Problem 32 - 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. $$\mathbf{v}=-3 \mathbf{i}+4 \mathbf{j}$$

Step-by-Step Solution

Answer

The magnitude of the vector is 5 and its direction, expressed as an angle in standard position, is approximately 126.87°.

Step 1 Title: Finding the Components of the Vector

A vector is represented in terms of i and j, which are the unit vectors along the x and y axes. The given vector is \(\mathbf{v}=-3 \mathbf{i}+4 \mathbf{j}\). Therefore, its x-component \(V_x\) equals -3 and its y-component \(V_y\) equals 4.

Step 2 Title: Calculating the Magnitude of the Vector

The magnitude of a vector is found by taking the square root of the sum of the squares of the components. For vector \(\mathbf{v}\), it is: \(\|\mathbf{v}\| = \sqrt{{V_x}^2 + {V_y}^2} = \sqrt{(-3)^2 + 4^2} = 5\)

Step 3 Title: Calculation of direction

The direction of the vector can be found using the tangent function. The angle in standard position, \(\theta\), is found by taking the arctan of \(V_y / V_x\). However, because arctan only gives results between -90 degrees and 90 degrees, some adjustment may be necessary depending on the quadrant. In this case, \(\theta = \text{arctan}(V_y / V_x) = \text{arctan}(4 / -3)\). As the x-component is negative, the vector is in the second quadrant and we need to add 180 degrees to get the correct angle. Thus, \(\theta = \text{arctan}(4 / -3) + 180° \approx 126.87° \) to two decimal places.

Key Concepts

Unit Vectors

Unit vectors are essential building blocks in vector calculus, denoted commonly by \(\mathbf{i}\) and \(\mathbf{j}\) for the x and y axes respectively. Their main utility is to provide a basis for any vector in a two-dimensional space.
As unit vectors have a magnitude of one, they are often used to express vectors in component form like \(\mathbf{v} = a\mathbf{i} + b\mathbf{j}\), where \(a\) and \(b\) are scalar coefficients that scale these unit vectors. This means every vector in 2D can be decomposed into these two parts:

• The \(a\mathbf{i}\) component runs parallel to the x-axis.
• The \(b\mathbf{j}\) component runs parallel to the y-axis.

The separation of a vector into these components allows us to analyze and manipulate vectors easily. For example, given \(\mathbf{v} = -3\mathbf{i} + 4\mathbf{j}\), the x-component is -3 and the y-component is 4, showing how the vector moves left (due to the negative x-component) and upward (due to the positive y-component). Understanding this decomposition is crucial for computations like finding magnitudes and directions.

Arctan Function

The arctan function, also known as the inverse tangent function, helps us determine angles from given trigonometric ratios. When dealing with vectors, especially when determining their direction, the arctan function is instrumental.
Consider a vector \(\mathbf{v} = a\mathbf{i} + b\mathbf{j}\), the direction \(\theta\) is the angle measured counterclockwise from the positive x-axis to the vector. This angle can be found via \(\theta = \text{arctan} \left( \frac{b}{a} \right)\).
It is important to note that the arctan function can only provide angles between -90 and 90 degrees, leading to the need for quadrant adjustments. In our example, applying \(\text{arctan}(\frac{4}{-3})\) reveals an angle in the second quadrant due to a negative x-component. Without adjustments, it would not reflect the actual direction.

Quadrant Adjustment

To correctly determine the direction of a vector, one must consider which quadrant the vector lies in. This understanding begins with recognizing the signs of the x and y components.
For four possible quadrants in a standard Cartesian plane:

• First Quadrant: Both components are positive.
• Second Quadrant: Negative x-component and positive y-component.
• Third Quadrant: Both components are negative.
• Fourth Quadrant: Positive x-component and negative y-component.

The process involves:- Identifying which signs the vector components have- Adjusting the angle calculated by the arctan function. As seen in the solution, if a vector like \(-3\mathbf{i} + 4\mathbf{j}\) lies in the second quadrant, adding 180 degrees to the initial arctan result \(\theta\) ensures the angle correctly reflects the vector’s orientation.
Thus, properly performing quadrant adjustments defines the correct angle \(\theta = 126.87^\circ\) in our example.