Problem 21 - Essential Calculus Early Transcendentals

Question

\(21-32=\) Find an equation of the plane. The plane through the point \(\left(-1, \frac{1}{2}, 3\right)\) and with normal vector \(\mathbf{i}+4 \mathbf{j}+\mathbf{k}\)

Step-by-Step Solution

Answer

The equation of the plane is \( x + 4y + z = 4 \).

Step 1: Identify Components

To find the equation of a plane, we need a point on the plane and a normal vector to the plane. We have: \( Point = (-1, \frac{1}{2}, 3) \) and \( Normal\ vector = \mathbf{i} + 4\mathbf{j} + \mathbf{k} \equiv (1, 4, 1) \).

Step 2: General Plane Equation

The general equation of a plane is given by: \( ax + by + cz = d \), where \((a, b, c)\) is the normal vector.

Step 3: Substitute Normal Vector

Using the normal vector \((1, 4, 1)\), the equation of the plane becomes: \( 1x + 4y + 1z = d \).

Step 4: Calculate Constant 'd'

Substitute the point \((x, y, z) = (-1, \frac{1}{2}, 3)\) into the equation to find \(d\). This gives us \( 1(-1) + 4(\frac{1}{2}) + 1(3) = d \).

Step 5: Solve for 'd'

Compute \(-1 + 2 + 3 = d\), which simplifies to \(d = 4\).

Step 6: Final Equation of Plane

Substitute \(d = 4\) into the equation to get the plane: \( x + 4y + z = 4 \).

Key Concepts

Normal Vector

In geometry, a normal vector is a vector that is perpendicular to a surface or a plane. Imagine standing straight with your arms stretched sideward; you are perpendicular to the ground in the same way the normal vector is perpendicular to its plane. The normal vector is key in defining a plane because it dictates its orientation in space.

For instance, consider the vector \( \mathbf{i} + 4\mathbf{j} + \mathbf{k} \), which represents the normal vector \((1, 4, 1)\) in the exercise. This vector indicates the direction that is perpendicular to every point on the plane. With the normal vector, you can easily find the unique plane by utilizing a point on that plane. Understanding this vector allows you to frame the equation of the plane and provides insights into the spatial properties of the plane.

Point on the Plane

A point on a plane provides an anchor to help define the complete equation of that plane. Having a point is essential because although the normal vector gives us the direction perpendicular to the plane, we need a specific point to root the plane in space.

In this exercise, the point \((-1, \frac{1}{2}, 3)\) is used. This means that the plane passes through this particular set of coordinates in three-dimensional space. The presence of this point ensures that we can compute the constant part of the general plane equation. This point, when combined with the normal vector, helps to establish a fixed position and orientation for the plane. It ensures that the equation accurately reflects the plane that encompasses this point.

General Plane Equation

The general equation of a plane is a straightforward yet powerful expression. Typically, it is given as \( ax + by + cz = d \). Here, \((a, b, c)\) represents the components of the normal vector, while \(d\) is a constant that adjusts based on a specific point through which the plane passes.

For example, using the normal vector \((1, 4, 1)\) and the point \((-1, \frac{1}{2}, 3)\), we substitute these into the general equation. The coefficients \(a\), \(b\), and \(c\) are directly derived from the normal vector, and substituting the point helps compute \(d\). This gives the plane equation as \(x + 4y + z = 4\). Calculating \(d\) ensures that the plane encompasses the specified point, balancing the equation. Understanding the general plane equation is essential for solving various geometric problems where defining planes are crucial.