Problem 35 - Thomas Calculus

Question

In Exercises 33–38, find the distance from the point to the line. $$ (2,1,3) ; \quad x=2+2 t, \quad y=1+6 t, \quad z=3 $$

Step-by-Step Solution

Answer

The distance from the point to the line is 0.

Step 1: Identify Given Information

We are given a point \((2, 1, 3)\) and a line described by the parametric equations \(x = 2 + 2t\), \(y = 1 + 6t\), and \(z = 3\). We need to find the distance from the point to this line.

Step 2: Determine Direction Vector of the Line

The direction vector of the line can be determined from the coefficients of \(t\) in the line's parametric equations. Here, the direction vector is \(\mathbf{d} = \langle 2, 6, 0 \rangle\).

Step 3: Formulate the Point-to-Line Vector

Find a vector from any point on the line to the given point. Using \(t = 0\) for simplicity, a point on the line is \((2, 1, 3)\). The point-to-line vector \( \mathbf{v} \), from \((2, 1, 3)\) to itself, is \(\langle 2 - 2, 1 - 1, 3 - 3 \rangle = \langle 0, 0, 0 \rangle\).

Step 4: Use the Cross Product to Find the Perpendicular Component

The cross product of the direction vector \( \mathbf{d} = \langle 2, 6, 0 \rangle \) and the vector \( \mathbf{v} = \langle 0, 0, 0 \rangle \) gives \( \mathbf{v} \times \mathbf{d} = \langle 0, 0, 0 \rangle \).

Step 5: Calculate the Magnitude of the Cross Product

The magnitude of the cross product \( \mathbf{v} \times \mathbf{d} \) is \( \| \langle 0, 0, 0 \rangle \| = 0 \).

Step 6: Compute the Magnitude of the Direction Vector

Find \( \| \mathbf{d} \| = \sqrt{2^2 + 6^2 + 0^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10} \).

Step 7: Apply the Distance Formula

The distance \( D \) from the point to the line is given by the formula \( D = \frac{\| \mathbf{v} \times \mathbf{d} \|}{\| \mathbf{d} \|} \). Substitute the known values: \( D = \frac{0}{2\sqrt{10}} = 0 \).

Step 8: Conclusion

Since the calculated distance is zero, the point \((2, 1, 3)\) lies on the line.

Key Concepts

Understanding Parametric Equations

Parametric equations are a powerful way to represent a line in three-dimensional space. They express the coordinates of the points on the line as functions of a parameter, often denoted by \( t \). In this specific problem, we see the equations:

• \( x = 2 + 2t \)
• \( y = 1 + 6t \)
• \( z = 3 \)

Each of these equations defines how the x, y, and z coordinates change as \( t \) varies. By altering the value of \( t \), we can find every point on the line. The constant term in each equation provides the coordinates of a point through which the line passes when \( t = 0 \). Understanding these equations helps identify the line's trajectory in space.

Defining the Direction Vector

The direction vector is crucial as it tells us in which direction the line extends in space. It is derived from the coefficients of the parameter \( t \) in the parametric equations. In this problem:

• From \( x = 2 + 2t \), the x-component is 2.
• From \( y = 1 + 6t \), the y-component is 6.
• The z-component is 0 because the z equation doesn't depend on \( t \).

So, the direction vector \( \mathbf{d} \) is \( \langle 2, 6, 0 \rangle \). This vector shows the direction the line moves as \( t \) increases. Understanding this vector helps you quickly visualize how the line stretches through space.

Utilizing the Cross Product

The cross product is a vital tool in finding the perpendicular distance from a point to a line. For vectors \( \mathbf{a} \) and \( \mathbf{b} \), the cross product \( \mathbf{a} \times \mathbf{b} \) results in a vector that is perpendicular to both. To determine the distance from a point to a line, the cross product helps in finding a vector that is perpendicular to the line. In our exercise:

• We use \( \mathbf{v} = \langle 0, 0, 0 \rangle \) and the direction vector \( \mathbf{d} = \langle 2, 6, 0 \rangle \).
• The cross product results in \( \langle 0, 0, 0 \rangle \), indicating the vectors are actually not forming a triangle area, hence the point lies on the line.

Understanding how the cross product works solidifies the concept of finding perpendicular vectors, crucial for distance calculations.

Exploring Magnitude Calculation

Magnitude is essentially the length of a vector. Calculating the magnitude is necessary not only for normalizing direction vectors but also for understanding how far certain points or objects are from one another. To calculate the magnitude of a vector \( \mathbf{d} = \langle d_1, d_2, d_3 \rangle \), you use the formula:\[ \| \mathbf{d} \| = \sqrt{d_1^2 + d_2^2 + d_3^2} \]In our exercise, we calculate the magnitude of the direction vector \( \mathbf{d} = \langle 2, 6, 0 \rangle \):

• Calculate each square: \( 2^2 = 4, \), \( 6^2 = 36 \).
• Sum and take the square root: \( \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10} \).

This magnitude is essential when applying the distance formula, allowing you to correctly scale the direction vector for comparing lengths in space.