Question

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

Short answer

The distance from the point (7,1,-2) to the plane \(2x - y - z = 13\) is \( \frac{\sqrt{6}}{3} \)

Step-by-step solution

- Identify the given plane equation

The given plane equation is \(2x - y - z = 13\).

- Find a point on the plane

Choose values for two variables and solve for the third. For instance, let’s choose \(x = 0\) and \(y = 0\). Substituting these into the plane equation \(2(0) - 0 - z = 13\), we get \(-z = 13\), which simplifies to \(z = -13\). So, the point \((0, 0, -13)\) lies on the plane.

- Find the normal vector of the plane

The normal vector \(\mathbf{n}\) of the plane \(2x - y - z = 13\) is \<2, -1, -1\>.

- 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 \[d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}\]. Translate the given plane equation to the form \(2x - y - z - 13 = 0\). The coefficients are \(A = 2\), \(B = -1\), \(C = -1\), and \(D = -13\). Substituting the point \((7, 1, -2)\) into the formula: \[d = \frac{|2(7) - 1(1) - 1(-2) - 13|}{\sqrt{2^2 + (-1)^2 + (-1)^2}}\].

- Calculate the numerator

Substitute the coordinates and constants into the numerator: \[2(7) - 1(1) - 1(-2) - 13 = 14 - 1 + 2 - 13 = 2.\]

- Calculate the denominator

Substitute the constants into the denominator and simplify: \[\sqrt{2^2 + (-1)^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}.\]

- Combine and simplify the formula

The distance is \[d = \frac{2}{\sqrt{6}} = \frac{2\sqrt{6}}{6} = \frac{\sqrt{6}}{3}.\]

Key concepts

Plane Equation

A plane equation represents a flat, two-dimensional surface that extends infinitely in three-dimensional space. The general form of the plane equation is given as \( Ax + By + Cz + D = 0 \). Here, \(A\), \(B\), \(C\), and \(D\) are constants, and \(x\), \(y\), and \(z\) are the coordinates of any point on the plane.

In this exercise, the plane equation provided is \(2x - y - z = 13\). To use this form in common formulas, it is often rewritten as \(2x - y - z - 13 = 0\), making it easier to identify the values for \(A\), \(B\), \(C\), and \(D\).

To find a specific point on this plane, you can choose values for two of the variables and solve for the third. For example, choosing \(x = 0\) and \(y = 0\), we substitute into the plane equation:
\[2(0) - 0 - z = 13\]
Solving this, we find \(z = -13\), giving us the point \((0, 0, -13)\) that lies on the plane.

Normal Vector

The normal vector to a plane is a vector that is perpendicular to every line lying on the plane. For a plane given by the equation \(Ax + By + Cz + D = 0\), the normal vector \( \mathbf{n} \) can be directly derived from the coefficients of \(x\), \(y\), and \(z\).

In this exercise, the plane equation is \(2x - y - z = 13\). From this, we identify the coefficients: \(A = 2\), \(B = -1\), and \(C = -1\). Thus, the normal vector \( \mathbf{n} \) is given by:
\[\mathbf{n} = \<2, -1, -1>\]
The normal vector is significant because it is used in various calculations, including finding the distance from a point to the plane.

Distance Formula

The distance from a point to a plane in three-dimensional space can be calculated using a specific formula. The formula for the distance \(d\) from a point \((x_1, y_1, z_1)\) to a plane \(Ax + By + Cz + D = 0\) is:
\[d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}\]

Let's apply this formula to the given problem. Here, the plane equation is \(2x - y - z - 13 = 0\), and the point is \((7, 1, -2)\).
First, identify the coefficients \(A = 2\), \(B = -1\), \(C = -1\), and \(D = -13\).
Next, substitute the point \((x_1, y_1, z_1) = (7, 1, -2)\) into the numerator of the distance formula:
\[2(7) - 1(1) - 1(-2) - 13 = 14 - 1 + 2 - 13 = 2\]
For the denominator, calculate the square root of the sum of the squares of the coefficients:
\[\sqrt{2^2 + (-1)^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}\]
Finally, substitute these values into the distance formula:
\[d = \frac{2}{\sqrt{6}} = \frac{2\sqrt{6}}{6} = \frac{\sqrt{6}}{3}\]
Thus, the distance from the point \((7, 1, -2)\) to the plane \(2x - y - z = 13\) is \( \frac{\sqrt{6}}{3} \).