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

Distance is 0; the point is on the plane.

Step 1: Identify the point and the plane equation

We need to find the distance from the point \((2,6,3)\) to the plane described by the equation \(-3x + 2y + z = 9\).

Step 2: Recall the formula for distance from a point to a plane

The distance \(D\) from a point \((x_1, y_1, z_1)\) to the plane \(ax + by + cz + d = 0\) is given by:\[D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}\]

Step 3: Convert plane equation to standard form

Convert the plane equation \(-3x + 2y + z = 9\) into the form \(ax + by + cz + d = 0\) by moving 9 to the left side of the equation:-3x + 2y + z - 9 = 0.

Step 4: Substitute point and coefficients into the distance formula

Substitute \((x_1, y_1, z_1) = (2, 6, 3)\) and \(a = -3\), \(b = 2\), \(c = 1\), and \(d = -9\) into the formula:\[D = \frac{|-3(2) + 2(6) + 1(3) - 9|}{\sqrt{(-3)^2 + 2^2 + 1^2}}\]

Step 5: Simplify the expression

Compute the numerator:\[-3(2) + 2(6) + 1(3) - 9 = -6 + 12 + 3 - 9 = 0\]Compute the denominator:\[\sqrt{(-3)^2 + 2^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14}\]

Step 6: Evaluate the distance

Since the numerator is 0, the distance \(D = \frac{0}{\sqrt{14}} = 0\). This means the point \((2, 6, 3)\) lies on the plane.

Key Concepts

Distance Formula

The distance formula is a key element in coordinate geometry. It helps in finding the shortest path between a point and a plane. To find the distance from a point

1. Identify the point and the coefficients of the plane equation.
2. Use the formula:
   \[D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}\]
   where \((x_1, y_1, z_1)\) is the point and \(a, b, c, d\) are from the plane equation.
3. Substitute the numbers and simplify.

The numerator represents the perpendicular distance, while the denominator is the length of the normal vector to the plane. The key idea is to express how far a point is out of alignment with a plane. This formula is derived from the Pythagorean theorem used in three-dimensional space.

Plane Equation

Understanding a plane equation helps grasp how a plane is mathematically represented in space. A standard form equation of a plane is given by:\[ax + by + cz + d = 0\].The coefficients \(a, b,\) and \(c\) describe the direction of the plane's normal vector:

• \(a\), \(b\), and \(c\) indicate the tilt of the plane in each dimension.
• The term \(d\) helps the plane position relative to the origin.

Using these coefficients, you can position and tilt a plane in three-dimensional space. When transitioning the equation into standard form, move constant terms to one side for easy application of formulas. This sets the stage for analyzing distances and geometry involving planes.

Coordinate Geometry

Coordinate geometry serves as a bridge between algebra and geometry.To position and analyze objects within a plane and space:

• Points are specified using coordinates, like \((x, y, z)\).
• Lines and planes follow linear equations with coefficients dictating their slope and position.

It allows mathematical rules to define geometric shapes using equations. By doing so, it enables finding distances, angles, and intersections efficiently. In higher dimensions, particularly, it assists in understanding how different objects relate spatially. Especially in three-dimensional geometry, it becomes a powerful tool for visualizing solutions and calculating precise distances.

Mathematics Education

Mathematics education involves conveying core concepts in ways that are accessible and engaging. Distance from a point to a plane is pivotal in learning more about spatial reasoning. Here are some educational approaches:

• Relate topics to real-world applications to promote understanding.
• Use visual aids such as graphs and models to demonstrate ideas.
• Incorporate interactive problem-solving to encourage hands-on learning.

Providing step-by-step breakdowns is effective for learning complex ideas. It builds confidence and skills to apply concepts in varied scenarios. This specific exercise strengthens mathematical thinking by assigning a practical task of computing the alignment of geometrical elements.