Question

Find the minimum distance between the origin and the plane \(x+3 y-2 z=4\)

Short answer

The minimum distance is \( \frac{2\sqrt{14}}{7} \).

Step-by-step solution

Understand the Formula for Distance to a Plane

To find the distance from a point to a plane, we can use the formula for the distance from a point \( (x_1, y_1, z_1) \) to a plane \( Ax + By + Cz + D = 0 \) which is given by: \[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \] In this problem, the point is the origin \((0, 0, 0)\) and the plane equation is \(x + 3y - 2z = 4\). First, rearrange the equation to: \(x + 3y - 2z - 4 = 0\). Here, \(A = 1\), \(B = 3\), \(C = -2\), and \(D = -4\).

Substitute Values into the Distance Formula

Substitute the values into the distance formula: \[ d = \frac{|1\cdot 0 + 3\cdot 0 - 2\cdot 0 - 4|}{\sqrt{1^2 + 3^2 + (-2)^2}} = \frac{|0 - 4|}{\sqrt{1 + 9 + 4}} = \frac{4}{\sqrt{14}} \]

Simplify the Expression

Simplify the expression to get the final answer. The distance is: \[ d = \frac{4}{\sqrt{14}} \] Conventionally, distillation may require rationalizing the denominator: \[ d = \frac{4 \cdot \sqrt{14}}{14} = \frac{2\sqrt{14}}{7} \]

Key concepts

Distance Formula in 3D

The formula for calculating the distance from a point to a plane in 3D space is quite intuitive when you break it down.
It allows us to determine the shortest distance between a fixed point in space, such as the origin, and a plane defined by its equation.
The standard form of the plane equation is given by \( Ax + By + Cz + D = 0 \), where \( A \), \( B \), \( C \), and \( D \) are constants.
To find the distance \( d \) from a point \((x_1, y_1, z_1)\) to this plane, you use:

• Calculate the numerator: This involves taking the absolute value of \( Ax_1 + By_1 + Cz_1 + D \). This gives the perpendicular "height" from the point to the plane in a sense.
• Calculate the denominator: This is the magnitude of the vector normal to the plane, \( \sqrt{A^2 + B^2 + C^2} \), representing its "tilt" and affecting the distance measure.

Apply these principles, and the formula becomes a straightforward tool to find the required distance, ensuring you're always calculating perpendicular to the plane.

Plane Equation in 3D

Understanding the equation of a plane in 3D is crucial to solving geometric problems, such as finding distances.
A plane can be imagined as a flat, infinite sheet and can be described using an equation involving three variables: \( x \), \( y \), and \( z \).
The standard form of a plane equation is expressed as \( Ax + By + Cz + D = 0 \).- **Coefficients \( A \), \( B \), and \( C \):** These determine the orientation of the plane in space. They are essentially the components of the vector normal (perpendicular) to the plane.- **Constant \( D \):** This shifts the plane along its normal vector with respect to the origin.Here, the plane equation \( x + 3y - 2z = 4 \) was rearranged to \( x + 3y - 2z - 4 = 0 \).
It’s critical to always express it in this standard form before applying distance formulas or other operations.
Grasping this concept aids in visualizing and solving spatial problems by properly contextualizing the plane's position relative to the origin or any other point.

Vector Algebra Applications

Vector algebra plays a pivotal role in 3D geometry, especially when dealing with planes and distances.
The application of vectors allows for intuitive physical interpretations and solutions to geometric problems.
Vectors aid in describing both the direction and magnitude in space.
Here’s how they are useful:

• **Normal Vector:** In the plane equation \( Ax + By + Cz + D = 0 \), the normal vector to the plane is \( \langle A, B, C \rangle \). This vector is fundamental in finding angles between planes, line-plane intersections, and computing distances.
• **Vector Magnitude:** The magnitude of this normal vector \( \sqrt{A^2 + B^2 + C^2} \) is used in the denominator of the point-to-plane distance formula. It normalizes the perpendicular "height" obtained from the numerator, yielding the direct distance.

By utilizing vector algebra, one can not only solve specific problems like finding distances but also explore further into concepts such as projections, angles between entities, and intersections.