91Ó°ÊÓ

A plane in three-dimensional space can be visualized as a flat, infinitely expansive surface. To mathematically describe this surface, we use a plane equation, which typically takes the form of \( ax + by + cz = d \). Here, \( a \), \( b \), and \( c \) are the coefficients that define the plane's orientation in space, while \( d \) represents the plane's offset from the origin. For instance, in the given problem, the plane equation is \( 2x + y + 2z = 4 \). This tells us:

• Parallel alignment with the standard 3D axes is determined by the coefficients \( a = 2 \), \( b = 1 \), and \( c = 2 \).
• The value \( d = 4 \) signifies that the plane is shifted four units from the origin along the normal vector defined by these coefficients.

Understanding the equation helps visualize how a plane cuts through three-dimensional space and defines how to measure perpendicular distance from a point to this plane.

To find the shortest distance from a point to a plane, we use a specialized distance formula. This formula is essential in various fields such as computational geometry and physics. The formula to calculate the distance \( D \) from a point \((x_0, y_0, z_0)\) to a plane \( ax + by + cz = d \) is:

• \[ D = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}} \]

The numerator, \( |ax_0 + by_0 + cz_0 - d| \), represents the absolute difference between the weighted sum of coordinates and the plane's offset. The denominator, \( \sqrt{a^2 + b^2 + c^2} \), ensures the calculation is with respect to the plane's normal vector's length, providing the true perpendicular distance. By plugging in the values of the point \( (2, 2, 3) \) and the given plane equation coefficients, we simplify and solve for \( D \). This formula underscores the importance of understanding plane coefficients and their geometric implications.

Calculating the distance from a point to a plane involves obtaining values for both the numerator and the denominator in the distance formula. Let's break down each part to understand better:

Numerator Calculation

The numerator of the distance formula, \( |ax_0 + by_0 + cz_0 - d| \), calculates the direct perpendicular offset of a point from the plane. Substituting our point \((2, 2, 3)\) and the plane coefficients \( a = 2 \), \( b = 1 \), \( c = 2 \), \( d = 4 \), we calculate:

• Substituting in: \( 2(2) + 1(2) + 2(3) - 4 \)
• Simplified as: \( 4 + 2 + 6 - 4 = 8 \)
• Result is \( |8| = 8 \)

Denominator Calculation

For the denominator, \( \sqrt{a^2 + b^2 + c^2} \), this quantifies the length of the plane's normal vector. It is important for normalization to ensure the distance measurement is accurate:

• Substituting in: \( \sqrt{2^2 + 1^2 + 2^2} \)
• Simplified calculation: \( \sqrt{4 + 1 + 4} = \sqrt{9} \)
• Result is \( 3 \)

By dividing the numerator by the denominator, \( \frac{8}{3} \), we find the precise distance from the point to the plane, ensuring correct interpretation of these calculations in three-dimensional space.