Problem 41 - Advanced Engineering Mathematics

Question

In Problems, find an equation of the plane that contains the given point and is perpendicular to the indicated vector. $$ (6,10,-7) ;-5 i+3 k $$

Step-by-Step Solution

Answer

The plane's equation is \\-5x + 3z + 51 = 0\\.

Step 1: Identify the Components

The given problem provides a point on the plane \(6, 10, -7\) and a vector \(-5\mathbf{i} + 3\mathbf{k}\) which is perpendicular to the plane. From the vector, identify the normal vector components as \(-5, 0, 3\).

Step 2: Equation of Plane

The equation of a plane with a point \(x_0, y_0, z_0\) and a normal vector \(a, b, c\) is \a(x-x_0) + b(y-y_0) + c(z-z_0) = 0\.

Step 3: Substitute Values

Substitute the point \(6, 10, -7\) and the normal vector \(-5, 0, 3\) into the plane equation:\[-5(x-6) + 0(y-10) + 3(z+7) = 0\].

Step 4: Simplify the Equation

Simplify the equation by expanding:\[-5x + 30 + 3z + 21 = 0\].

Step 5: Combine Like Terms

Combine terms to reach the simplest form:\[-5x + 3z + 51 = 0\].

Key Concepts

Normal Vector

The concept of a normal vector is crucial in understanding the geometry of planes in three-dimensional space. A normal vector is essentially a vector that is orthogonal, or perpendicular, to the plane in question.

• The normal vector defines the orientation of the plane.
• It is represented by its components, which are coefficients of the vector's basis components typically written as \(a, b, c\).

When a plane is described by the general equation \(ax + by + cz = d\), the components \(a, b, c\) are those of the normal vector.
In this exercise, the vector given is \(-5\mathbf{i} + 3\mathbf{k}\). Notice that it lacks a \(j\)-component, meaning the \(b\) term is zero. This results in the normal vector components being \(-5, 0, 3\). These values dictate the plane's slope and position in 3D space.

Perpendicular Vector

Vectors that are perpendicular to planes or other vectors must maintain a dot product of zero with the vectors or normals said to be orthogonal to them.

• Perpendicular vectors help define angles and determine orthogonality in space.
• They are used to derive planes and extras like projections or cross products in vector calculus.

In the context of the exercise, the vector given, \(-5\mathbf{i} + 3\mathbf{k}\), is perpendicular to the plane we need to define. By being perpendicular, it implies that any movement along this vector would not increase or decrease distance from the plane itself. This property is why it serves directly to construct the plane's equation.

Point on Plane

Identifying a specific point on a plane is fundamental when trying to find its equation. Using a known point ensures that the plane is accurately anchored in space.

• A point on the plane is provided as \((x_0, y_0, z_0)\).
• In the plane's equation \(a(x-x_0) + b(y-y_0) + c(z-z_0) = 0\), \(x_0, y_0, z_0\) represent this anchoring point.

In this case, the point \((6, 10, -7)\) is given. This point is crucial as it helps define the position of the plane by fixing a known coordinate while the normal vector determines its orientation. Once plugged into the equation, in conjunction with the normal vector, it leads directly to the correct plane equation. This combination grounds the abstract concept of a plane into the tangible three-dimensional space we are examining.