Problem 65 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 65-68, find the equation of the plane having the given normal vector \(\mathbf{n}\) and passing through the given point \(P\). $$ \mathbf{n}=2 \mathbf{i}-4 \mathbf{j}+3 \mathbf{k} ; P(1,2,-3) $$

Step-by-Step Solution

Answer

The plane's equation is \(2x - 4y + 3z + 15 = 0.\)

Step 1: Understand the Problem

To find the equation of a plane, we need a point on the plane and a normal vector to the plane. We are given the normal vector \(\mathbf{n} = 2\mathbf{i} - 4\mathbf{j} + 3\mathbf{k}\) and a point \(P(1,2,-3)\) on the plane.

Step 2: General Equation of a Plane

The general equation of a plane with normal vector \(\mathbf{n} = (a, b, c)\) and passing through a point \(P(x_0, y_0, z_0)\) is \(a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\).

Step 3: Substitute Given Values

Substitute the given normal vector \((2, -4, 3)\) and point \((1, 2, -3)\) into the general equation: \[ 2(x - 1) - 4(y - 2) + 3(z + 3) = 0. \]

Step 4: Expand and Simplify

Expand and simplify the equation:\[ 2x - 2 - 4y + 8 + 3z + 9 = 0. \]Combine like terms to get:\[ 2x - 4y + 3z + 15 = 0. \]

Step 5: Final Equation of the Plane

The simplified equation of the plane is:\[ 2x - 4y + 3z + 15 = 0. \] This is the equation of the plane with the given normal vector, passing through the specified point.

Key Concepts

Understanding the Normal Vector

The normal vector, often denoted as \(\mathbf{n}\), is a crucial element in understanding planes in three-dimensional space. This vector is perpendicular to the plane, meaning it points directly away from the surface of the plane. A normal vector for plane equations can be expressed as \(\mathbf{n} = (a, b, c)\), where \(a\), \(b\), and \(c\) are constants that define the orientation of the plane.

• The components \(a\), \(b\), and \(c\) of the normal vector impact the slope and tilt of the plane.
• A change in any component of \(\mathbf{n}\) results in a different plane orientation, making these values critical for the plane's identity.

Understanding the normal vector helps in determining how the plane cuts through the axes in geometry. It assures us that any change in its direction alters the plane's position in space.

Point on a Plane and Its Role

A specific point on a plane is essential to define its location in three-dimensional space. When we have a point \(P(x_0, y_0, z_0)\), it signifies a location through which the plane passes. This point is combined with the normal vector to form the equation of the plane.

• The point \(P\) helps anchor the plane in space.
• It allows us to precisely establish the plane's position, distinct from other parallel planes, which might have the same normal vector but pass through different points.

In the problem we discussed, the point \((1, 2, -3)\) ensures the calculated equation represents a unique plane in space, reinforcing how important a reference point is in determining the plane’s equation.

General Equation of a Plane

The general equation of a plane is a widely used formula in vector mathematics. Given a normal vector \(\mathbf{n} = (a, b, c)\) and a point \((x_0, y_0, z_0)\) on the plane, the equation is expressed as:\[a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\]This expression summarizes the relationship between the plane's orientation and its position.

• \(a(x - x_0)\): Describes how the plane moves along the x-axis.
• \(b(y - y_0)\): Reflects movement along the y-axis.
• \(c(z - z_0)\): Indicates how the plane's position changes along the z-axis.

All combined, these components illustrate the unique space the plane occupies. By substituting the specific normal vector and point values, we derive the exact location and angles of the plane in space. Hence, the general equation provides a powerful tool for easily defining any plane given the necessary elements.