Problem 70 - University Calculus: Early Transcendentals

Question

Find equations for the line in the plane \(z=3\) that makes an angle of \(\pi / 6\) rad with \(i\) and an angle of \(\pi / 3\) rad with \(j .\) Describe the reasoning behind your answer.

Step-by-Step Solution

Answer

The line's equation is given by: \(x = \sqrt{3}t, y = t, z = 3\) with \(t\) as a parameter.

Step 1: Understand the Line Direction

The line you're looking for should lie in the plane where the z-coordinate is always 3, meaning it has direction vector \(v = (a, b, 0)\). This means the z-component of the direction vector is zero.

Step 2: Use Angle with i-Vector

The line's direction makes an angle of \(\frac{\pi}{6}\) with the vector \(i = (1, 0, 0)\). The cosine of the angle between direction vector \(v\) and \(i\) is given by \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\). Thus, we have:\[\frac{a}{\sqrt{a^{2} + b^{2}}} = \frac{\sqrt{3}}{2}\].Solving for \(a\) using the identity gives you a relationship between \(a\) and \(b\).

Step 3: Use Angle with j-Vector

Similarly, the line makes an angle of \(\frac{\pi}{3}\) with the vector \(j = (0, 1, 0)\). The cosine of this angle is \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\). Thus, we have:\[\frac{b}{\sqrt{a^{2} + b^{2}}} = \frac{1}{2}\].This gives a second relationship between \(a\) and \(b\).

Step 4: Solve System of Equations

From Steps 2 and 3, you have two equations:1. \(\frac{a}{\sqrt{a^{2} + b^{2}}} = \frac{\sqrt{3}}{2}\)2. \(\frac{b}{\sqrt{a^{2} + b^{2}}} = \frac{1}{2}\)By solving these equations, you can find the values of \(a\) and \(b\). These yield \(a = \sqrt{3}\) and \(b = 1\), or vice versa.

Step 5: Write Parametric Equation

Given \(a = \sqrt{3}\) and \(b = 1\), the direction vector of the line is \(\langle \sqrt{3}, 1, 0 \rangle\). Any point on the line can be described using the parametric equations for the line:\[ x = x_0 + \sqrt{3}t, \]\[ y = y_0 + t, \]\[ z = 3. \]Take any initial point \((x_0, y_0, 3)\). For instance, the line can pass through \((0, 0, 3)\).

Step 6: Finalize Line Equation

Using the initial point \((0, 0, 3)\), the parametric equations for the line are:\[ x = \sqrt{3}t, \]\[ y = t, \]\[ z = 3. \]It's important to note that similar results could be obtained by using \((-\sqrt{3}, -1, 0)\) if the orientation is reversed.

Key Concepts

Parametric Equations

Parametric equations provide a way to describe a line in three-dimensional space using a parameter, usually denoted as "t". These equations express each coordinate of the points on the line in terms of this parameter, making it easy to trace out the line's path in space. For example, if a line passes through a point \( (x_0, y_0, z_0) \,\) and has a direction vector \( \langle a, b, c \rangle \,\), the parametric equations for the line can be written as:

• \( x = x_0 + at \)
• \( y = y_0 + bt \)
• \( z = z_0 + ct \)

To find the parametric equations in the given problem, we need to find the direction vector that lies on the plane \(z=3\) and use it to form such equations.

Direction Vectors

A direction vector is essential in determining the orientation of lines and planes in space. It appears in the parametric equation of a line and defines the line's direction. In our example, we're dealing with a line that lies in the \(z=3\) plane. This imposes a restriction—the direction vector cannot have a \(z\)-component other than zero, thus taking the form \( \langle a, b, 0 \rangle \,\). To define the line precisely, the direction vector must also satisfy angle conditions with the standard basis vectors \(i\) and \(j\), as given in the problem:

• The angle with vector \(i = \langle 1, 0, 0 \rangle \) must be \( \frac{\pi}{6} \).
• The angle with vector \(j = \langle 0, 1, 0 \rangle \) must be \( \frac{\pi}{3} \).

Thus, by solving corresponding cosine equations, the direction vector is determined.

Angles in Vector Analysis

Analyzing angles between vectors is an essential part of understanding spatial relationships in geometry and physics. The angle between two vectors can determine collinearity, orthogonality, or any other geometric configuration.The cosine of the angle \(\theta\) between vectors \(\mathbf{u}\) and \(\mathbf{v}\) can be found using the formula:\[\cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\|\|\mathbf{v}\|}\],where \(\mathbf{u} \cdot \mathbf{v}\) is the dot product, and \(\|\mathbf{u}\|\) and \(\|\mathbf{v}\|\) are the magnitudes of the vectors.For the given exercise, the line’s direction vector forms specific angles with basis vectors that are important clues for finding its components. Solving equations using these angle conditions leads us to determine the direction vector that fulfills all requirements.

Plane

A plane in three-dimensional space can be described by an equation of the form \(Ax + By + Cz = D\). All points \( (x, y, z) \) that satisfy this equation lie on the plane.The problem specifies the plane \(z=3\), which is a special case of such a plane where\( C = 1 \) and all \( z \) values are constant at 3.
Thus, directions in this plane can only have variations in \(x\) and \(y\) coordinates.
This constraint was crucial to developing the direction vector \( \langle a, b, 0 \rangle \), ensuring the line remains confined to the plane \(z=3\).
Understanding how a line behaves in relation to a given plane is fundamental in many applications, from physics to computer graphics.

Cosine Angle Identity

The cosine angle identity is a powerful tool in vector analysis, especially when you know angles between vectors but not their explicit form. It helps relate the angle to a direct calculation between vector components and magnitudes.In our scenario, the cosine of the angles between the line's direction vector and the unit vectors \(i\) and \(j\) were crucial.

• For \(\frac{\pi}{6}\) radians with \(i\), \(\cos(\theta) = \frac{\sqrt{3}}{2}\).
• For \(\frac{\pi}{3}\) radians with \(j\), \(\cos(\theta) = \frac{1}{2}\).

These constants helped formulate equations that were solved to find the relationships between the components \(a\) and \(b\) of the direction vector. This approach simplifies initially complex geometric problems into straightforward algebraic solutions.