Problem 43 - Thomas Calculus

Question

In Exercises \(39-44,\) find the distance from the point to the plane. $$ (0,-1,0), \quad 2 x+y+2 z=4 $$

Step-by-Step Solution

Answer

The distance from the point to the plane is \(\frac{5}{3}\).

Step 1: Identify the point and the plane equation

We have the point \((0, -1, 0)\) and the plane equation given as \(2x + y + 2z = 4\). Our goal is to find the distance between this point and the plane.

Step 2: Use the distance formula

The distance \(D\) from a point \((x_0, y_0, z_0)\) to a plane \(ax + by + cz = d\) is given by \[D = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}.\] In this case, \(a=2\), \(b=1\), \(c=2\), \((x_0, y_0, z_0) = (0, -1, 0)\), and \(d=4\).

Step 3: Substitute the values into the formula

Substitute the values \(a = 2\), \(b = 1\), \(c = 2\), \(d = 4\), and the coordinates of the point \((x_0, y_0, z_0) = (0, -1, 0)\) into the distance formula: \[D = \frac{|2 \cdot 0 + 1 \cdot (-1) + 2 \cdot 0 - 4|}{\sqrt{2^2 + 1^2 + 2^2}}.\]

Step 4: Calculate the numerator of the formula

Calculate the expression inside the absolute value: \(2 \cdot 0 + 1 \cdot (-1) + 2 \cdot 0 - 4 = -1 - 4 = -5\).Thus, the numerator is \(|-5| = 5\).

Step 5: Calculate the denominator of the formula

Compute the sum of the squares: \(2^2 + 1^2 + 2^2 = 4 + 1 + 4 = 9\).Then, calculate the square root: \(\sqrt{9} = 3\).

Step 6: Compute the distance

Now, divide the absolute value by the square root: \[D = \frac{5}{3}.\]This is the distance from the point to the plane.

Key Concepts

Distance Formula

The distance formula is a crucial tool when determining the shortest distance between a point and a plane. It's a specific instance of a more generalized understanding of distances in geometry.

For a point \(x_0, y_0, z_0\) and a plane characterized by the equation \(ax + by + cz = d\), the distance \(D\) is computed as:

• First, calculate the dot product of the plane's normal vector, \((a, b, c)\), and the point's coordinates.
• Adjust by subtracting the plane's constant, \(d\), resulting in \((ax_0 + by_0 + cz_0 - d)\).
• Take the absolute value of this result to ensure a positive distance.
• Divide by the magnitude \(\sqrt{a^2 + b^2 + c^2}\) of the plane's normal vector.

By following these steps, we find the shortest distance from the specified point to the plane.

Plane Equation

The plane equation, given in the format \(ax + by + cz = d\), represents a flat, two-dimensional surface in three-dimensional space. Each coefficient (

• \(a\), \(b\), and \(c\)) signifies the directions normal to the plane.
• \(d\) is the constant that determines the plane's position in relation to the origin.

A vector normal to the plane provides significant insight into its orientation, which is crucial for many calculations, including distance problems.

Absolute Value

Absolute value is a mathematical operation that measures the magnitude of a real number without regard to its sign. Simply put, it transforms negative numbers into positive ones, while positive numbers remain unchanged:

• For example, \(|-5| = 5\).
• The absolute value ensures the distance is always a positive measurement, signifying the shortest path without any directional biases.

This operation is essential in distance formulas to maintain the integrity of the calculation.

Square Root

Taking the square root is an integral part of distance calculations. It involves finding a number that, when multiplied by itself, gives the original value. In our context:

• Upon computing the sum of squared coefficients, such as \(2^2 + 1^2 + 2^2 = 9\), we determine its square root.
• Here, \(\sqrt{9} = 3\), which serves as the divisor in our formula.

This process aids in standardizing distance by relating the value back to the plane's normal vector's magnitude.

Numerator and Denominator Computation

In the distance formula, computations for the numerator and denominator are pivotal. This breakdown simplifies the task into manageable steps:

• The numerator involves substituting the point's coordinates into the plane equation and evaluating the result, then adjusting using the absolute value.
• The numerator comes from \(2 \times 0 + 1 \times (-1) + 2 \times 0 - 4 = -5\), simplified to \(|-5| = 5\).
• For the denominator, compute the sum of squares of the coefficients and take the square root.
• Here, \((2^2 + 1^2 + 2^2)\) results in \(\sqrt{9} = 3\).

Putting all together, the distance becomes \(\frac{5}{3}\). Each component's precise calculation confirms the shortest distance.