Question

Find the distance from the origin to the plane \(3 x-2 y-6 z=7\).

Short answer

The distance from the origin to the plane is 1 unit.

Step-by-step solution

Identify the Plane Equation

The given plane equation is: \[3x - 2y - 6z = 7\]

Determine the Point on the Plane

The point from which to measure the distance is the origin, which has coordinates (0, 0, 0).

Use the Distance Formula

The distance from a point \((x_1, y_1, z_1)\) to a plane \(Ax + By + Cz + D = 0\) is given by the formula: \[d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}\]Here, the point is (0, 0, 0), and the plane equation needs to be rewritten in the form \(Ax + By + Cz + D = 0\).

Transform the Plane Equation

Rewrite the plane equation:\[3x - 2y - 6z - 7 = 0\] Here, \(A = 3\), \(B = -2\), \(C = -6\), and \(D = -7\)

Substitute Values and Calculate

Substitute \(A = 3\), \(B = -2\), \(C = -6\), \(D = -7\), \((x_1, y_1, z_1) = (0, 0, 0)\) into the formula to get:\[d = \frac{|3(0) - 2(0) - 6(0) - 7|}{\sqrt{3^2 + (-2)^2 + (-6)^2}} = \frac{|-7|}{\sqrt{9 + 4 + 36}} = \frac{7}{\sqrt{49}} = \frac{7}{7} = 1\]

Key concepts

Plane Equation

A plane equation is a way to describe a flat surface in three-dimensional space. It's typically written in the form: \[Ax + By + Cz + D = 0\]where:

• A, B, and C are coefficients that determine the orientation of the plane.
• x, y, and z are the Cartesian coordinates of any point on the plane.
• D is a constant that shifts the plane along the normal vector.

In the problem, the given plane equation is \[3x - 2y - 6z = 7\]. By rewriting it as \[3x - 2y - 6z - 7 = 0\], we align it with the generic form. This makes it easier for us to apply various calculations, such as finding the distance from a point to the plane.

Distance Formula

To find the distance from a point to a plane, we use a specific distance formula. This formula helps measure the shortest distance between a point and a flat surface. The formula is given by: \[d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\text{sqrt}(A^2 + B^2 + C^2)}\]. Here's a quick breakdown of the variables:

• (x₁, y₁, z₁) are the coordinates of the point from which you're measuring the distance.
• A, B, C are coefficients from the plane equation.
• D is the constant term from the plane equation.

For our problem, the point is the origin: (0, 0, 0), and the plane equation is \[3x - 2y - 6z - 7 = 0\]. Substituting these values into the distance formula and simplifying helps us find the required distance.

Cartesian Coordinates

Cartesian coordinates are a system for specifying points in a plane through ordered pairs (x, y) or in three-dimensional space through ordered triples (x, y, z). This system uses perpendicular axes—the x-axis, y-axis, and z-axis—to define the location of a point. Each coordinate represents a distance along an axis from the origin point (0, 0, 0).

• The x-coordinate indicates horizontal position.
• The y-coordinate indicates vertical position.
• The z-coordinate indicates depth in three dimensions.

In the exercise, the point of interest is the origin (0, 0, 0). This simplifies calculations because all coordinates are zero, making the distance formula straightforward to apply.

Vector Projection

Vector projection is a way of projecting one vector onto another. In simpler terms, it helps find the shadow or footprint of one vector on another. While not directly used in the problem given, understanding vector projection can be crucial when dealing with distances in higher dimensions.To project a vector \(\textbf{a}\) onto another vector \(\textbf{b}\), the formula is: \[\text{proj}_{\textbf{b}} \textbf{a} = \frac{\textbf{a} \bullet \textbf{b}}{\textbf{b} \bullet \textbf{b}} \textbf{b}\]

• The dot product \(\textbf{a} \bullet \textbf{b}\) calculates the magnitude of the projection.
• Dividing by \(\textbf{b} \bullet \textbf{b}\) normalizes it relative to the vector \(\textbf{b}\).
• Multiplying by \(\textbf{b}\) scales the unit projection back to vector form.

Vector projection helps visualize concepts like shortest distance, as the perpendicular drop from a point to a plane can be seen as the projection of the point's position vector onto the normal vector of the plane.