Problem 66 - Thomas Calculus

Question

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.

Step-by-Step Solution

Answer

The parametric equations are: \(x = \sqrt{3}t\), \(y = t\), \(z = 3\).

Step 1: 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.

Step 2: 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: \(\cos(\pi/3) = \frac{b}{\sqrt{a^2 + b^2 + c^2}}\).

Step 3: 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.

Step 4: 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.

Step 5: 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.

Step 6: 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 \(\mathbf{j}\).

Key Concepts

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.

When solving problems involving direction vectors, it is essential to apply trigonometric relationships and scalar multiples to fit the scenario, as evidenced by setting the direction vector to \(\langle \sqrt{3}k, k, 0 \rangle \) from the given angles with \(\mathbf{i}\) and \(\mathbf{j}\).

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.

Using these dot product properties, direction vectors can be adjusted to find the exact line orientation that meets the conditions of the problem.

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.

The solution's parametric equations \(x = \sqrt{3}t\), \(y = t\), \(z = 3\) embody these concepts, showing how plane geometry limits and defines the line.

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.

Understanding these angles and their trigonometric relationships is crucial for solving complex vector problems and facilitates the equation-forming process for the line within the plane \(z=3\).