Problem 49 - University Calculus: Early Transcendentals

Question

Find the acute angles between the intersecting lines. $$x=t, y=2 t, z=-t \quad \text { and } \quad x=1-t, y=5+t, z=2 t$$

Step-by-Step Solution

Answer

The acute angle between the lines is approximately 80.41 degrees.

Step 1: Identify the Direction Vectors

To find the acute angle between two lines in 3D, we need their direction vectors. From the lines' parametric equations: The first line has a direction vector \( \mathbf{d_1} = (1, 2, -1) \). The second line has a direction vector \( \mathbf{d_2} = (-1, 1, 2) \).

Step 2: Compute the Dot Product between Direction Vectors

The dot product of two vectors \( \mathbf{d_1} \) and \( \mathbf{d_2} \) is calculated as: \[ \mathbf{d_1} \cdot \mathbf{d_2} = 1(-1) + 2(1) + (-1)(2) = -1 + 2 - 2 = -1 \]

Step 3: Find the Magnitudes of the Vectors

Calculate the magnitudes of \( \mathbf{d_1} \) and \( \mathbf{d_2} \):\[ \| \mathbf{d_1} \| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \] \[ \| \mathbf{d_2} \| = \sqrt{(-1)^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \]

Step 4: Apply Dot Product to Find Cosine of the Angle

Using the dot product formula, \( \cos(\theta) \) is found using:\[ \cos(\theta) = \frac{\mathbf{d_1} \cdot \mathbf{d_2}}{\| \mathbf{d_1} \| \| \mathbf{d_2} \|} = \frac{-1}{\sqrt{6} \cdot \sqrt{6}} = \frac{-1}{6} \]

Step 5: Compute the Angle From the Cosine Value

Calculate the angle \( \theta \) using the inverse cosine function:\[ \theta = \cos^{-1}\left(\frac{-1}{6}\right) \] Since the angle needed is acute, you continue with the positive angle, approximately \( \theta \approx 99.59^{\circ} \), but for the acute angle, use \( \theta = 180^{\circ} - 99.59^{\circ} \approx 80.41^{\circ} \).

Key Concepts

Direction Vectors

Direction vectors provide important information about lines in 3D space. They indicate the direction in which a line moves. From the parametric equations, each coordinate (x, y, z) represents a function of the parameter \( t \). By examining these equations, we can extract the direction vectors of the lines.
For instance, in the exercise provided, the direction vector for the first line \( \mathbf{d_1} \) is derived directly from its coefficients: \( (1, 2, -1) \). Meanwhile, the second line produces the direction vector \( \mathbf{d_2} = (-1, 1, 2) \), also from its coefficients.

• Direction vectors are crucial for determining angles between lines.
• They simplify the 3D relationships into manageable mathematical components.
• Extracting these vectors is the initial step in many 3D geometry problems.

By understanding the "direction" indicated by these vectors, we gain the ability to analyze the spatial relationship between lines.

Dot Product

The dot product is a powerful tool in vector calculations. It helps to find the angle between vectors, determine vector similarity, and more.
The mathematical formula for the dot product of two vectors \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \) is:
\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]
In our exercise, the dot product was calculated as:
\( \mathbf{d_1} \cdot \mathbf{d_2} = 1(-1) + 2(1) + (-1)(2) = -1 \). This value plays a key role in the determination of the angle between the two lines.

• The dot product helps us find whether the angle is acute by indicating its sign (positive or negative).
• If the dot product is negative, the angle between the vectors is obtuse.
• Conversely, if it is positive, the angle is acute.

Utilizing the dot product simplifies complex spatial problems, making it an essential concept in 3D geometry.

3D Parametric Equations

3D parametric equations describe lines in three-dimensional space by expressing the x, y, and z coordinates as functions of a common parameter, often denoted as \( t \).
In these parametric equations, the coefficients of \( t \) reveal the directions of lines, while constants adjust their positions in space.

• Each equation is essentially a function of \( t \), mapping values of \( t \) to points in 3D space.
• These equations are instrumental in defining both straight lines and curves in 3D geometry.
• They ease the process of visualizing and calculating the properties of geometric figures.

Understanding 3D parametric equations allows us to analyze line intersections, compute distances, and examine angle relationships, making them a cornerstone of spatial mathematical studies.