Problem 37 - Thomas Calculus

Question

In Exercises \(33-38,\) find the distance from the point to the line. $$ (3,-1,4) ; \quad x=4-t, \quad y=3+2 t, \quad z=-5+3 t $$

Step-by-Step Solution

Answer

The distance is \( \frac{9 \sqrt{42}}{7} \).

Step 1: Identify the components of the line and point

The given line is represented parametrically as \( x = 4 - t, \; y = 3 + 2t, \; z = -5 + 3t \). We are tasked with finding the distance from the point \((3, -1, 4)\) to this line.

Step 2: Find the direction vector of the line

The direction vector \( \mathbf{d} \) of the line corresponds to the coefficients of \(t\) in the parametric equations. Hence, \( \mathbf{d} = (-1, 2, 3) \).

Step 3: Find a point on the line

A point on the line can be found by setting \( t = 0 \). When \( t = 0 \), the point on the line is \((x, y, z) = (4, 3, -5)\).

Step 4: Calculate the vector from the point on the line to the given point

This vector \( \mathbf{p} \) is \( (3, -1, 4) - (4, 3, -5) = (-1, -4, 9) \).

Step 5: Use the formula for the distance from a point to a line in space

The distance \( D \) from a point to a line is given by the formula:\[ D = \frac{\| \mathbf{p} \times \mathbf{d} \|}{\| \mathbf{d} \|} \]Calculate \( \mathbf{p} \times \mathbf{d} \), where \( \mathbf{p} = (-1, -4, 9) \) and \( \mathbf{d} = (-1, 2, 3) \).

Step 6: Compute the cross product \( \mathbf{p} \times \mathbf{d} \)

Find the cross product:\( \mathbf{p} \times \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -1 & -4 & 9 \ -1 & 2 & 3 \end{vmatrix} = \mathbf{i} \, ((-4)(3) - (9)(2)) - \mathbf{j} \, ((-1)(3) - (9)(-1)) + \mathbf{k} \, ((-1)(2) - (-4)(-1)) \)Simplifying gives: \( ( -12 - 18 )\mathbf{i} - ( -3 + 9 )\mathbf{j} + ( -2 - 4 )\mathbf{k} = -30\mathbf{i} + 6\mathbf{j} - 6\mathbf{k} \).

Step 7: Find the magnitudes \( \| \mathbf{p} \times \mathbf{d} \| \) and \( \| \mathbf{d} \| \)

Calculate the magnitude \( \| \mathbf{p} \times \mathbf{d} \| = \sqrt{(-30)^2 + 6^2 + (-6)^2} = \sqrt{900 + 36 + 36} = \sqrt{972} = \sqrt{4 \times 243} = 6\sqrt{27} = 6\sqrt{3 \times 9} = 18\sqrt{3} \).\(\| \mathbf{d} \| = \sqrt{(-1)^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \).

Step 8: Calculate the distance

Substitute the magnitudes into the distance formula:\[ D = \frac{18 \sqrt{3}}{\sqrt{14}} \approx \frac{18}{\sqrt{14/3}} \]Simplifying gives:\[ D = \frac{18 \sqrt{3}}{\sqrt{14}} = \frac{18 \sqrt{3} \cdot \sqrt{14}}{14} = \frac{18 \sqrt{42}}{14} = \frac{9 \sqrt{42}}{7} \].

Step 9: Provide the final distance value

The distance from the point \((3, -1, 4)\) to the given line is \( \frac{9 \sqrt{42}}{7} \).

Key Concepts

Vector Cross Product

The vector cross product is an important tool in calculating angles and distances in three-dimensional space. It allows us to derive a vector that is orthogonal (perpendicular) to two given vectors.
Imagine two vectors represented in the same three-dimensional space. The cross product results in a new vector that perfectly aligns perpendicularly to the plane formed by the original vectors.
In our exercise, we use the cross product to find the perpendicular distance of a point from a line. We had two vectors: vector \( \mathbf{p} \) from point to line, and the direction vector of the line \( \mathbf{d} \). The cross product \( \mathbf{p} \times \mathbf{d} \) helped us determine a vector that is perpendicular to both \( \mathbf{p} \) and \( \mathbf{d} \).

For computing the vector cross product:

• Use the determinant of a 3x3 matrix, with the top row having unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \). These represent directions in the 3D space.
• The second and third rows contain the components of the two vectors.

The calculated cross product then allows us to measure the magnitude necessary for the distance calculation.

Parametric Equations

Parametric equations are incredibly useful when representing lines and curves in both two and three-dimensional spaces. They express a set of related quantities as explicit functions of a common variable, typically time \( t \).
These equations provide a way to calculate coordinates \( (x, y, z) \) for every value of \( t \). In our problem, they define the position along the line as \( t \) changes.

Through parametric equations:

• We can quickly find points on the line by plugging in different values for \( t \).
• This is especially helpful when determining a starting point or direction vector for calculations.

For instance, setting \( t = 0 \) gives us a point on the line: \( (4, 3, -5) \). This step is crucial when determining the distance from a separate point to the line defined by these equations.

Distance Formula in Space

The distance formula in space extends the familiar distance formula from the two-dimensional plane into three dimensions. It calculates the shortest path or line between a point and a line or plane.
In the context of calculating distance from a point to a line, understanding the vector relationships is critical.

The specific formula used is:\[D = \frac{\| \mathbf{p} \times \mathbf{d} \|}{\| \mathbf{d} \|}\]This formula incorporates:

• \( \| \mathbf{p} \times \mathbf{d} \| \): The magnitude of the cross product of the vector from the line to the point and the direction vector of the line.
• \( \| \mathbf{d} \| \): The magnitude of the direction vector.

The resulting value \( D \) is the perpendicular distance from the point to the line. This method provides an efficient means of finding distances in three-dimensional space using vectors.

Magnitude of a Vector

The magnitude of a vector provides a measure of its length in space, just like distance in everyday life. For any vector \( \mathbf{v} = (a, b, c) \), the magnitude \( \| \mathbf{v} \| \) is calculated as:
\[\| \mathbf{v} \| = \sqrt{a^2 + b^2 + c^2}\]
Calculating the magnitude is vital in normalization and understanding vector properties, which is crucial in our particular problem solving.

The magnitude was significant in two parts of the exercise:

• Measuring the length of the cross product vector, \( \| \mathbf{p} \times \mathbf{d} \| \).
• Determining the length of the direction vector \( \| \mathbf{d} \| \).

By calculating these magnitudes, we obtain essential components for producing final distance calculations, highlighting how vector properties underpin more complex geometric reasoning.