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Chapter 12: Problem 52

Find the direction angles of the vector represented by \(P Q\). \(P(0.0 .0), Q(5,12,13)\)

Short Answer

Expert verified
The direction angles are \( \cos^{-1} \left( \frac{5}{\sqrt{338}} \right), \cos^{-1} \left( \frac{12}{\sqrt{338}} \right), \cos^{-1} \left( \frac{13}{\sqrt{338}} \right) \).

Step by step solution

01

Understand the Concept

Direction angles are the angles a vector makes with the positive x, y, and z axes. To find them, we first need the direction cosines, which are the cosines of these angles, and they are proportional to the vector's components.
02

Determine Vector Components

The vector \( \overrightarrow{PQ} \) is determined by subtracting the coordinates of point \( P \) from point \( Q \). Thus, \( \overrightarrow{PQ} = (5-0, 12-0, 13-0) = (5, 12, 13) \).
03

Calculate the Magnitude of the Vector

Calculate the magnitude of vector \( \overrightarrow{PQ} \) using the formula \( \| \overrightarrow{PQ} \| = \sqrt{5^2 + 12^2 + 13^2} \). Work this out to get \( \| \overrightarrow{PQ} \| = \sqrt{25 + 144 + 169} = \sqrt{338} \).
04

Find Direction Cosines

The direction cosines of the vector are \( \cos \alpha = \frac{5}{\sqrt{338}} \), \( \cos \beta = \frac{12}{\sqrt{338}} \), \( \cos \gamma = \frac{13}{\sqrt{338}} \), where \( \alpha, \beta, \gamma \) are the angles between \( \overrightarrow{PQ} \) and the x, y, z axes, respectively.
05

Calculate Direction Angles

Determine the direction angles using the inverse cosine function: \( \alpha = \cos^{-1} \left( \frac{5}{\sqrt{338}} \right) \), \( \beta = \cos^{-1} \left( \frac{12}{\sqrt{338}} \right) \), \( \gamma = \cos^{-1} \left( \frac{13}{\sqrt{338}} \right) \). Compute these values to get the final angles.

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Key Concepts

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

Direction Cosines
To understand direction cosines, think about how a vector aligns with the x, y, and z axes in 3D space. Direction cosines are specifically the cosines of these alignment angles. They are useful because they tell us how much of the vector is pointing in each of the axis directions. When a vector aligns more closely with an axis, the direction cosine for that axis will be closer to 1, and if it's perpendicular, closer to 0.

Here's how they work:
  • The cosine of the angle between the vector and the x-axis is denoted as \( \cos \alpha \).
  • The angle with the y-axis is represented by \( \cos \beta \).
  • And for the z-axis, it is \( \cos \gamma \).
For our given vector \( \overrightarrow{PQ} = (5, 12, 13) \), these direction cosines can be calculated by dividing each component by the vector's magnitude. This shows each component's proportionate contribution to the vector's direction.
Vector Components
Vector components break down a vector into parts that correspond to each axis. Think of them as the building blocks of the vector. They indicate how much of the vector travels in each direction: x, y, and z.

To find the components of a vector given two points, subtract the coordinates of the initial point from the terminal point. For example:
  • Start with point \( P(0, 0, 0) \) and point \( Q(5, 12, 13) \).
  • Subtract the coordinates of \( P \) from \( Q \): \( (5-0, 12-0, 13-0) = (5, 12, 13) \).
This result, \( (5, 12, 13) \), is the vector \( \overrightarrow{PQ} \). Each number represents how far the vector moves along that particular axis.
Magnitude of a Vector
The magnitude of a vector is essentially its length. It's the distance from the vector's tail to its head in space. Calculating the magnitude involves using the Pythagorean Theorem in three dimensions.

For the vector \( \overrightarrow{PQ} = (5, 12, 13) \), the magnitude is found with the formula:\[ \| \overrightarrow{PQ} \| = \sqrt{5^2 + 12^2 + 13^2}\]
  • Compute each square: \( 5^2 = 25 \), \( 12^2 = 144 \), \( 13^2 = 169 \).
  • Sum these values: \( 25 + 144 + 169 = 338 \).
  • Take the square root of this sum: \( \sqrt{338} \).
The result, \( \sqrt{338} \), is the magnitude of vector \( \overrightarrow{PQ} \). This value helps in calculating the direction cosines and consequently the direction angles, as it's needed to normalize the vector's components.

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Most popular questions from this chapter

In Problems 7 through 12 , find the rectangular coordinates of the points with the given spherical coordinates \((\rho, \phi, \theta)\). \((2,0, \pi)\)

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Determine whether the graph of the given equation is a paraboloid or a hyperboloid. Check your answer graphically if you have access to a computer algebra system with a "contour plotting" facility. \(6 x y+8 y z=14\)

Describe the traces of the given surfaces in planes of the indicated type. \(z=4 x^{2}+9 y^{2} ;\) in planes parallel to the \(y z\) -plane

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