Problem 39 - Calculus Early Transcendentals

Question

Find the shortest distance from the point \((2,1,-1)\) to the plane \(x+y-z=1 .\)

Step-by-Step Solution

Answer

The shortest distance from the point \((2,1,-1)\) to the plane is \(\sqrt{3}\).

Step 1: Understand the Problem

We need to find the shortest distance from the point \((2,1,-1)\) to the plane given by the equation \(x+y-z=1\). The shortest distance from a point to a plane is the perpendicular distance.

Step 2: Identify the Plane's Normal Vector

The given plane equation is \(x+y-z=1\). The coefficients of \(x\), \(y\), and \(z\) give us the normal vector \(\mathbf{n} = \langle 1, 1, -1 \rangle\) to the plane.

Step 3: Use the Distance Formula

The formula to calculate the distance \(d\) from a point \((x_1, y_1, z_1)\) to a plane \(ax + by + cz + d = 0\) is:\[d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}\]For the plane equation \(x + y - z = 1\), we rearrange it to \(x + y - z - 1 = 0\). Here, \(a = 1\), \(b = 1\), \(c = -1\), and \(d = -1\).

Step 4: Substitute Point Coordinates into the Formula

Substitute the coordinates \((2, 1, -1)\) of the point and the plane parameters into the formula:\[d = \frac{|1\cdot2 + 1\cdot1 - 1\cdot(-1) - 1|}{\sqrt{1^2 + 1^2 + (-1)^2}}\]

Step 5: Calculate the Numerator

Calculate the expression in the numerator:\[|1\cdot2 + 1\cdot1 - 1\cdot(-1) - 1| = |2 + 1 + 1 - 1| = |3|\]

Step 6: Calculate the Denominator

Calculate the expression in the denominator:\[\sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}\]

Step 7: Complete the Distance Calculation

Plug the results from steps 5 and 6 into the distance formula:\[d = \frac{|3|}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}\]

Step 8: Simplify Result if Necessary

Since \( \sqrt{3} \) is already in its simplest form, no further simplification is needed. The shortest distance is \( \sqrt{3} \).

Key Concepts

Normal Vector

In the context of geometry and vectors, a **normal vector** is a crucial concept. It’s a vector that is perpendicular to a given surface, such as a plane. When dealing with planes, the normal vector is pivotal because it defines the plane's orientation in space. To find the normal vector of a plane from its equation, you can simply look at the coefficients of its variables.

For example, consider the plane equation:

• \(x + y - z = 1\).

Here, the coefficients of \(x\), \(y\), and \(z\) are 1, 1, and -1 respectively. These coefficients form the components of the normal vector:

• \(\mathbf{n} = \langle 1, 1, -1 \rangle\).

This vector is perpendicular to every line that lies on the plane, making it a handy tool for numerous calculations, including finding distances.

Distance Formula

The **distance formula** is used in algebra and geometry to find the shortest path from a point to a plane. This formula involves the normal vector of the plane and the coordinates of the point in question. When dealing with a plane equation of the form \(ax + by + cz + d = 0\), the formula for finding the distance \(d\) from a point \((x_1, y_1, z_1)\) to the plane is:

• \[d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}\]

This equation breaks down neatly:

• The numerator is the absolute value of the plane equation with the point coordinates substituted in.
• The denominator is the magnitude of the normal vector, which gives a scale of the steepness.

By applying this formula, one can calculate the perpendicular distance effectively.

Perpendicular Distance

The **perpendicular distance** in three-dimensional space is the shortest distance from a point to a plane. This concept is fundamental in understanding distances in a geometric context. The key characteristic is that it forms a right angle (90 degrees) with the surface it meets, i.e., the plane.
To find this distance, it's necessary to use the point-to-plane distance formula. This method ensures that the distance calculated is the smallest possible, as it travels along the plane's normal vector. Additionally, this perpendicular path is not only mathematically precise but also geometrically optimal.

Plane Equation

The **plane equation** is vital in specifying a plane in three-dimensional space. It’s typically given in the form \(ax + by + cz = d\), or rearranged for calculations as \(ax + by + cz + d = 0\).

• Each term coefficient \(a\), \(b\), and \(c\) relate to the components of the normal vector to the plane.
• \(d\) adjusts the plane's position relative to the origin.

The equation not only sets out how the plane appears in space but also is central to deriving other related properties, such as the normal vector. In practical problems, rearranging the terms can facilitate easier computations, like determining distances. The understanding of plane equations is therefore crucial for navigating and solving geometric problems in a 3D space.