Question

Find a point on the plane \(2 x-y-z=13 .\) Find the distance from (7,1,-2) to the plane.

Short answer

\(\frac{\text{sqrt}(6)}{3}\)

Step-by-step solution

- Identify plane equation and known point

The given plane equation is \[2x - y - z = 13\].Identify the known point from which the distance is to be calculated, which is (7, 1, -2).

- Use the distance formula

The distance from a point \((x_1, y_1, z_1)\) to a plane \[Ax + By + Cz + D = 0\]is calculated using the formula: \[\text{Distance} = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\text{sqrt}(A^2 + B^2 + C^2)}\].

- Rewrite the plane equation in standard form

Rewriting the plane equation \[2x - y - z = 13\] in the form \[Ax + By + Cz + D = 0\],we have: \[2x - y - z - 13 = 0\]. Thus, \[A = 2, B = -1, C = -1, D = -13\].

- Apply the distance formula

Plug the values into the distance formula: \[\text{Distance} = \frac{|2(7) - 1(1) - 1(-2) - 13|}{\text{sqrt}(2^2 + (-1)^2 + (-1)^2)}\]. Simplify the numerator and denominator separately.

- Simplify the expression

First, simplify the numerator: \[|14 - 1 + 2 - 13| = |2| = 2\]. Next, simplify the denominator: \[\text{sqrt}(4 + 1 + 1) = \text{sqrt}(6).\]

- Calculate the final distance

The distance is: \[\frac{2}{\text{sqrt}(6)}\]. Rationalize the denominator to get: \[\frac{2}{\text{sqrt}(6)} \times \frac{\text{sqrt}(6)}{\text{sqrt}(6)} = \frac{2 \text{sqrt}(6)}{6} = \frac{\text{sqrt}(6)}{3}.\]

Key concepts

Plane Equation

The plane equation is given in the form \(Ax + By + Cz + D = 0 \). This represents a flat, two-dimensional surface extending infinitely in three-dimensional space.
In the exercise, the plane equation is specified as \(2x - y - z = 13 \). To transform this into the standard form, we need to move the constant term to the left side of the equation. This gives us:\(2x - y - z - 13 = 0 \). Now, we can easily identify the coefficients: \(A = 2, B = -1, C = -1, D = -13 \).

Distance Formula

The distance formula calculates the shortest path between a point and a plane. The formula we use is: \[ \text{Distance} = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\text{sqrt}(A^2 + B^2 + C^2)} \].
The components \(A, B, C, \) and \(D \) come from the plane’s equation, while \(x_1, y_1, \text{ and } z_1 \) are the coordinates of the given point. In this exercise, our point is \((7, 1, -2) \) and the plane values are \(A=2, B=-1, C=-1, D=-13 \).
Plugging these into the formula: \[\frac{ | 2(7) - 1(1) - 1(-2) - 13 |}{ \text{sqrt}(2^2 + (-1)^2 + (-1)^2) } = \frac{|2 \cdot 7 - 1 \cdot 1 - 1 \cdot -2 - 13|}{\text{sqrt}(2^2 + (-1)^2 + (-1)^2)}\].

Coordinate Geometry

Coordinate geometry allows us to determine positions and distances in space using algebra. Here’s what you need to know:

• Points are described by coordinates: \( (x, y, z) \).
• Planes are described by equations including parameters \(x, y, z \) and constants.
• Distances between points and planes can be computed using specific formulas, like the distance formula we saw earlier.

In our problem, we'll use coordinate geometry to find the distance between the point \((7, 1, -2) \) and the plane \(2x - y - z = 13 \). By identifying the coefficients from the plane equation and the coordinates of the point, we can plug them into the distance formula and solve.

Rationalizing Denominators

Rationalizing the denominator is a technique we use to remove radicals from the bottom of fractions. Here’s a step-by-step:

• Identify the denominator: \( \text{sqrt}(6) \) in our problem.
• Multiply both the numerator and the denominator by \( \text{sqrt}(6)\) to get rid of the radical in the denominator.
• So we transform \( \frac{2}{\text{sqrt}(6)} \) into \[ \frac{2 \text{sqrt}(6)}{ \text{sqrt}(6) \times \text{sqrt}(6)} = \frac{2 \text{sqrt}(6)}{6} \].
• Finally, simplify the fraction \( \frac{2 \text{sqrt}(6)}{6} \) as \( \frac{\text{sqrt}(6)}{3} \).

This ensures our final answer is simplified and free from radicals in the denominator.