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Chapter 7: Problem 31

Find the magnitude and direction angle of each vector. $$\langle 8,-8 \sqrt{3}\rangle$$

Short Answer

Expert verified
Magnitude: 16, Direction angle: 300掳

Step by step solution

01

Finding the Magnitude

First, calculate the magnitude of the vector \(\langle 8, -8 \sqrt{3}\rangle\). The magnitude \(\|\vec{v}\|\) of a vector \(\vec{v} = \langle a, b \rangle\) is found using the formula: \[\|\vec{v}\| = \sqrt{a^2 + b^2}\]Substitute the given values into the formula:\[\|\vec{v}\| = \sqrt{8^2 + (-8 \sqrt{3})^2} \|\vec{v}\| = \sqrt{64 + 192}\|\vec{v}\| = \sqrt{256}\|\vec{v}\| = 16\]Thus, the magnitude of the vector is 16.
02

Finding the Direction Angle

Next, compute the direction angle \(\theta\) of the vector. The direction angle is given by the formula: \[\theta = \tan^{-1}\left(\frac{b}{a}\right)\]Using the vector's components, substitute \(a = 8\) and \(b = -8\sqrt{3}\):\[\theta = \tan^{-1}\left(\frac{-8\sqrt{3}}{8}\right)\theta = \tan^{-1}(-\sqrt{3})\]The angle whose tangent is \( -\sqrt{3} \) is \( -\frac{\pi}{3} \) radians or \( -60^\circ\). Since direction angles are generally measured from the positive x-axis, add \(360^\circ \) if needed to get a positive equivalent:\[-60^\circ + 360^\circ = 300^\circ\]Thus, the direction angle is \( 300^\circ\).

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Key Concepts

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

Vector Magnitude
The magnitude of a vector is a measure of its length. Think of it as the distance from the origin to the point defined by the vector. To find the magnitude of a vector \(\vec{v} = \langle a, b \rangle\), we use the Pythagorean theorem: \[ \|\vec{v}\| = \sqrt{a^2 + b^2} \] In our example with the vector \(\langle 8, -8 \sqrt{3} \rangle\), we substituted \(a = 8\) and \(b = -8 \sqrt{3}\). Performing the calculations, we found that the magnitude is 16. This tells us that the vector has a 'length' of 16 units.
Direction Angle
The direction angle of a vector points towards where the vector aims in relation to the positive x-axis. It's important in describing the vector's orientation. To find this angle, we use the inverse tangent or tangent-inverse function, which comes from trigonometry. For the vector \(\langle 8, -8 \sqrt{3} \rangle\), the formula we used is: \[ \theta = \tan^{-1}\left(\frac{b}{a}\right) \] Plugging in the values, \(\theta = \tan^{-1}(-\sqrt{3}) \), we get \(\theta = -60^\circ \). To convert this into a positive angle, we add \(360^\circ\), resulting in \(300^\circ\). So, the direction angle is \(300^\circ\).
Tangent-Inverse Function
The tangent-inverse function, written as \(\tan^{-1}(x)\), is the reverse operation of the tangent function. It helps find angles when the tangent of the angle is known. For example, if \(\tan(\theta) = x\), then \(\theta = \tan^{-1}(x)\). In our problem, this means using \(\tan^{-1}\) to identify the angle \(\theta\) from the ratio \(\frac{b}{a}\). This is handy in various vector calculations and helps translate between different forms of representing directions.
Trigonometry in Vectors
Trigonometry plays a crucial role in working with vectors, especially in finding magnitudes and direction angles. We often use trigonometric ratios like sine, cosine, and tangent. For the vector \(\langle a, b \rangle\), if we need to find the angle \(\theta\) it makes with the x-axis, we use:
  • \(\tan(\theta) = \frac{b}{a}\) (for the direction angle)
  • Pythagorean theorem \(\sqrt{a^2 + b^2}\) (for magnitude)
Our example demonstrated using \(\tan^{-1}\) to find the direction angle, which is a direct application of trigonometry principles to vector problems.

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