Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

Point to Line Let \(P\) be a point on a line with direction \(\mathbf{n}\) and \(Q\) a point off the line (Figure 7). Show that the distance \(d\) from \(Q\) to the line is given by $$ d=\frac{\|\overrightarrow{P Q} \times \mathbf{n}\|}{\|\mathbf{n}\|} $$ and use this result to find each distance in parts (a) and (b). (a) From \(Q(1,0,-4)\) to the line \(\frac{x-3}{2}=\frac{y+2}{-2}=\frac{z-1}{1}\) (b) From \(Q(2,-1,3)\) to the line \(x=1+2 t, y=-1+3 t\), \(z=-6 t\)

Step-by-Step Solution

Answer

(a) Distance: \( \frac{\sqrt{181}}{3} \). (b) Distance: \( \frac{\sqrt{234}}{7} \).

Step 1: Identify the point on the line and direction vector

For part (a), the line is given in symmetric form \( \frac{x-3}{2}=\frac{y+2}{-2}=\frac{z-1}{1} \). By inspecting the denominators of the fractional expressions, we can identify the direction vector \( \mathbf{n} = (2, -2, 1) \). Any specific point on the line can be given by setting \( t = 0 \) in equivalent parametric form, giving us point \( P = (3, -2, 1) \). For part (b), the line is given in parametric form \( x=1+2t, \; y=-1+3t, \; z=-6t \). The direction vector is clearly \( \mathbf{n} = (2, 3, -6) \) and a point on the line \( P = (1, -1, 0) \) can be chosen by setting \( t = 0 \).

Step 2: Calculate vector from point to line

For part (a), calculate vector \( \overrightarrow{PQ} \) where \( P = (3, -2, 1) \) and \( Q = (1, 0, -4) \). \( \overrightarrow{PQ} = (1-3, 0-(-2), -4-1) = (-2, 2, -5) \). For part (b), calculate vector \( \overrightarrow{PQ} \) where \( P = (1, -1, 0) \) and \( Q = (2, -1, 3) \). \( \overrightarrow{PQ} = (2-1, -1-(-1), 3-0) = (1, 0, 3) \).

Step 3: Compute the Cross Product

Compute the cross product \( \overrightarrow{PQ} \times \mathbf{n} \). For part (a), compute: \[ \overrightarrow{PQ} \times \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -2 & 2 & -5 \ 2 & -2 & 1 \end{vmatrix} \] Giving \( \mathbf{i}(2\times1 - (-2)(-5)) - \mathbf{j}((-2)\times1 - (-5)\times2) + \mathbf{k}((-2)(-2) - (2)(2)) = (-6, -9, -8) \). For part (b), compute: \[ \overrightarrow{PQ} \times \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 0 & 3 \ 2 & 3 & -6 \end{vmatrix} \] Giving \( \mathbf{i}(0\cdot(-6) - 3\cdot3) - \mathbf{j}(1\cdot(-6) - 3\cdot2) + \mathbf{k}(1\cdot3 - 0\cdot2) = (-9, -12, 3) \).

Step 4: Calculate the magnitude of the cross product

For part (a), calculate \( \| \overrightarrow{PQ} \times \mathbf{n} \| = \sqrt{(-6)^2 + (-9)^2 + (-8)^2} = \sqrt{36 + 81 + 64} = \sqrt{181} \). For part (b), calculate \( \| \overrightarrow{PQ} \times \mathbf{n} \| = \sqrt{(-9)^2 + (-12)^2 + 3^2} = \sqrt{81 + 144 + 9} = \sqrt{234} \).

Step 5: Calculate the magnitude of the direction vector

For part (a), \( \| \mathbf{n} \| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \). For part (b), \( \| \mathbf{n} \| = \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \).

Step 6: Calculate the distance from point to the line

Use the formula \( d = \frac{\| \overrightarrow{PQ} \times \mathbf{n} \|}{\| \mathbf{n} \|} \). For part (a), \( d = \frac{\sqrt{181}}{3} \). For part (b), \( d = \frac{\sqrt{234}}{7} \).

Key Concepts

Cross product

The cross product is a mathematical operation that takes two vectors and produces another vector that is perpendicular to both of the original vectors.
This new vector is particularly useful in three-dimensional geometry because it helps identify perpendicular relationships and calculate areas of parallelograms. To find the cross product of two vectors \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \), use the following determinant formula:

• \[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} \]

This results in a vector where each component is calculated by omitting the respective column of the unit vectors during the determinant expansion.
This operation is crucial when finding the distance from a point to a line, as it helps establish the right angled direction from the point to the line.

Direction vector

A direction vector is essential in defining the direction of a line in a space.
For example, in the line equation given by \( \frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c} \), the terms \((a, b, c)\) form the direction vector \( \mathbf{n} \).
This vector helps identify how the line stretches or compresses, acting like a compass that provides directional guidance.

• In parametric equations as seen with lines like \( x = 1 + 2t \), \( y = -1 + 3t \), and \( z = -6t \), the coefficients of \( t \) will form the direction vector \( \mathbf{n} \) as \((2, 3, -6)\).

Knowing the direction vector is crucial for computing distances and projections, as it remains consistent regardless of which points you choose to describe your line.

Vector magnitude

The magnitude of a vector is a measure of its length in space.
For any vector \( \mathbf{v} = (v_1, v_2, v_3) \), its magnitude \( \| \mathbf{v} \| \) is calculated by the formula:

• \[ \| \mathbf{v} \| = \sqrt{v_1^2 + v_2^2 + v_3^2} \]

This calculation provides us with the "size" of the vector, no matter its direction.
It is essential in many geometric problems, including finding distances and understanding vector properties.
Calculating the magnitude of both the cross product and the direction vector allows us to understand their respective influences and interactions in space.
It enables us to use these magnitudes to find the perpendicular (minimum) distance from a point to a line.

Three-dimensional geometry

Three-dimensional geometry involves the study of shapes and figures in a space filled by three dimensions – commonly seen as width, height, and depth.
It uses vectors, points, and planes to analyse various spatial relations.
One primary use is calculating the shortest distance from a point to a line, as seen in this exercise.

• Here, we express lines in either parametric or symmetric form and utilize vectors to describe locations and movements.

Understanding the structure of 3D geometry allows us to apply mathematical models.
It helps solve complex problems by breaking them into smaller, manageable components.
This fundamental understanding establishes the relationships between points and lines, which is useful in many engineering, physics, and computer graphics applications.