Problem 390 - Calculus Volume 3

Question

Find the point on the plane \(4 x+3 y+z=2\) that is closest to the point (1,-1,1) .

Step-by-Step Solution

Answer

The closest point on the plane is (1, -1, 1).

Step 1: Understand the Problem

We need to find the closest point on the plane \(4x + 3y + z = 2\) to the point \((1, -1, 1)\). The shortest distance from a point to a plane is along the line perpendicular to the plane.

Step 2: Identify the Plane's Normal Vector

The normal vector to the plane can be found from the coefficients of \(x\), \(y\), and \(z\). For the plane equation \(4x + 3y + z = 2\), the normal vector is \((4, 3, 1)\).

Step 3: Write the Parametric Equation of the Line

The line through the point \((1, -1, 1)\) perpendicular to the plane has the direction vector equal to the normal vector \((4, 3, 1)\). Thus, the line's parametric equations are: \[ x = 1 + 4t, \quad y = -1 + 3t, \quad z = 1 + t \]

Step 4: Substitute the Line's Equations into the Plane Equation

Substitute \(x = 1 + 4t\), \(y = -1 + 3t\), and \(z = 1 + t\) into the plane equation \(4x + 3y + z = 2\): \[\begin{align*} 4(1 + 4t) + 3(-1 + 3t) + (1 + t) &= 2 \ 4 + 16t - 3 + 9t + 1 + t &= 2 \ 26t + 2 &= 2 \end{align*}\]

Step 5: Solve for Parameter t

Simplify the equation: \[ 26t + 2 = 2 \] To solve for \(t\): \[ 26t = 0 \] \[ t = 0 \]

Step 6: Find the Closest Point on the Plane

Substitute \(t = 0\) back into the parametric equations of the line: \[ x = 1 + 4(0) = 1 \] \[ y = -1 + 3(0) = -1 \] \[ z = 1 + 0 = 1 \] Thus, the closest point on the plane is \((1, -1, 1)\).

Key Concepts

Normal Vector

A normal vector is an important concept when dealing with planes in three-dimensional space. It is a vector that is perpendicular to a particular surface or plane. In simpler terms, if you think of a plane as a flat sheet of paper, the normal vector is like a stick pointing straight up from the surface of the paper.

To find the normal vector of a plane, you can use the coefficients from the plane's equation. For example, in the plane equation given as \(4x + 3y + z = 2\), the coefficients are 4, 3, and 1. These coefficients directly form the normal vector \((4, 3, 1)\).

Normal vectors are crucial because they help determine the orientation of the plane in space. Furthermore, when finding the distance from a point to a plane, the shortest path is along this normal direction. This is what makes the normal vector a valuable tool in geometry.

Parametric Equation of a Line

The parametric equation of a line provides a way to express the line using a parameter, typically denoted as \(t\). This helps in representing all the points on the line as functions of \(t\).

In our exercise, we have a line passing through the point \((1,-1,1)\) and perpendicular to the plane \(4x + 3y + z = 2\). The direction vector of this line is equal to the normal vector of the plane, which is \((4, 3, 1)\).

This gives us the parametric equations:

• \(x = 1 + 4t\)
• \(y = -1 + 3t\)
• \(z = 1 + t\)

These equations tell us that as we change \(t\), we move along the line in its direction. By substituting different values for \(t\), you can find different points along the line.

Plane Equation

The equation of a plane in three-dimensional space is usually written in the form \(ax + by + cz = d\), where \(a\), \(b\), and \(c\) are constants that determine the plane's orientation, and \(d\) is a constant that shifts the plane up or down. This type of equation uniquely describes every point \((x, y, z)\) that lies on the plane.

For example, consider the plane equation \(4x + 3y + z = 2\). Here, 4, 3, and 1 are the coefficients of \(x\), \(y\), and \(z\), which are also used to form the normal vector of the plane, \((4, 3, 1)\). These coefficients are crucial because they define how the plane is oriented in space.

Understanding the concept of a plane equation is essential not only for solving problems related to geometry but also for applications in physics and engineering, where planes often represent boundaries or surfaces of objects.