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Chapter 9: Problem 51

Find the magnitude and direction (in degrees) of the vector. $$\mathbf{v}=\mathbf{i}+\sqrt{3} \mathbf{j}$$

Short Answer

Expert verified
Magnitude: 2, Direction: 60掳

Step by step solution

01

Identify the Components

The vector \( \mathbf{v} \) is given as \( \mathbf{i} + \sqrt{3} \mathbf{j} \). This means its components are: \( x = 1 \) and \( y = \sqrt{3} \).
02

Calculate the Magnitude

The magnitude of a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} \) is calculated using the formula \( |\mathbf{v}| = \sqrt{a^2 + b^2} \). Substituting the values, we have: \( |\mathbf{v}| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \).
03

Find the Direction Angle

The direction \( \theta \) of the vector in degrees is given by \( \theta = \tan^{-1}\left(\frac{y}{x}\right) \). Substituting the known values, \( \theta = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) \). This simplifies to \( \theta = 60^\circ \), because \( \tan 60^\circ = \sqrt{3} \).

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Key Concepts

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

Vector Components
A vector is a mathematical entity with both a magnitude and a direction. It is often represented in a two-dimensional space with components along the x-axis (horizontal) and y-axis (vertical). These components help in understanding the vector's influence in each direction.
The vector in our exercise is given by \( \mathbf{v} = \mathbf{i} + \sqrt{3} \mathbf{j} \), where \( \mathbf{i} \) and \( \mathbf{j} \) are unit vectors along the x and y axes, respectively.
  • Here, the vector component along the x-axis is \( x = 1 \).
  • The component along the y-axis is \( y = \sqrt{3} \).

These components are vital as they provide the necessary values for further calculations, such as magnitude and direction. Understanding vector components can greatly help in visualizing the vector in a coordinate system.
Magnitude Formula
The magnitude of a vector is a measure of its length and is always a positive quantity. To find the magnitude, we use the Pythagorean theorem as a basis for our formula.
For a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} \), the magnitude is calculated using:\[|\mathbf{v}| = \sqrt{a^2 + b^2}\]
This formula is derived from the properties of a right-angled triangle, where the vector forms the hypotenuse.
In the exercise above, substituting the given components \( a = 1 \) and \( b = \sqrt{3} \):
  • Calculate \( 1^2 + (\sqrt{3})^2 = 1 + 3 = 4 \).
  • Thus, \( |\mathbf{v}| = \sqrt{4} = 2 \).

The magnitude helps us understand how far the vector extends in the coordinate plane.
Direction Angle
Once you know a vector's components, calculating its direction angle is crucial to fully understanding its orientation in space. The direction angle, often denoted as \( \theta \), measures the angle between the positive x-axis and the vector.
The angle \( \theta \) is found using the formula:\[\theta = \tan^{-1}\left(\frac{b}{a}\right)\]
where \( b \) is the y-component and \( a \) is the x-component of the vector.
  • In our example, \( \theta = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) \).
  • This expression simplifies to \( \theta = \tan^{-1}(\sqrt{3}) \).
  • From trigonometric tables or calculators, we know \( \tan 60^\circ = \sqrt{3} \).

Thus, the direction angle is \( 60^\circ \). Understanding this angle helps in both the graphical representation and practical applications of vector analysis.
Inverse Tangent Function
The inverse tangent function, denoted as \( \tan^{-1} \) or sometimes as \( \arctan \), is used to determine the angle whose tangent is a given number. In vector problems, it is particularly important for calculating angles when you have the ratio of the two components.
You apply \( \tan^{-1} \) when you are given the opposite and adjacent sides of a right triangle, which in vector terminology means the y and x components.
For the vector discussed:
  • You need \( \tan^{-1}(\sqrt{3}) \).
  • Consult trigonometric tables or a calculator to find that \( \tan^{-1}(\sqrt{3}) \) equals \( 60^\circ \).

Using the inverse tangent function effectively allows accurate determination of direction angles, thus being an essential tool in the analysis of vector quantities.

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Most popular questions from this chapter

The lengths of two vectors a and \(b\) and the angle \(\theta\) between them are given. Find the length of their cross product, \(|\mathbf{a} \times \mathbf{b}|\). $$|\mathbf{a}|=4, \quad|\mathbf{b}|=5, \quad \theta=30^{\circ}$$

Find the area of the parallelogram determined by the given vectors. $$\mathbf{u}=\langle 3,2,1\rangle, \quad \mathbf{v}=\langle 1,2,3\rangle$$

Given three vectors \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w},\) their scalar triple product can be performed in six different orders: $$\begin{array}{lll} \mathbf{u} \cdot(\mathbf{v} \times \mathbf{w}), & \mathbf{u} \cdot(\mathbf{w} \times \mathbf{v}), & \mathbf{v} \cdot(\mathbf{u} \times \mathbf{w}) \\ \mathbf{v} \cdot(\mathbf{w} \times \mathbf{u}), & \mathbf{w} \cdot(\mathbf{u} \times \mathbf{v}), & \mathbf{w} \cdot(\mathbf{v} \times \mathbf{u}) \end{array}$$ (a) Calculate each of these six triple products for the vectors: $$\mathbf{u}=\langle 0,1,1\rangle, \quad \mathbf{v}=\langle 1,0,1\rangle, \quad \mathbf{w}=\langle 1,1,0\rangle$$ (b) On the basis of your observations in part (a), make a conjecture about the relationships between these six triple products. (c) Prove the conjecture you made in part (b).

Find the area of \(\triangle P Q R\). $$P(1,0,1), Q(0,1,0), R(2,3,4)$$

Find the horizontal and vertical components of the vector with given length and direction, and write the vector in terms of the vectors i and j. $$|\mathbf{v}|=800, \quad \theta=125^{\circ}$$
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