Problem 73 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the distance from \((1,-1,2)\) to the plane \(x+3 y+z=7 .\)

Step-by-Step Solution

Answer

The distance from the point to the plane is \(\frac{7}{\sqrt{11}}\).

Step 1: Identify Given Points and Plane Equation

We are given the point \((1, -1, 2)\) and the equation of the plane \(x + 3y + z = 7\). We need to find the distance from the point to this plane.

Step 2: Recall the Formula for Distance from Point to Plane

The formula to find the distance \(d\) from a point \((x_0, y_0, z_0)\) to a plane \(Ax + By + Cz + D = 0\) is:\[d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}.\]

Step 3: Rewrite the Plane Equation in Standard Form

Rearrange the plane equation \(x + 3y + z = 7\) into the standard form of \(Ax + By + Cz + D = 0\). This becomes \(x + 3y + z - 7 = 0\). Therefore, \(A = 1\), \(B = 3\), \(C = 1\), and \(D = -7\).

Step 4: Substitute Values into the Distance Formula

Substitute \(A = 1\), \(B = 3\), \(C = 1\), \(D = -7\), and the point \((x_0, y_0, z_0) = (1, -1, 2)\) into the distance formula:\[d = \frac{|1\cdot1 + 3\cdot(-1) + 1\cdot2 - 7|}{\sqrt{1^2 + 3^2 + 1^2}}.\]

Step 5: Calculate the Numerator of the Distance Formula

Calculate the numerator: \[1\cdot1 + 3\cdot(-1) + 1\cdot2 - 7 = 1 - 3 + 2 - 7 = -7.\]Take the absolute value: \[|{-7}| = 7.\]

Step 6: Calculate the Denominator of the Distance Formula

Calculate the denominator: \[\sqrt{1^2 + 3^2 + 1^2} = \sqrt{1 + 9 + 1} = \sqrt{11}.\]

Step 7: Divide to Find the Distance

Finally, divide the absolute value of the numerator by the denominator to find the distance:\[d = \frac{7}{\sqrt{11}}.\]This is the distance from the point to the plane.

Key Concepts

distance formula

The distance formula is a powerful tool in geometry, especially when dealing with finding distances in three-dimensional space. It helps us determine how far a point is from a plane. In general, the distance from a point \((x_0, y_0, z_0)\) to a plane described by the equation \(Ax + By + Cz + D = 0\) is calculated using the formula:\[d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}.\]This formula focuses on two key components:

• The numerator: It represents the perpendicular distance from the point to the plane and includes the coefficients from the plane equation and the point's coordinates.
• The denominator: This is the magnitude of the normal vector to the plane and is derived from the coefficients \(A, B, \text{and } C\).

The distance formula gives us a direct way to measure this perpendicular distance, crucial in various mathematical and engineering applications.

3D geometry

3D geometry extends the ideas of 2D geometry into three-dimensional space, adding an extra layer of complexity and interest. In 3D, objects have width, depth, and height, exemplified by points, lines, planes, and more complex shapes such as spheres or cubes. Understanding 3D geometry involves:

• Points: Defined by a set of three coordinates \(x, y, z\) in space.
• Lines: Represented as the intersection of two planes or by a vector and a point.
• Planes: Flat, two-dimensional surfaces that extend infinitely within three-dimensional space.

In our case, we are focusing on the interaction between points and planes in 3D. Measuring distances, determining intersections, and analyzing spatial relationships are fundamental uses of 3D geometry, providing insight into the spatial structure and the relative placement of objects.

plane equation

Plane equations form the basis of descriptions and analyses of planes in 3D geometry. A plane in three-dimensional space can be described using the general equation:\[Ax + By + Cz + D = 0.\]Each coefficient in this equation plays a role:

• \(A, B, \text{and } C\) are the coefficients of the normal vector perpendicular to the plane.
• \(D\) is a constant that affects the plane's position relative to the origin.

In our exercise, the given plane equation is \(x + 3y + z - 7 = 0\), indicating a plane with normal vector components \(1, 3, \text{and } 1\). Understanding plane equations is essential in solving numerous geometric problems, allowing us to find intersections, measure distances, and evaluate the orientation of surfaces in various contexts.

vectors and planes

Vectors play a crucial role in understanding and analyzing planes within three-dimensional space, as they help define directions and magnitudes. A plane can be described by its normal vector, which is perpendicular to every line lying on the plane. This normal vector has components \(A, B, \text{and } C\) from the plane's standard equation.

• A normal vector dictates the plane's orientation. In our problem, the normal vector is \(\mathbf{n} = \langle 1, 3, 1 \rangle\).
• Vectors also describe positions and directions, essential for performing operations like vector addition, scalar multiplication, and dot products.
• Understanding these vector operations aids in perceiving how objects leap between dimensions, rotate, or move within the confines of a given space.

Grasping how vectors relate to planes is vital for various real-world applications, including engineering, physics simulations, and graphics programming.