Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 12: Problem 66
Find equations for the line in the plane \(z=3\) that makes an angle of \(\pi / 6\) rad with \(\mathbf{i}\) and an angle of \(\pi / 3\) rad with \(\mathbf{j}\) . Describe the reasoning behind your answer.
Short Answer
Expert verified
The parametric equations are: \(x = \sqrt{3}t\), \(y = t\), \(z = 3\).
Step by step solution
01
Understanding the Problem
The problem involves finding equations for a line in the plane defined by the equation \(z=3\) that makes specific angles with the \(x\)-axis (\(\mathbf{i}\)) and \(y\)-axis (\(\mathbf{j}\)). The angles with \(\mathbf{i}\) and \(\mathbf{j}\) are \(\pi/6\) and \(\pi/3\) radians, respectively.
02
Determining the Direction Vector
The line's direction vector \(\mathbf{d} = \langle a, b, c \rangle\) should satisfy certain dot product relationships. For angle \(\theta = \pi/6\) with \(\mathbf{i}\), use \(\cos(\theta) = \frac{\mathbf{d} \cdot \mathbf{i}}{\|\mathbf{d}\|}\), giving the equation: \(\cos(\pi/6) = \frac{a}{\sqrt{a^2 + b^2 + c^2}}\). Similarly, for angle \(\theta = \pi/3\) with \(\mathbf{j}\), use: \(\cos(\pi/3) = \frac{b}{\sqrt{a^2 + b^2 + c^2}}\).
03
Solving for Direction Vector Components
Solve the equations from the previous step: \(\cos(\pi/6) = \frac{a}{\sqrt{a^2 + b^2 + c^2}} = \frac{\sqrt{3}}{2}\) and \(\cos(\pi/3) = \frac{b}{\sqrt{a^2 + b^2 + c^2}} = \frac{1}{2}\). Let \(c = 0\) since the line lies entirely in the plane \(z=3\). Set system of equations: \(a = \sqrt{3}k\) and \(b = k\), \(c = 0\), where \(k\) is any scalar.
04
Utilizing Plane Condition
Since the line is in the plane \(z=3\), any point \((x, y, z)\) on the line has \(z = 3\). Thus, parameterize \(z\) to be constant: \(z = 3 + 0t = 3\), where \(t\) represents the parameter.
05
Equation Formation
Given the direction vector components \(\langle \sqrt{3}k, k, 0 \rangle\), the parametric equations for the line can be expressed as: \(x = \sqrt{3}kt + x_0\), \(y = kt + y_0\), \(z = 3\). Pick any point \((x_0, y_0, 3)\) as the line's starting point, commonly \((0, 0, 3)\) is used for simplicity.
06
Finalizing Parametric Equations
Selecting \(k = 1\) for simplicity, the parametric equations of the line are: \(x = \sqrt{3}t\), \(y = t\), \(z = 3\). These equations satisfy the condition of being in the \(z=3\) plane and make the given angles with \(\mathbf{i}\) and \(\mathbf{j}\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Direction Vectors
In the realm of geometry and vector mathematics, a direction vector is a crucial concept. It provides a way to represent the direction of a line in space. It is a vector that points in the direction the line is going. A direction vector for a line can be written as \( \mathbf{d} = \langle a, b, c \rangle \). Here, \(a\), \(b\), and \(c\) are the components that describe the line's direction in 3D space.
- For this particular line on the plane \(z = 3\), only the components \(a\) and \(b\) are relevant since \(c = 0\).
- This implies no movement perpendicular to the plane, maintaining the line strictly in the \(z=3\) level.
Dot Product
The dot product, or scalar product, is a fundamental operation in vector algebra. It helps us determine the angle between two vectors in space. For vectors \(\mathbf{u}\) and \(\mathbf{v}\), the dot product is given by \(\mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\| \|\mathbf{v}\| \cos(\theta)\), where \(\theta\) is the angle between them.
- In the exercise, the direction vector's dot product with \(\mathbf{i}\) and \(\mathbf{j}\) is used to find its components.
- The relationships derived, \(\cos(\pi/6) = \frac{a}{\| \mathbf{d} \|}\) and \(\cos(\pi/3) = \frac{b}{\| \mathbf{d} \|}\), help define the direction vector to satisfy given angles.
Plane Geometry
Plane geometry concerns with flat surfaces and shapes that lie flat in a plane. In this exercise, the plane is constant at \(z = 3\). This implies that any activity along the given line does not alter its height. The mathematical expression of this phenomenon is \(z = 3 + 0t = 3\), illustrating that the plane's height is unaffected by the parameter \(t\).
- Beginning by understanding the plane z=3 helps to constrain the movement and set the conditions under which the line operates.
- Lines within such a plane can be described with parametric equations that respect the invariability of the \(z\) coordinate.
Trigonometric Angles
Trigonometry plays a significant role in understanding and solving problems involving angles between vectors. Angles inform both direction and magnitude relationships.
- In this exercise, the angles \(\frac{\pi}{6}\) and \(\frac{\pi}{3}\) with \(\mathbf{i}\) and \(\mathbf{j}\), respectively, dictate the orientation of the direction vector.
- Using known cosine values at these angles, \(\cos(\pi/6) = \frac{\sqrt{3}}{2}\) and \(\cos(\pi/3) = \frac{1}{2}\), determines how to adjust the components of the direction vector.
