Problem 44 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The distance is \(\frac{3\sqrt{2}}{2}\).

Step 1: Understand the Problem

We need to find the distance from the point \((1, 0, -1)\) to the plane given by the equation \(-4x + y + z = 4\).

Step 2: Recall the Distance Formula from Point to Plane

The formula for the distance \(d\) from a point \((x_0, y_0, z_0)\) to a plane defined by the equation \(Ax + By + Cz + D = 0\) is:\[d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}\]

Step 3: Identify Coefficients and Re-write Plane Equation

The given plane equation is \(-4x + y + z = 4\). Re-write it as:\(-4x + y + z - 4 = 0\).Here, \(A = -4\), \(B = 1\), \(C = 1\) and \(D = -4\).

Step 4: Substitute Values into the Distance Formula

Substitute \((x_0, y_0, z_0) = (1, 0, -1)\) and \(A, B, C, D\) into the distance formula:\[d = \frac{|-4(1) + 1(0) + 1(-1) - 4|}{\sqrt{(-4)^2 + 1^2 + 1^2}}\]

Step 5: Simplify Numerator

Calculate the numerator:\[|-4(1) + 0 - 1 - 4| = |-4 - 1 - 4| = |-9| = 9\]

Step 6: Simplify Denominator

Calculate the denominator:\[\sqrt{(-4)^2 + 1^2 + 1^2} = \sqrt{16 + 1 + 1} = \sqrt{18} = 3\sqrt{2}\]

Step 7: Calculate the Distance

Substitute the simplified numerator and denominator back into the formula:\[d = \frac{9}{3\sqrt{2}} = \frac{3}{\sqrt{2}}\]Rationalize the denominator:\[d = \frac{3\sqrt{2}}{2}\]

Step 8: Write the Final Answer

The distance from the point \((1, 0, -1)\) to the plane \(-4x + y + z = 4\) is \(\frac{3\sqrt{2}}{2}\).

Key Concepts

Distance Formula in Geometry

In geometry, the distance formula is crucial for calculating the space between specific geometric entities, such as points, lines, and planes. The basic distance formula for two points in a Cartesian plane is derived from the Pythagorean theorem. It is expressed as:

• In 2D: The distance between points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
• In 3D: The formula is extended to calculate the distance between \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \). It becomes \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \).

In the context of distance from a point to a plane, this formula helps us understand how to measure the shortest path to a flat surface in three-dimensional space. This version isn't directly used in point-to-plane problems but underpins the reasoning behind such calculations.

Plane Equation in 3D Space

The equation of a plane in three-dimensional space is typically written in the form \( Ax + By + Cz + D = 0 \). Each component plays a role:

• \(A, B, \text{ and } C\) are the coefficients of the normal vector to the plane. This vector is perpendicular to the plane and fundamental in defining its orientation.
• \(D\) is the constant term. It adjusts the position of the plane relative to the origin without affecting its orientation.

To formulate the given plane equation \(-4x + y + z = 4\), we rewrite it as \(-4x + y + z - 4 = 0\). This matches the standard form, making it easier to identify the coefficients for plugging into other formulas. Understanding the plane's equation in this format is crucial for various calculations, including distances and intersections.

Point-to-Plane Distance Calculation

Calculating the distance from a point to a plane involves applying a specialized version of the distance formula. Given a point \( (x_0, y_0, z_0) \) and a plane \( Ax + By + Cz + D = 0 \), the formula for distance \( d \) is:

• \[\ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

This formula calculates the perpendicular distance, ensuring it's the shortest possible distance between the point and the plane. Let's break it down:

• The numerator \(|Ax_0 + By_0 + Cz_0 + D|\) represents the absolute value of the plane equation evaluated at the point.
• The denominator \(\sqrt{A^2 + B^2 + C^2}\) normalizes the direction vector, maintaining consistency in units and scale.

In our problem, substituting the values and simplifying yields the distance from the point \( (1,0,-1) \) to the plane \(-4x + y + z = 4\): \(\frac{3\sqrt{2}}{2}\). The calculated distance maintains key properties of geometry, confirming it's the shortest linear path to the plane.