathbf{B} = c\mathbf{i} + d\mathbf{j} \), then \( \mathbf{A} + \mathbf{B} = (a+c)\mathbf{i} + (b+d)\mathbf{j} \).
• Vector Subtraction: This is adding a negative vector. For example, subtracting \( \mathbf{B} \) from \( \mathbf{A} \) is like adding \( -\mathbf{B} \) to \( \mathbf{A} \).

Vector operations are visually intuitive because they can be represented graphically, aiding in understanding the physical phenomena they describe.

Vectors in Trigonometry

Vectors are deeply interconnected with trigonometry, as they often involve angles and magnitudes. A vector in two-dimensional space can be represented by its magnitude and direction, often expressed using trigonometric functions.

Consider a vector \( \mathbf{V} \). It can be decomposed into its horizontal and vertical components using trigonometry:

• Magnitude: The length of the vector, computed as \( \sqrt{a^2 + b^2} \) for a vector \( a\mathbf{i} + b\mathbf{j} \).
• Direction: The angle \( \theta \) it makes with the positive x-axis, typically found using \( \tan^{-1}\left(\frac{b}{a}\right) \).

When vectors are broken down into these components, we use the cosine and sine of the angle to resolve the vector into x and y components. This application is prevalent in physics problems, such as projectile motion, where understanding components helps break down complex motions into simpler parts.