Problem 354 - Calculus Volume 3

Question

Find the point in the plane \(2 x-y+2 z=16\) that is closest to the origin.

Step-by-Step Solution

Answer

The closest point to the origin is \(\left( \frac{32}{9}, -\frac{16}{9}, \frac{32}{9} \right)\).

Step 1: Express the Plane in General Form

The given plane is expressed as \(2x - y + 2z = 16\). This equation defines a plane in a 3-dimensional coordinate system, where \(x\), \(y\), and \(z\) represent the coordinates of any point on the plane.

Step 2: Understand the Closest Point to the Origin

To find the point on the plane that is closest to the origin, we need to find the perpendicular from the origin \(0, 0, 0\) to the plane. The perpendicular direction is determined by the plane's normal vector \(\mathbf{n} = \langle 2, -1, 2 \rangle\).

Step 3: Set the Point Along the Normal Vector

Any point along this normal vector from the origin can be expressed as \((2t, -t, 2t)\), where \(t\) is a scalar that scales the normal vector.

Step 4: Substitute into the Plane Equation

Substitute the coordinates \((2t, -t, 2t)\) into the plane equation: \(2(2t) - (-t) + 2(2t) = 16\). Simplify this to find the value of \(t\): \[4t + t + 4t = 16\]

Step 5: Solve for the Scalar t

Simplifying further gives \(9t = 16\), which yields \(t = \frac{16}{9}\).

Step 6: Calculate the Coordinates of the Closest Point

Substitute \(t = \frac{16}{9}\) back into the expressions for \(x\), \(y\), and \(z\). The coordinates of the closest point are \(\left( 2 \times \frac{16}{9}, -\frac{16}{9}, 2 \times \frac{16}{9} \right) = \left( \frac{32}{9}, -\frac{16}{9}, \frac{32}{9} \right)\).

Key Concepts

Perpendicular Distance

When trying to determine the shortest distance from a point to a plane in 3D geometry, the focus is on finding the perpendicular distance. This distance is the most direct path to the plane, akin to the concept of standing directly above a point and dropping a line straight down to it. To achieve this, you need to use the normal vector to the plane. By understanding the shortest distance as \(d\), you can begin by identifying a direction vector from the given point to the plane, aligned along this normal vector, and using it to formulate the distance equation:

• The perpendicular line to the plane is the one along the normal vector.
• This line will also intersect the plane at the closest point, providing the perpendicular distance.

In essence, the perpendicular distance is about finding that specific line segment, perfectly aligned to the normal vector, minimizing the path from point to plane.
In our exercise, this is achieved along the direction defined by the plane's normal vector, ensuring the distance is truly the "perpendicular" or shortest one.

Normal Vector

The normal vector plays a fundamental role in geometry, particularly when dealing with planes. A normal vector is perpendicular to every line lying on the plane, acting as a kind of invisible stick pointing straight out from the surface. In 3D coordinate geometry, this vector helps in identifying essential geometric properties, especially in finding distances and angles.
Going back to the original exercise, the normal vector is crucial because:

• It provides the direction for the shortest distance from the point (origin).
• The components of the normal vector \(\langle 2, -1, 2 \rangle\) correspond directly to the coefficients in the plane's equation, \(2x - y + 2z = 16\).

By acknowledging the significance of the normal vector, we can determine the specific line that crosses through both the origin and the closest point on the plane. This vector is essentially a roadmap that guides the solution process, making it indispensable for solving problems involving planes.

3D Coordinate Geometry

3D coordinate geometry is like the study of shapes in a space that adds depth to flat figures, allowing you to explore not just length and width but also height. It involves plotting points on a space defined by three axes: the x-axis, the y-axis, and the z-axis.
In this particular problem, the plane is defined using a standard equation format, \( Ax + By + Cz = D \), where each letter represents a constant. It takes three dimensions to define planes reliably:

• The x-axis, representing the width.
• The y-axis, representing the height.
• The z-axis, representing the depth.

This representation allows for accurate depiction of spatial relationships, such as determining the location of a point in space relative to a plane. In our exercise, considering the coordinates of the closest point, \( \left( \frac{32}{9}, -\frac{16}{9}, \frac{32}{9} \right) \), helps illustrate how points are resolved in a 3D environment. With 3D coordinate geometry, you can visualize complex structures clearly, even when they extend into spatial dimensions beyond the paper or screen.