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Chapter 12: Problem 37

Find the direction cosines and direction angles of the vector. (Give the direction angles correct to the nearest degree.) $$ \langle c, c, c\rangle, \quad \text { where } c>0 $$

Short Answer

Expert verified
Direction cosines are \( \frac{1}{\sqrt{3}} \) and direction angles are approximately 55掳.

Step by step solution

01

Understand the Vector and Magnitude

The vector given is \( \langle c, c, c \rangle \). To find the direction cosines and angles, we first need the magnitude of this vector. The magnitude \( |\mathbf{v}| \) is given by \( \sqrt{c^2 + c^2 + c^2} = \sqrt{3c^2} = c\sqrt{3} \). Understanding the magnitude is crucial for finding the direction cosines.
02

Find the Direction Cosines

The direction cosines are derived from dividing each component of the vector by the magnitude. For \( \mathbf{v} = \langle c, c, c \rangle \), each direction cosine is \( \frac{c}{c\sqrt{3}} = \frac{1}{\sqrt{3}} \). Thus, the direction cosines are \( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \).
03

Calculate the Direction Angles

The direction angles \( \alpha, \beta, \gamma \) can be found using \( \cos(\alpha)=\cos(\beta)=\cos(\gamma)=\frac{1}{\sqrt{3}} \). Therefore, \( \alpha = \beta = \gamma = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \). Calculating this gives a common angle: \( \alpha = \beta = \gamma \approx 55^{\circ} \) when rounded to the nearest degree.

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Key Concepts

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

Vector Magnitude
When dealing with vectors, understanding the magnitude is the foundation. It tells you how long a vector is in space.
The magnitude of a vector \( \mathbf{v} = \langle a, b, c \rangle \) is calculated using the formula:\[ |\mathbf{v}| = \sqrt{a^2 + b^2 + c^2} \]In our specific problem, the vector is \( \langle c, c, c \rangle \). By substituting the components into the magnitude formula, we find:\[ |\mathbf{v}| = \sqrt{c^2 + c^2 + c^2} = \sqrt{3c^2} = c\sqrt{3} \]This tells us that no matter the value of \( c \), the vector's magnitude is proportional to \( c\sqrt{3} \).
This measurement is crucial because it helps us understand the direction of the vector, which leads us to the direction cosines.
Direction Angles
Direction angles are angles that help to describe the orientation of a vector in relation to the coordinate axes.
They are closely related to the vector's direction cosines. To find the direction angles \( \alpha, \beta, \gamma \), you use the cosines of these angles.The vector \( \langle c, c, c \rangle \) has components that can be aligned with the coordinate axes. Thus, the direction angles are derived from the direction cosines:\[ \cos(\alpha) = \cos(\beta) = \cos(\gamma) = \frac{1}{\sqrt{3}} \]To find the angles themselves, we compute these using the inverse cosine function:\[ \alpha = \beta = \gamma = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \]After calculation, each angle comes out to approximately \( 55^{\circ} \), providing a clear understanding of how the vector spreads out equally among all three axes.
Cosine Inverse
The cosine inverse function, often denoted as \( \cos^{-1} \), is an essential part of trigonometry.
It allows us to find the angle associated with a given cosine value.
In this problem, each direction cosine is \( \frac{1}{\sqrt{3}} \).
The task is to find the angle whose cosine is \( \frac{1}{\sqrt{3}} \). By applying the cosine inverse function:\[ \alpha = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \]This computation tells us that the angle \( \alpha \) is approximately \( 55^{\circ} \) when rounded to the nearest degree.
This process helps us convert between trigonometric functions and angle measures, revealing the geometric interpretation of the vector's direction.

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Most popular questions from this chapter

Find an equation of the plane. The plane through the point \((3,-2,8)\) and parallel to the plane \(z=x+y\)

If \(\mathbf{v}_{1}, \mathbf{v}_{2},\) and \(\mathbf{v}_{3}\) are noncoplanar vectors, let \(\mathbf{k}_{1}=\frac{\mathbf{v}_{2} \times \mathbf{v}_{3}}{\mathbf{v}_{1} \cdot\left(\mathbf{v}_{2} \times \mathbf{v}_{3}\right)} \quad \mathbf{k}_{2}=\frac{\mathbf{v}_{3} \times \mathbf{v}_{1}}{\mathbf{v}_{1} \cdot\left(\mathbf{v}_{2} \times \mathbf{v}_{3}\right)}\) \(\mathbf{k}_{3}=\frac{\mathbf{v}_{1} \times \mathbf{v}_{2}}{\mathbf{v}_{1} \cdot\left(\mathbf{v}_{2} \times \mathbf{v}_{3}\right)}\) (These vectors occur in the study of crystallography. Vectors of the form \(n_{1} \mathbf{v}_{1}+n_{2} \mathbf{v}_{2}+n_{3} \mathbf{v}_{3},\) where each \(n_{i}\) is an integer, form a lattice for a crystal. Vectors written similarly in terms \(\mathbf{k}_{1}, \mathbf{k}_{2},\) and \(\mathbf{k}_{3}\) form the reciprocal lattice.) (a) Show that \(\mathbf{k}_{i}\) is perpendicular to \(\mathbf{v}_{j}\) if \(i \neq j\) (b) Show that \(\mathbf{k}_{i} \cdot \mathbf{v}_{i}=1\) for \(i=1,2,3\) (c) Show that \(\mathbf{k}_{1} \cdot\left(\mathbf{k}_{2} \times \mathbf{k}_{3}\right)=\frac{1}{\mathbf{v}_{1} \cdot\left(\mathbf{v}_{2} \times \mathbf{v}_{3}\right)}\)

Graph the surfaces \(z=x^{2}+y^{2}\) and \(z=1-y^{2}\) on a common screen using the domain \(|x| \leqslant 1.2,|y| \leqslant 1.2\) and observe the curve of intersection of these surfaces. Show that the projection of this curve onto the \(x y\) -plane is an ellipse.

A cooling tower for a nuclear reactor is to be constructed in the shape of a hyperboloid of one sheet (see the photo on page 839 ). The diameter at the base is \(280 \mathrm{m}\) and the minimum diameter, \(500 \mathrm{m}\) above the base, is \(200 \mathrm{m}\). Find an equation for the tower.

(a) Find all vectors \(\mathbf{v}\) such that \(\langle 1,2,1\rangle \times \mathbf{v}=\langle 3,1,-5\rangle\) (b) Explain why there is no vector \(\mathbf{v}\) such that \(\langle 1,2,1\rangle \times \mathbf{v}=\langle 3,1,5\rangle\)
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