Question

Find the distance between the point and the plane. $$ \begin{array}{l} (3,2,1) \\ x-y+2 z=4 \end{array} $$

Short answer

The distance between the point (3,2,1) and the plane \( x - y + 2z = 4 \) is \( \frac{1}{\sqrt{6}} units.

Step-by-step solution

Rewrite the plane equation in normal form

Any plane equation of the form ax + by + cz = d can be rewritten in normal form as \( Ax + By + Cz + D = 0 \), where A, B and C are the coefficients of x, y, and z in the plane equation. Therefore, rewrite the plane equation \( x - y + 2z = 4 \) in normal form. This gives \( x - y + 2z - 4 = 0 \)

Identify the normal vector and the coordinates of the given point

From the rewritten plane equation, the normal vector (A,B,C) of the plane equation can be identified as (1, -1, 2). The given point can be written in coordinates as P(3,2,1).

Apply the distance formula

The distance d from a point P(x0, y0, z0) to a plane Ax + By + Cz + D = 0 is given by \( d = \frac{|Ax0 + By0 + Cz0 + D|}{\sqrt{A^2 + B^2 + C^2}} \). Substitute A, B, C, D from the plane equation and x0, y0, z0 from the point into the distance formula. This gives \( d = \frac{|(1*3) + (-1*2) + (2*1) - 4|}{\sqrt{(1^2) + (-1^2) + (2^2)}} \)

Calculate the distance

Solve the expressions in the formula: \( d = \frac{|3 - 2 + 2 - 4|}{\sqrt{1 + 1 + 4}} = \frac{1}{\sqrt{6}} \). The distance \( d \) is equal to \( \frac{1}{\sqrt{6}} \)

Key concepts

Plane Equation in Normal Form

Understanding the plane equation in its normal form is crucial for solving various geometrical problems. The normal form of a plane equation is a way of expressing the equation of a plane such that the coefficients of the variables represent a normal vector to the plane. A normal vector is a vector that is perpendicular to the surface of the plane.

To rewrite any plane equation into the normal form, you start with the standard equation of a plane, typically presented as \( ax + by + cz = d \). Then, this is converted to \( Ax + By + Cz + D = 0 \), where \( A \), \( B \), and \( C \) are the coefficients of \( x \), \( y \), and \( z \) respectively, and \( D \) is the negative of \( d \). This neatly arranges the equation so the normal vector to the plane can be directly identified and the equation is ready for further calculations, such as finding the distance from a point to the plane.

Normal Vector of a Plane

The normal vector is a fundamental concept when working with planes in three-dimensional space. It is a vector that is perpendicular to the plane's surface and is usually denoted as \( \vec{n} \). In the context of the equation of a plane in normal form (\( Ax + By + Cz + D = 0 \)), the coefficients \( A \), \( B \), and \( C \) form the components of the normal vector, which can be represented as \( \vec{n} = (A, B, C) \).

For example, if the plane equation is \( x - y + 2z = 4 \), the normal vector \( \vec{n} \) would be \( (1, -1, 2) \), simply extracted from the coefficients of the variables in the equation. This vector will be crucial for calculating distances as it provides a direction that is orthogonal to every point on the plane.

Distance Formula for Point to Plane

Calculating the perpendicular distance from a point to a plane is a standard problem in geometry and can be accomplished using the distance formula for a point to a plane. The formula is derived from the properties of dot products and the definition of a plane in 3D space.

For a point \( P(x_{0}, y_{0}, z_{0}) \) and a plane with the normal form equation \( Ax + By + Cz + D = 0 \), the distance \( d \) can be calculated using the formula:
\[ d = \frac{|Ax_{0} + By_{0} + Cz_{0} + D|}{\sqrt{A^{2} + B^{2} + C^{2}}} \]
This formula takes the absolute value of the dot product of the normal vector and the point's position vector added to the constant \( D \) from the plane's equation, and normalizes it by the magnitude of the normal vector. This calculation yields the shortest distance from the point to the plane, which is always measured along the line perpendicular to the plane.

Calculating Distance with Coordinates

When you have the coordinates of a point and the normal form of a plane's equation, you can compute the point-to-plane distance using substitution into the distance formula. By inputting the coordinates of the point into the place of \( x_{0}, y_{0}, z_{0} \) in the formula, you can determine the numerator's value. For the plane's equation \( Ax + By + Cz + D = 0 \), substitute the coefficients \( A, B, C \) and \( D \) along with the point's coordinates into the distance formula to find the precise distance.

As a practical example, consider a point \( (3, 2, 1) \) and the plane equation \( x - y + 2z - 4 = 0 \). By plugging the values into the distance formula, \( d = \frac{|3 - 2 + 2 - 4|}{\sqrt{1 + 1 + 4}} \), you perform the arithmetic to find the shortest distance between the point and the plane. In this case, the calculated distance is \( \frac{1}{\sqrt{6}} \) units. By understanding these steps and the underlying concepts, you can tackle any similar problems in geometry with confidence.