Problem 44 - Thomas Calculus

Question

In Exercises 39–44, 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: Review the formula for distance from a point to a plane

The formula to find the distance from a point \((x_1, y_1, z_1)\) to a plane defined by the equation \(Ax + By + Cz + D = 0\) is:\[D = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}\] This formula gives the perpendicular distance from the point to the plane.

Step 2: Identify values from the problem

Your point is \((1, 0, -1)\) and the plane is given by the equation \(-4x + y + z = 4\). Rewriting the plane equation, we have \(A = -4\), \(B = 1\), \(C = 1\), and \(D = -4\) after rearranging to \(-4x + y + z - 4 = 0\).

Step 3: Substitute the values into the distance formula

Insert the values identified:\[A = -4,\ B = 1,\ C = 1,\ D = -4,\ x_1 = 1,\ y_1 = 0,\ z_1 = -1\]Substitute these into the formula:\[D = \frac{|-4(1) + 1(0) + 1(-1) - 4|}{\sqrt{(-4)^2 + 1^2 + 1^2}}\]

Step 4: Simplify the expression

Calculate the numerator:\[|-4 - 1 - 4| = |-9|\]Which simplifies to 9.Calculate the denominator:\[\sqrt{16 + 1 + 1} = \sqrt{18}\]Therefore, the distance \(D\) becomes:\[D = \frac{9}{\sqrt{18}}\]

Step 5: Simplify the final answer

Simplify \(\frac{9}{\sqrt{18}}\). The \(\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}\), hence:\[D = \frac{9}{3\sqrt{2}} = \frac{3}{\sqrt{2}}\]Rationalize the denominator:\[\frac{3}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{3\sqrt{2}}{2}\]

Key Concepts

Distance Formula

The distance formula is a mathematical expression used to find the shortest distance between a point and a plane in space. This formula is crucial in problems involving 3D geometry, where you need to calculate the perpendicular distance from a given point to a plane.

To use the distance formula, you start with a point \(x_1, y_1, z_1\) and a plane defined by the equation \(Ax + By + Cz + D = 0\). The formula is:

• \(D = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}\)

This formula involves:

• The numerator, which calculates the absolute value of the linear combination of point coordinates substituted into the plane equation.
• The denominator, which is the magnitude of the normal vector \(
  A, B, C )\) of the plane.

Substituting these values into the formula gives you the perpendicular or shortest distance from the point to the plane.

Plane Equation

A plane in coordinate geometry is a flat, two-dimensional surface that extends infinitely in all directions. To describe a plane mathematically, we use the plane equation: \(Ax + By + Cz + D = 0\).

This equation is defined by:

• Coefficients \(
  A, B, C )\) which together form the normal vector. This vector is perpendicular to every point on the plane.
• \(D\) is a constant that adjusts the plane’s position relative to the origin.

In exercises involving plane equations, it's often necessary to rewrite the plane equation in standard form. For example, rearranging \(-4x + y + z = 4\) to \(-4x + y + z - 4 = 0\). Doing so allows for easy identification of \(A, B, C,\) and \(D\) for further calculations.

Coordinate Geometry

Coordinate geometry, or analytic geometry, is a mathematical discipline that represents geometric figures using a coordinate system. This field connects algebra and geometry, allowing for the representation and analysis of geometric concepts via algebraic equations.

In 3D coordinate geometry, every point is defined by three coordinates \(x, y, z\). Key concepts in this space include:

• Points, which are locations represented by their coordinates.
• Lines, which can be represented by equations depending on their orientation in space.
• Planes, represented by equations such as \(Ax + By + Cz + D = 0\).
• Vectors, which define direction and magnitude between points and are used for operations such as finding distances.

Understanding coordinate geometry is essential in solving problems involving the distance from a point to a plane as it encompasses the principles of point locations and plane orientations.

Vector Distances

Vector distances refer to distances calculated using vectors, which are mathematical entities with both magnitude and direction. In the context of the distance from a point to a plane, vectors help define the shortest path without moving through angles or curves, directly contributing to the distance formula.

Vectors often represent the direction of a plane or the line perpendicular to a plane (normal vector).

• A normal vector like \(-4, 1, 1\) in our given plane equation is crucial for calculating distances, as it provides the direction perpendicular to the plane.
• The length of this vector, \(\sqrt{A^2 + B^2 + C^2}\), appears in the distance formula's denominator and represents the vector's magnitude.

This magnitude normalizes the distance, ensuring it's accurately scaled relative to the plane's orientation in space. Therefore, understanding vector distances simplifies problems like calculating the closest distance from a point to a plane.