Problem 31 - Precalculus : Building Concepts and Connections

Question

Find the magnitude and direction of each of the given vectors. Express the direction as an angle \(\theta\) in standard position, where \(0^{\circ} \leq \theta<360^{\circ},\) to tuo decimal places. $$\mathbf{v}=4 \mathbf{i}-2 \mathbf{j}$$

Step-by-Step Solution

Answer

The magnitude of the vector \( \mathbf{v}=4\mathbf{i}-2\mathbf{j} \) is \(2\sqrt{5}\) and its direction, expressed as an angle in standard position, is approximately \(306.87^{\circ}\) to two decimal places.

Step 1: Compute the magnitude of the vector

The magnitude of the vector, denoted as \(|| \mathbf{v} || \), is given by the formula \( || \mathbf{v} ||= \sqrt{x^2+y^2} \). Here, \(x\) and \(y\) are the components of the vector. For the given vector \( \mathbf{v}=4\mathbf{i}-2\mathbf{j} \), \(x=4\) and \(y=-2\). So, \( || \mathbf{v} ||= \sqrt{(4)^2+(-2)^2} = \sqrt{16+4} = \sqrt{20}=2\sqrt{5} \). So, the magnitude of the vector is \(2\sqrt{5}\).

Step 2: Compute the angle in radians of the vector

The angle \(\theta\) is given by the formula \(\theta = \arctan(\frac{y}{x})\). Here, \(x=4\) and \(y=-2\). So, \(\theta = \arctan(\frac{-2}{4}) = \arctan(-0.5) \).

Step 3: Convert the angle to degrees and adjust to lie within the correct quadrant

Since the \(x\) component of the vector is positive and the \(y\) component is negative, the vector lies in the fourth quadrant. Due to the way the arctan function works, it will have given an angle in the first quadrant (between \( -90^{\circ} \) and \( 0^{\circ} \)), so we need to adjust this by adding 360 to give an angle in the fourth quadrant. First, convert \(\theta\) from radians to degrees by multiplying by \( \frac{180}{\pi} \). Then, add 360 to this result if it's negative. After performing these operations, we get \(\theta \approx 306.87^{\circ}\). This is the angle of the vector in standard position to two decimal places.

Key Concepts

Vector Analysis

Vector analysis is a crucial aspect of understanding physical quantities that have both magnitude and direction. A vector is represented in mathematics by a directed line segment with an initial point and a terminal point. For instance, the vector \(\mathbf{v}=4\mathbf{i}-2\mathbf{j}\) can be thought of as an arrow pointing from one point to another in a coordinate system.

The analysis of such a vector requires us first to determine its magnitude, or 'length'. This scalar quantity expresses the size or extent of the vector, irrespective of its direction. For \(\mathbf{v}\), we calculate its magnitude using the Pythagorean theorem which states that the sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse. In terms of vector components, the formula becomes \(|| \mathbf{v} || = \sqrt{x^2+y^2}\). For our example \(\mathbf{v}\), the components are \(x=4 \) and \(y=-2\), resulting in a magnitude of \(2\sqrt{5}\).

Understanding this principle is significant not just for abstract mathematical computation but also for practical implications like physics where vectors are used to represent forces, velocities, and other directional quantities.

Trigonometry in Vectors

Trigonometry is often used in conjunction with vector analysis to determine the direction of a vector. The direction is usually given by an angle \(\theta \) which indicates its orientation relative to a predefined axis. For the vector \(\mathbf{v}\), the direction can be found using the arctangent function. This function relates the angle \(\theta \) to the ratio of the vector's \(y\) component to its \(x\) component, so \(\theta = \arctan(\frac{y}{x}) \).

Since \(\mathbf{v}\) has components \(x=4\) and \(y=-2\), we find that \(\theta \approx \arctan(-0.5)\). The angle obtained from the arctangent function needs to be interpreted carefully since it only yields values between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\). It's imperative to adjust \(\theta\) to fall in the correct quadrant by taking into account the signs of the vector components. For \(\mathbf{v}\), this means we locate it in the fourth quadrant, which gives us \(\theta \approx 306.87^\circ \) after conversion to degrees and necessary adjustments.

Trigonometric relationships in vectors are foundational for translating between the algebraic and geometric representations of vectors, especially when dealing with components and directions.

Standard Position Angle

The standard position angle of a vector is the angle \(\theta \) it makes with the positive x-axis, measured counterclockwise from that axis. It's an essential notion in both mathematics and physics because it provides a standardized way of expressing the direction of a vector. The angle is usually measured in degrees, where a full circle represents \(360^\circ\) and is within the range of \(0^\circ \) to \(360^\circ\), though sometimes radians are used instead.

In the context of the given vector \(\mathbf{v}\), the standard position angle was found to be approximately \(306.87^\circ\). This measurement considers that the vector lies in the fourth quadrant and, therefore, \(\theta\) must be between \(270^\circ\) and \(360^\circ\). In general, determining the standard position angle involves trigonometry and knowledge of the coordinate system the vector resides in. This standardized approach aids in the clear and consistent communication of directional information across various scientific and engineering disciplines.