Chapter 1: Problem 19
Finding the Component Form of a Vector Using Trigonometry Find the component form of a vector with magnitude 4 that forms an angle of \(-45^{\circ}\) with the \(x\) -axis.
Short Answer
Expert verified
The component form of the vector is \((2\sqrt{2}, -2\sqrt{2})\).
Step by step solution
01
Understand the Components of the Vector
A vector can be broken into two components: an x-component and a y-component. For a vector with magnitude \( |\mathbf{v}| \) and angle \( \theta \), it can be expressed as \( \mathbf{v} = (v_x, v_y) \). To find these components, you will use trigonometry.
02
Use Cosine for the x-component
The x-component of the vector \( v_x \) can be found using the cosine of the angle: \[v_x = |\mathbf{v}| \cdot \cos(\theta) = 4 \cdot \cos(-45^{\circ})\]Cosine of \(-45^{\circ}\) is the same as cosine of \(45^{\circ}\), which is \(\frac{\sqrt{2}}{2}\). Thus, \[v_x = 4 \cdot \frac{\sqrt{2}}{2} = 2\sqrt{2}\]
03
Use Sine for the y-component
The y-component of the vector \( v_y \) can be found using the sine of the angle: \[v_y = |\mathbf{v}| \cdot \sin(\theta) = 4 \cdot \sin(-45^{\circ})\]Sine of \(-45^{\circ}\) is \(-\frac{\sqrt{2}}{2}\). Thus, \[v_y = 4 \cdot \left(-\frac{\sqrt{2}}{2}\right) = -2\sqrt{2}\]
04
Combine the Components
Now that you have both components, you can express the vector in its component form: \[\mathbf{v} = (v_x, v_y) = (2\sqrt{2}, -2\sqrt{2})\] Therefore, the component form of the vector is \((2\sqrt{2}, -2\sqrt{2})\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Trigonometry in Vector Decomposition
Trigonometry plays a vital role in breaking down vectors into their components. A vector can be thought of as having a direction and a magnitude—kind of like an arrow pointing somewhere with a certain length.
To split this into x and y components, we use trigonometric functions. Here are the main functions to remember when dealing with right triangles in this context:
To split this into x and y components, we use trigonometric functions. Here are the main functions to remember when dealing with right triangles in this context:
- The cosine function, which tells us about the adjacent side in a right-angled triangle.
- The sine function, which tells us about the opposite side.
- The tangent function, which is the ratio of the opposite to the adjacent side, but isn’t typically used for component breakdown.
- The x-component formula is: \(v_x = |\mathbf{v}| \cdot \cos(\theta)\)
- The y-component formula is: \(v_y = |\mathbf{v}| \cdot \sin(\theta)\)
Calculating the Magnitude of a Vector
The magnitude of a vector is essentially "how long" the vector is. Imagine it as the length of an arrow. It's a scalar quantity, which means it has only magnitude and no direction.
If you have a vector in component form, let's say \(\mathbf{v} = (v_x, v_y)\), its magnitude can be found using the Pythagorean theorem:
When given the magnitude and the angle, as in our problem with a magnitude of 4, this length is already known. But using trigonometry with our previously-found components lets us confirm it matches, ensuring our understanding is nice and solid.
If you have a vector in component form, let's say \(\mathbf{v} = (v_x, v_y)\), its magnitude can be found using the Pythagorean theorem:
- \(|\mathbf{v}| = \sqrt{v_x^2 + v_y^2}\)
When given the magnitude and the angle, as in our problem with a magnitude of 4, this length is already known. But using trigonometry with our previously-found components lets us confirm it matches, ensuring our understanding is nice and solid.
Positioning the Vector with Respect to the x-axis
The angle \(\theta\) between the vector and the x-axis is crucial because it tells us the vector's direction.
This isn't just any angle; it's always measured counterclockwise from the positive x-axis, like a compass for vectors. However, negative angles indicate clockwise measurement. For example, an angle of \(-45^{\circ}\) indicates the vector is directed downward and to the right relative to the positive x-axis.
To find out the vector's orientation, you can:
This isn't just any angle; it's always measured counterclockwise from the positive x-axis, like a compass for vectors. However, negative angles indicate clockwise measurement. For example, an angle of \(-45^{\circ}\) indicates the vector is directed downward and to the right relative to the positive x-axis.
To find out the vector's orientation, you can:
- Use cosine to find the x-component: this reflects how "right" or "left" the vector points.
- Use sine for the y-component: this tells us how "up" or "down" it extends.