Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

Line to Line Let \(P\) and \(Q\) be points on nonintersecting skew lines with directions \(\mathbf{n}_{1}\) and \(\mathbf{n}_{2}\), and let \(\mathbf{n}=\mathbf{n}_{1} \times \mathbf{n}_{2}\) (Figure 8). Show that the distance \(d\) between these lines is given by $$ d=\frac{|\overrightarrow{P Q} \cdot \mathbf{n}|}{\| \mathbf{n} \mid} $$ and use this result to find the distance between each pair of lines in parts (a) and (b). (a) \(\frac{x-3}{1}=\frac{y+2}{1}=\frac{z-1}{2}\) and \(\frac{x+4}{3}=\frac{y+5}{4}=\frac{z}{5}\) (b) \(x=1+2 t, y=-2+3 t, z=-4 t\) and \(x=3 t, y=1+t\), \(z=-5 t\)

Step-by-Step Solution

Answer

(a) Distance is \(\frac{23}{\sqrt{11}}\); (b) Distance is \(\frac{77}{\sqrt{594}}\).

Step 1: Direction Vectors

For each line, extract the direction vectors \(\mathbf{n}_1\) and \(\mathbf{n}_2\). From part (a), line 1 has direction vector \(\mathbf{n}_1 = [1, 1, 2]\) and line 2 has direction vector \(\mathbf{n}_2 = [3, 4, 5]\). In part (b), line 1 has direction vector \(\mathbf{n}_1 = [2, 3, -4]\) and line 2 has \(\mathbf{n}_2 = [3, 1, -5]\).

Step 2: Compute Cross Product

Compute the cross product \(\mathbf{n} = \mathbf{n}_1 \times \mathbf{n}_2\). For part (a), \(\mathbf{n} = [1, 1, 2] \times [3, 4, 5] = [-3, 1, 1]\). For part (b), \(\mathbf{n} = [2, 3, -4] \times [3, 1, -5] = [-11, 22, -7]\).

Step 3: Select Points P and Q

Identify points \(P\) and \(Q\) on each line. From part (a), pick \(P = [3, -2, 1]\) from line 1 and \(Q = [-4, -5, 0]\) from line 2. From part (b), choose \(P = [1, -2, 0]\) from line 1 and \(Q = [0, 1, 0]\) from line 2.

Step 4: Vector PQ

Find the vector from point \(P\) to point \(Q\), denoted by \(\overrightarrow{PQ} = Q - P\). For part (a), \(\overrightarrow{PQ} = [-4-3, -5+2, 0-1] = [-7, -3, -1]\). For part (b), \(\overrightarrow{PQ} = [0-1, 1+2, 0-0] = [-1, 3, 0]\).

Step 5: Dot Product PQ·n

Calculate the dot product \(\overrightarrow{PQ} \cdot \mathbf{n}\). For part (a), the dot product is \([-7, -3, -1] \cdot [-3, 1, 1] = 23\). For part (b), the dot product is \([-1, 3, 0] \cdot [-11, 22, -7] = 77\).

Step 6: Magnitude of Cross Product

Compute the magnitude of the cross product \(\|\mathbf{n}\|\). For part (a), \(\|\mathbf{n}\| = \sqrt{(-3)^2 + 1^2 + 1^2} = \sqrt{11}\). For part (b), \(\|\mathbf{n}\| = \sqrt{(-11)^2 + 22^2 + (-7)^2} = \sqrt{594}\).

Step 7: Calculate Distance

Use the formula \(d = \frac{|\overrightarrow{PQ} \cdot \mathbf{n}|}{\| \mathbf{n} \|}\) to find the distance between the lines. For part (a), \(d = \frac{|23|}{\sqrt{11}} = \frac{23}{\sqrt{11}}\). For part (b), \(d = \frac{|77|}{\sqrt{594}}\). Simplify both expressions if possible.

Key Concepts

Direction Vectors

When dealing with lines in three-dimensional space, direction vectors play an essential role. A direction vector essentially describes the orientation of a line. It tells you in which direction the line is moving from a certain point. For instance, look at part (a) given in the textbook example: line 1 has direction vector \(\mathbf{n}_1 = [1, 1, 2]\) and line 2 has \(\mathbf{n}_2 = [3, 4, 5]\). These vectors indicate the path you'd follow to remain on these lines starting from any point on them.

• Direction vectors are typically derived from the parametric equations of the lines.
• You can think of them as arrows that give you precise navigation across the space on their respective lines.

Grasping direction vectors is a primary step when calculating distance between skew (non-intersecting) lines, as they serve as fundamental components in further calculations, such as cross products.

Cross Product

The cross product of two direction vectors is a technique used to find a vector that is perpendicular to both. This perpendicular vector is especially useful in determining the distance between skew lines.In the original exercise, the cross product \(\mathbf{n} = \mathbf{n}_1 \times \mathbf{n}_2\) is computed for each pair of lines. For example, in part (a), if you take the vectors \([1, 1, 2]\) and \([3, 4, 5]\), their cross product is \([-3, 1, 1]\).

• The cross product is calculated using the determinant of a matrix formed by the standard unit vectors \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) and the given two vectors.
• This operation helps when determining the area across vectors as well as finding orthogonal vectors.

This resulting vector \(\mathbf{n}\) simplifies many complex spatial calculations, playing a key role in finding the shortest distance between the skew lines.

Dot Product

The dot product is a crucial concept, especially when it comes to measuring angles between vectors or projecting one vector onto another. It is calculated by multiplying corresponding components of two vectors and summing them up.In the exercise, the dot product \(\overrightarrow{PQ} \cdot \mathbf{n}\) involves calculating the dot product between the vector that connects points \(P\) and \(Q\) and the cross product of the direction vectors. In part (a), this calculation yields a result of \(23\) from \([-7, -3, -1] \cdot [-3, 1, 1]\).

• The dot product is a scalar, which means it results in a single number rather than another vector.
• Its absolute value, when divided by the magnitude of another vector, provides a key component in some distance measurements.

Understanding how the dot product works is essential for effectively solving problems involving projection and applying that knowledge to find distances in vector spaces.

Magnitude of a Vector

The magnitude of a vector provides a measure of its 'length' or 'size'. It's a single number that tells us how far the vector projects in space from its origin.To find the magnitude of a vector \(\mathbf{n}\) that resulted from a cross product, you use the formula \(\|\mathbf{n}\| = \sqrt{n_x^2 + n_y^2 + n_z^2}\). For instance, in part (a) of the exercise, the vector \([-3, 1, 1]\) has a magnitude of \(\sqrt{11}\).

• You obtain this by squaring each component of the vector, summing them up, and taking the square root of the result.
• Magnitude helps us normalize vectors or compare their lengths regardless of direction.

In problems involving distances like the distance between skew lines, the magnitude of the cross product of direction vectors directly influences the denominator in the distance formula. Understanding magnitude is critical in vector mathematics as it adds depth to spatial perceptions and aids in dimensional analysis.