Problem 15
Question
Find the component form of the vector \(\vec{v}\) using the information given about its magnitude and direction. Give exact values. \(\|\vec{v}\|=4\); when drawn in standard position \(\vec{v}\) lies in Quadrant II and makes a \(30^{\circ}\) angle with the negative \(x\) -axis
Step-by-Step Solution
Verified Answer
The component form of the vector is \((-2\sqrt{3}, 2)\).
1Step 1: Understanding the Problem
We need to find the component form of the vector \( \vec{v} \). We know its magnitude is 4, and it is positioned such that it lies in Quadrant II, making a 30-degree angle with the negative x-axis.
2Step 2: Determine the Vector's Direction
In Quadrant II, a vector making a 30-degree angle with the negative x-axis means it is 180° - 30° from the positive x-axis. Thus, the direction angle \( \theta \) is 150°.
3Step 3: Finding the x-component
The x-component of the vector is calculated using the formula, \( \vec{v}_x = \| \vec{v} \| \cos(\theta) \). Here, \( \theta = 150^\circ \), and \( \| \vec{v} \| = 4 \). Calculate: \( \vec{v}_x = 4 \cos(150^\circ) = 4 \times \left(-\frac{\sqrt{3}}{2}\right) = -2 \sqrt{3} \).
4Step 4: Finding the y-component
The y-component of the vector is calculated using the formula, \( \vec{v}_y = \| \vec{v} \| \sin(\theta) \). Here, \( \theta = 150^\circ \). Calculate: \( \vec{v}_y = 4 \sin(150^\circ) = 4 \times \frac{1}{2} = 2 \).
5Step 5: Write the Component Form
The component form of \( \vec{v} \) is formed as \( (\vec{v}_x, \vec{v}_y) \). From the calculations: \( \vec{v}_x = -2\sqrt{3} \) and \( \vec{v}_y = 2 \). Thus, the component form is \( (-2\sqrt{3}, 2) \).
Key Concepts
Magnitude and DirectionVector in Quadrant IITrigonometric Functions
Magnitude and Direction
When talking about vectors, two important aspects to understand are their magnitude and direction. The magnitude of a vector indicates its 'length' or 'size', which in practical terms means how far it reaches. The direction, on the other hand, tells us where the vector is pointing.
Imagine throwing a ball or shooting an arrow; the magnitude would represent how hard you throw it, while the direction shows the path it's going to take. In mathematical terms, the magnitude of a vector \( \vec{v} \) is represented by \( \|\vec{v}\| \). For the vector given in the problem, the magnitude is 4.
The direction is given by the angle the vector makes with the reference axis. In this specific case, we have a vector making a 30° angle with the negative x-axis. Understanding the proper calculation is essential since it defines how the vector is oriented in the Cartesian plane.
Imagine throwing a ball or shooting an arrow; the magnitude would represent how hard you throw it, while the direction shows the path it's going to take. In mathematical terms, the magnitude of a vector \( \vec{v} \) is represented by \( \|\vec{v}\| \). For the vector given in the problem, the magnitude is 4.
The direction is given by the angle the vector makes with the reference axis. In this specific case, we have a vector making a 30° angle with the negative x-axis. Understanding the proper calculation is essential since it defines how the vector is oriented in the Cartesian plane.
Vector in Quadrant II
The Cartesian plane is divided into four quadrants, each determining the general direction of vectors or points. Quadrant II, which is important for our discussion, is located between the positive y-axis and the negative x-axis.
For a vector like \( \vec{v} \) to be in Quadrant II, it must have a negative x-component and a positive y-component. This positioning is significant for calculating the vector's actual direction. In this case, the vector's angle with the negative x-axis is 30°. However, to find the vector's standard position angle, we need to think of it as an angle with the positive x-axis, which is calculated as 180° - 30° = 150°. This tells us that our vector aims backwards to some extent, headed more towards the vertical (y) direction than the horizontal (x) one.
Knowing where a vector lies in the plane helps recognize how both of its components (x and y directions) act and contribute to its total movement and orientation.
For a vector like \( \vec{v} \) to be in Quadrant II, it must have a negative x-component and a positive y-component. This positioning is significant for calculating the vector's actual direction. In this case, the vector's angle with the negative x-axis is 30°. However, to find the vector's standard position angle, we need to think of it as an angle with the positive x-axis, which is calculated as 180° - 30° = 150°. This tells us that our vector aims backwards to some extent, headed more towards the vertical (y) direction than the horizontal (x) one.
Knowing where a vector lies in the plane helps recognize how both of its components (x and y directions) act and contribute to its total movement and orientation.
Trigonometric Functions
Trigonometric functions are crucial when it comes to breaking down vectors into their components. They help us in using angles to determine the exact horizontal and vertical distances of a vector.
The primary functions we use here are sine (\( \sin \)) and cosine (\( \cos \)). They relate the angles to the sides of a right triangle, which is fundamental in dividing a vector into parts.
A vector's x-component \( \vec{v}_x \) is calculated with the cosine of the angle: \( \vec{v}_x = \| \vec{v} \| \cos(\theta) \). For example, using the 150° angle in our vector \( \vec{v} \), we find that \( \cos(150^\circ) = -\frac{\sqrt{3}}{2} \), leading to an x-component of \( -2\sqrt{3} \).
The y-component \( \vec{v}_y \) uses the sine function: \( \vec{v}_y = \| \vec{v} \| \sin(\theta) \). With \( \sin(150^\circ) = \frac{1}{2} \), our vector's y-component turns out to be 2.
These calculations provide a complete breakdown of the vector into workable parts, clearly showing its overall direction and extent.
The primary functions we use here are sine (\( \sin \)) and cosine (\( \cos \)). They relate the angles to the sides of a right triangle, which is fundamental in dividing a vector into parts.
A vector's x-component \( \vec{v}_x \) is calculated with the cosine of the angle: \( \vec{v}_x = \| \vec{v} \| \cos(\theta) \). For example, using the 150° angle in our vector \( \vec{v} \), we find that \( \cos(150^\circ) = -\frac{\sqrt{3}}{2} \), leading to an x-component of \( -2\sqrt{3} \).
The y-component \( \vec{v}_y \) uses the sine function: \( \vec{v}_y = \| \vec{v} \| \sin(\theta) \). With \( \sin(150^\circ) = \frac{1}{2} \), our vector's y-component turns out to be 2.
These calculations provide a complete breakdown of the vector into workable parts, clearly showing its overall direction and extent.
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