Problem 74 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the distance from \((2,6,3)\) to the plane \(-3 x+2 y+z=9\).

Step-by-Step Solution

Answer

The distance from (2,6,3) to the plane is 0.

Step 1: Understand the Problem

We need to find the perpendicular distance from the point \((2, 6, 3)\) to the plane given by the equation \(-3x + 2y + z = 9\). This is done using the formula for distance from a point to a plane.

Step 2: Use the Distance Formula

The distance \(d\) from a point \((x_1, y_1, z_1)\) to a plane \(Ax + By + Cz + D = 0\) is given by the formula:\[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]For the given plane \(-3x + 2y + z = 9\), rearrange to \(-3x + 2y + z - 9 = 0\), where \(A = -3\), \(B = 2\), \(C = 1\), and \(D = -9\).

Step 3: Substitute the Point into the Equation

Substitute the point \((2, 6, 3)\) into the plane equation which gives: \[-3(2) + 2(6) + 1(3) - 9\].

Step 4: Calculate the Numerator

Simplify the expression from the previous step:\(-3(2) + 2(6) + 1(3) - 9 = -6 + 12 + 3 - 9\) = \(0\).

Step 5: Compute the Denominator

Compute the denominator using \(A^2 + B^2 + C^2\):\[\sqrt{(-3)^2 + (2)^2 + (1)^2} = \sqrt{9 + 4 + 1} = \sqrt{14}\]

Step 6: Calculate the Distance

Since the numerator is 0, the distance is:\[ d = \frac{0}{\sqrt{14}} = 0 \]

Step 7: Interpret the Result

A distance of 0 indicates that the point is directly on the plane.

Key Concepts

Perpendicular Distance

In geometry, when we talk about the perpendicular distance from a point to a plane, we are referring to the shortest distance between that point and the plane. This distance is measured along a perpendicular line from the point to the plane.
Understanding the concept of perpendicular distance is essential, especially when dealing with planes in 3D space. This is because, in three-dimensional geometry, many problems involve measuring distances where a direct line isn't possible, or where multiple paths could be taken from the point to the plane. By considering the perpendicular distance, we ensure that we are always measuring the shortest possible path.

• This concept is critical in various fields, such as physics, engineering, and computer graphics, as it helps in optimizing parts placement and aligning elements in 3D space.

When applied to our example, the perpendicular distance from the point (2, 6, 3) to the plane (-3x + 2y + z = 9) is found by essentially "dropping" a vertical line from our point straight to the plane, ensuring that the line strikes the plane at a right angle. Understanding and calculating this perpendicular distance can sometimes even reveal that the point lies on the plane, resulting in a distance of zero, as in our solved exercise.

Distance Formula

The distance formula used to calculate the perpendicular distance from a point to a plane in a 3D space is of great utility. This formula derives from the principle of vector projections and involves crucial components of the plane’s equation.
The formula is expressed as:\[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]Where

• (A, B, C) are the coefficients of the plane's equation.
• (x1, y1, z1) are the coordinates of the given point.
• D is the constant in the rearranged plane equation (Ax + By + Cz + D = 0).

In the context of our example, first, ensure the plane equation is set as \(-3x + 2y + z - 9 = 0.\)Substitute the point (2, 6, 3) into the formula. As shown in the original solution, the numerators equal zero, confirming that the point lies exactly on the plane.
This formula not only gives you a tool to calculate distance but also provides insight into spatial relationships, showing if points and planes intersect (distance = 0) or if they are apart.

Planes in 3D Space

Planes in 3D space can be visualized as flat, two-dimensional surfaces extending infinitely in all directions. These planes are characterized by their normal vector, a perpendicular line emanating from the plane's surface, and have an equation of the form Ax + By + Cz + D = 0.

• The normal vector (A, B, C) effectively describes the plane's tilt or orientation within 3D space.
• The constant D shifts the plane along the direction of the normal vector.

In our example, the plane defined by \(-3x + 2y + z = 9\) implies specific orientation and position within the 3D coordinate system.

• This definition allows for comprehensive analytical exploration, including determining parallelism, intersection with other planes or lines, and measuring distances to points, as we calculated.

Recognizing these planes in the spatial context is crucial for solving related problems and has practical applications in designing 3D models and simulations. Planes help us understand structural features in physical space, process geographical data and serve in a multitude of other areas in science and technology.