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Chapter 11: Problem 37

Find the direction angles of the vector. $$ \mathbf{u}=\langle-1,5,2\rangle $$

Short Answer

Expert verified
The direction angles of the vector \(\mathbf{u}=\langle -1,5,2 \rangle\) are given by \( \alpha = \cos^{-1}\left(-\frac{1}{\sqrt{30}}\right)\), \( \beta = \cos^{-1}\left(\frac{5}{\sqrt{30}}\right)\), and \( \gamma = \cos^{-1}\left(\frac{2}{\sqrt{30}}\right)\)

Step by step solution

01

Find the magnitude of the vector

Calculate the magnitude of the vector \(\mathbf{u}\) using the formula \(|\mathbf{u}| = \sqrt{x^2+y^2+z^2}\), where \(x=-1\), \(y=5\), and \(z=2\), so \(|\mathbf{u}| = \sqrt{(-1)^2+5^2+2^2} = \sqrt{30}\)
02

Find the direction cosines

Calculate the direction cosines using the formulas: \(\cos(\alpha) = \frac{x}{|\mathbf{u}|}\), \(\cos(\beta) = \frac{y}{|\mathbf{u}|}\) and \(\cos(\gamma) = \frac{z}{|\mathbf{u}|}\). This will give \( \cos(\alpha) = -\frac{1}{\sqrt{30}}, \cos(\beta) = \frac{5}{\sqrt{30}}, \text{ and } \cos(\gamma) = \frac{2}{\sqrt{30}}\)
03

Find the direction angles

Apply inverse cosine function to each direction cosine equation to find direction angles. This gives \( \alpha = \cos^{-1}\left(-\frac{1}{\sqrt{30}}\right)\), \( \beta = \cos^{-1}\left(\frac{5}{\sqrt{30}}\right)\), and \( \gamma = \cos^{-1}\left(\frac{2}{\sqrt{30}}\right)\)

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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 is a measure of its length. It's like figuring out the distance it covers in space, similar to measuring the length of a stick. For a 3-dimensional vector \(\mathbf{u} = \langle x, y, z \rangle\), the magnitude is found using the Pythagorean theorem. Here's the formula:
  • \(|\mathbf{u}| = \sqrt{x^2 + y^2 + z^2}\)
This formula combines all the components of the vector. In our example with \(\mathbf{u} = \langle -1, 5, 2 \rangle\), calculate it as:
  • \(|\mathbf{u}| = \sqrt{(-1)^2 + 5^2 + 2^2} = \sqrt{1 + 25 + 4} = \sqrt{30}\)

This tells us how far out the vector reaches in space, giving you a clear picture of its 'size' or extent.
Direction Cosines
Direction cosines are ratios that show how much a vector points in each of the three coordinate directions: x, y, and z. Think of them as telling you how the vector "leans" towards each axis.

For a vector \(\mathbf{u} = \langle x, y, z \rangle\), the direction cosines are obtained by dividing each component by the magnitude of the vector:
  • \(\cos(\alpha) = \frac{x}{|\mathbf{u}|}\)
  • \(\cos(\beta) = \frac{y}{|\mathbf{u}|}\)
  • \(\cos(\gamma) = \frac{z}{|\mathbf{u}|}\)
In our case, with \(|\mathbf{u}| = \sqrt{30}\) and vector components (\(-1, 5, 2\)), we get:
  • \(\cos(\alpha) = -\frac{1}{\sqrt{30}}\)
  • \(\cos(\beta) = \frac{5}{\sqrt{30}}\)
  • \(\cos(\gamma) = \frac{2}{\sqrt{30}}\)
These values help you understand how the vector lies in the 3D space, highlighting the angle's inclination toward each axis.
Inverse Cosine Function
The inverse cosine function, also known as \(\cos^{-1}(x)\) or arccos(x), helps find the angle when you know the cosine value. For direction angles of a vector, once you have the direction cosines, you apply the inverse cosine to get the angles \(\alpha\), \(\beta\), and \(\gamma\):

  • \(\alpha = \cos^{-1}\left(-\frac{1}{\sqrt{30}}\right)\)
  • \(\beta = \cos^{-1}\left(\frac{5}{\sqrt{30}}\right)\)
  • \(\gamma = \cos^{-1}\left(\frac{2}{\sqrt{30}}\right)\)
Using these formulas, you calculate the specific angles that the vector makes with each coordinate axis. This is important because it shows you exactly how the vector is oriented in space, offering a complete picture of its direction.

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Most popular questions from this chapter

The spherical coordinates of a point \((x, y, z)\) are unique.Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false.

Work A toy wagon is pulled by exerting a force of 25 pounds on a handle that makes a \(20^{\circ}\) angle with the horizontal (see figure in left column). Find the work done in pulling the wagon 50 feet.

Sketch the solid that has the given description in spherical coordinates. \(0 \leq \theta \leq 2 \pi, 0 \leq \phi \leq \pi / 6,0 \leq \rho \leq a \sec \phi\)

If the projection of \(\mathbf{u}\) onto \(\mathbf{v}\) has the same magnitude as the projection of \(\mathbf{v}\) onto \(\mathbf{u}\), can you conclude that \(\|\mathbf{u}\|=\|\mathbf{v}\|\) ? Explain.

Let \(A, B\), and \(C\) be vertices of a triangle. Find \(\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C A}\).
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