Find the direction cosines and direction angles of the vector. 21-j- 7k cos(x) = cos(3) 11 cos(y) = a= В Y = 11 000
Find the direction cosines and direction angles of the vector. 21-j- 7k cos(x) = cos(3) 11 cos(y) = a= В Y = 11 000
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Finding the Direction Cosines and Direction Angles of a Vector**
Let's determine the direction cosines and direction angles for the given vector \(2\mathbf{i} - 3\mathbf{j} - 7\mathbf{k}\). To obtain the direction angles, we must first understand the concept of direction cosines.
### Direction Cosines
Direction cosines are the cosines of the angles that a vector makes with the coordinate axes. For a vector \(\mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\), the direction cosines are given by:
\[
\cos(\alpha) = \frac{a}{|\mathbf{v}|}, \quad \cos(\beta) = \frac{b}{|\mathbf{v}|}, \quad \cos(\gamma) = \frac{c}{|\mathbf{v}|}
\]
where \( |\mathbf{v}| \) is the magnitude of the vector and \(\alpha\), \(\beta\), and \(\gamma\) are the angles between the vector and the x, y, and z axes, respectively.
### Given Vector:
The vector provided is \(2\mathbf{i} - 3\mathbf{j} - 7\mathbf{k}\).
### Magnitude of the Vector:
\[
|\mathbf{v}| = \sqrt{2^2 + (-3)^2 + (-7)^2} = \sqrt{4 + 9 + 49} = \sqrt{62}
\]
### Direction Cosines:
1. \(
\cos(\alpha) = \frac{2}{\sqrt{62}} = \frac{2}{\sqrt{62}}\)
2. \(
\cos(\beta) = \frac{-3}{\sqrt{62}} = \frac{-3}{\sqrt{62}}\)
3. \(
\cos(\gamma) = \frac{-7}{\sqrt{62}} = \frac{-7}{\sqrt{62}}\)
### Direction Angles:
The direction angles \(\alpha\), \(\beta\), and \(\gamma\) can be determined by taking the inverse cosine of the direction cosines.
1. \(\alpha = \cos^{-1}\left(\frac{2}{\sqrt{62}}\right)\)
2. \(\beta = \cos^{-1}\left(\frac{-3}{\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb29ff817-7b8d-4056-886e-42f43ca6da22%2Fea2a0583-a2cc-4d4e-95f1-5063f59f2373%2Fczm4exs_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Finding the Direction Cosines and Direction Angles of a Vector**
Let's determine the direction cosines and direction angles for the given vector \(2\mathbf{i} - 3\mathbf{j} - 7\mathbf{k}\). To obtain the direction angles, we must first understand the concept of direction cosines.
### Direction Cosines
Direction cosines are the cosines of the angles that a vector makes with the coordinate axes. For a vector \(\mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\), the direction cosines are given by:
\[
\cos(\alpha) = \frac{a}{|\mathbf{v}|}, \quad \cos(\beta) = \frac{b}{|\mathbf{v}|}, \quad \cos(\gamma) = \frac{c}{|\mathbf{v}|}
\]
where \( |\mathbf{v}| \) is the magnitude of the vector and \(\alpha\), \(\beta\), and \(\gamma\) are the angles between the vector and the x, y, and z axes, respectively.
### Given Vector:
The vector provided is \(2\mathbf{i} - 3\mathbf{j} - 7\mathbf{k}\).
### Magnitude of the Vector:
\[
|\mathbf{v}| = \sqrt{2^2 + (-3)^2 + (-7)^2} = \sqrt{4 + 9 + 49} = \sqrt{62}
\]
### Direction Cosines:
1. \(
\cos(\alpha) = \frac{2}{\sqrt{62}} = \frac{2}{\sqrt{62}}\)
2. \(
\cos(\beta) = \frac{-3}{\sqrt{62}} = \frac{-3}{\sqrt{62}}\)
3. \(
\cos(\gamma) = \frac{-7}{\sqrt{62}} = \frac{-7}{\sqrt{62}}\)
### Direction Angles:
The direction angles \(\alpha\), \(\beta\), and \(\gamma\) can be determined by taking the inverse cosine of the direction cosines.
1. \(\alpha = \cos^{-1}\left(\frac{2}{\sqrt{62}}\right)\)
2. \(\beta = \cos^{-1}\left(\frac{-3}{\
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