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Chapter 7: Problem 27

Find the magnitude and direction angle of each vector. $$\langle\sqrt{3}, 1\rangle$$

Short Answer

Expert verified
Magnitude is 2, direction angle is \(\frac{\pi}{6}\).

Step by step solution

01

- Identify the Components of the Vector

Recognize that the vector \(\<\sqrt{3}, 1 \>\) has components \((x, y) = (\sqrt{3}, 1)\)
02

- Calculate the Magnitude

The magnitude of a vector \(\\) is given by \(\sqrt{a^2 + b^2}\). Substitute \(a = \sqrt{3}\) and \(b = 1\): \[\|\<\sqrt{3}, 1 \>\| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2\]
03

- Find the Direction Angle

The direction angle \(\theta\) of a vector can be found using the arctangent function: \(\theta = \arctan\left(\frac{y}{x}\right)\). Substitute \(x = \sqrt{3}\) and \(y = 1\): \[\theta = \arctan\left(\frac{1}{\sqrt{3}}\right) = \arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}\]

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Key Concepts

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

magnitude of a vector
The magnitude of a vector, also known as the length or norm, is a measure of how long the vector is. It tells us the distance from the origin to the point represented by the vector. For any vector \(\text{\}\), the magnitude can be computed using the Pythagorean theorem, which gives us \(\text{\|}\text{\|} = \sqrt{a^2 + b^2}\)

Let's look at our example vector \(\text{\<}\text{\sqrt{3}, 1\>}\). To find the magnitude:
  • Identify components: \(a = \text{\sqrt{3}}\) and \(b = 1\).
  • Plug into the formula: \(\text{\|}<\text{\text{\sqrt{3}, 1}>\text{\|} = \sqrt{\text{(\text{\sqrt{3}})}^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2\)
    The magnitude of our vector is 2.
direction angle of a vector
The direction angle of a vector is the angle that the vector makes with the positive x-axis. This angle helps in understanding the vector's orientation in a plane. To find the direction angle \(\theta\) of a vector \(\text{\}\), we use the arctangent function: \(\theta = \text{arctan}\text{(y/x)}\)

For our vector \(\<\text{\sqrt{3}, 1\>}\), we can compute its direction angle as follows:
  • Identify components: \( a = \text{\sqrt{3}}\) and \( b = 1\).
  • Use the formula: \(\theta = \text{arctan}(\text{\frac{y}{x}}) = \text{arctan}(\text{\frac{1}{\text{\sqrt{3}}})\).
    Note: \(\text{arctan(1/\sqrt{3})}\) is a well-known value and equals \(\text{\frac{\pi}{6}}\) radians.

The direction angle of our vector is \(\frac{\pi}{6}\) radians.
arctangent function
The arctangent function, often written as \(\text{\arctan}\text{(x)}\), is used to find the angle whose tangent is \(x\). It's the inverse function of the tangent (tan) function from trigonometry. When you have a ratio like \(\text{\frac{y}{x}}\), the arctangent function helps determine the angle corresponding to that ratio.
  • For example, \(\text{\arctan(1/\sqrt{3})}\) gives us the angle whose tangent is \(\text{\frac{1}{\sqrt{3}}}\).
  • This value is \(\text{\frac{\pi}{6}}\), which is equivalent to 30 degrees.

The arctangent function is particularly useful when dealing with vectors and trigonometry problems because it converts a ratio of sides back into an angle, providing a way to understand directions and orientations in two-dimensional space.

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Most popular questions from this chapter

Find the product of the given complex number and its complex conjugate in trigonometric form. $$3\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)$$

Write each complex number in the form \(a+b i\). $$8.1(\cos \pi+i \sin \pi)$$

Solve each problem. Given that \(z=-3+3 i,\) find \(z^{2}\) by writing \(z\) in trigonometric form and computing \(z \cdot z\)

For each given complex number, determine its complex conjugate in trigonometric form. $$3\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)$$

Given that \(\mathbf{A}=\langle 3,1\rangle\) and \(\mathbf{B}=\langle- 2,3\rangle,\) find the magnitude and direction angle for each of the following vectors. $$5 \mathbf{B}$$
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