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Chapter 9: Problem 40

Find the magnitude and direction angle of the vector \(\boldsymbol{v}\). $$\mathbf{v}=4 \mathbf{i}-8 \mathbf{j}$$

Short Answer

Expert verified
Answer: The magnitude of the vector is \(4\sqrt{5}\), and its direction angle is approximately \(296.6^\circ\).

Step by step solution

01

Find the magnitude of the vector

To find the magnitude of the vector \(\mathbf{v}=4 \mathbf{i}-8 \mathbf{j}\), we will use the Pythagorean theorem: $$\|\mathbf{v}\| = \sqrt{(4^2) + (-8^2)} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}$$ So, the magnitude of the vector is \(4\sqrt{5}.\)
02

Find the direction angle of the vector

To find the direction angle \(\theta\) of the vector \(\mathbf{v}=4 \mathbf{i}-8 \mathbf{j}\), we will use the inverse tangent function: $$\theta = \tan^{-1}\left(\frac{-8}{4}\right) = \tan^{-1}(-2)$$ To find the angle in the correct quadrant, as our vector has a positive x-component and a negative y-component, we must remember to adjust the result based on the polar coordinates and the fact that the vector is pointing in the fourth quadrant. Using a calculator, we have: $$\tan^{-1}(-2) \approx -63.4^\circ$$ To adjust our result into the fourth quadrant, we can add 360 degrees: $$-63.4^\circ + 360^\circ = 296.6^\circ$$ Thus, the direction angle of the vector is approximately \(296.6^\circ\).
03

Present the final answer

The magnitude and direction angle of the vector \(\mathbf{v}=4 \mathbf{i}-8 \mathbf{j}\) are as follows: Magnitude: \(4\sqrt{5}\) Direction angle: \(296.6^\circ\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnitude of a Vector
Vectors are entities that have both magnitude and direction. Understanding the magnitude of a vector is crucial in vector analysis, as it represents the length or the size of the vector. In this exercise, we have the vector \( \mathbf{v} = 4 \mathbf{i} - 8 \mathbf{j} \).
To find the magnitude, we employ the formula:
  • \( \|\mathbf{v}\| = \sqrt{(4^2) + (-8^2)} \)
  • This results in \( \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5} \)
By calculating \( \sqrt{80} \), we find that the magnitude is \( 4\sqrt{5} \). This tells us how long the vector is from its starting point to its endpoint in coordinate space.
Direction Angle
The direction angle of a vector indicates its orientation with respect to a reference line, often the positive x-axis. For the vector \( \mathbf{v} = 4 \mathbf{i} - 8 \mathbf{j} \), we determine the angle by analyzing the components. The vector is positioned in the fourth quadrant because it has a positive x-component (4) and a negative y-component (-8).
We use the inverse tangent function to find the angle relative to the x-axis:
  • \( \theta = \tan^{-1}\left(\frac{-8}{4}\right) = \tan^{-1}(-2) \)
  • Typically, \( \tan^{-1}(-2) \approx -63.4^\circ \)
To place this angle correctly within the fourth quadrant, we add 360 degrees, resulting in approximately \( 296.6^\circ \). This is because direction angles are usually measured counterclockwise from the positive x-axis.
Pythagorean Theorem
The Pythagorean theorem is a fundamental principle often used to calculate vector magnitudes. It states that in a right-angled triangle, the square of the hypotenuse's length is equal to the sum of the squares of the other two sides. In vector terms, the components \( 4 \) and \( -8 \) are the sides, and the vector's magnitude is the hypotenuse.
The formula is:
  • \( \|\mathbf{v}\| = \sqrt{(x^2) + (y^2)} \)
  • For \( \mathbf{v} = 4 \mathbf{i} - 8 \mathbf{j} \), \( \sqrt{(4^2) + (-8^2)} \)
  • This gives the magnitude \( \sqrt{80} = 4\sqrt{5} \)
This principle simplifies the process of finding how "long" a vector is just as the hypotenuse represents the longest side in a triangle.
Inverse Tangent Function
The inverse tangent function, also known as \( \tan^{-1} \), is used to calculate the angle whose tangent is a given number. It's particularly useful in vector analysis to find the angle between a vector and the x-axis. For the vector \( \mathbf{v} = 4 \mathbf{i} - 8 \mathbf{j} \), we calculate as follows:
  • \( \tan^{-1}\left(\frac{-8}{4}\right) = \tan^{-1}(-2) \)
  • The angle \( \theta \approx -63.4^\circ \) corresponds with the backward sweep of the tangent across the x-axis
Because the function originally yields an angle in the negative direction, adding 360 degrees situates it properly within the standard angle measure from the positive x-axis. This adjustment ensures you get an angle that aligns with real-world directional standards.

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