Problem 8 - APEX Calculus

Question

Give the equation of the described plane in standard and general forms. Passes through (1,3,5) and has normal vector \(\vec{n}=\langle 0,2,4\rangle\)

Step-by-Step Solution

Answer

The plane in standard form is \(0x + 2y + 4z = 26\); in general form, \(2y + 4z - 26 = 0\).

Step 1: Understand the Plane Equation

The general form of a plane equation is given by \(Ax + By + Cz = D\), where \(\vec{n} = \langle A, B, C \rangle\) is the normal vector of the plane. Here, \(A = 0\), \(B = 2\), and \(C = 4\). The plane passes through the point \((1, 3, 5)\).

Step 2: Plug Point into Plane Equation

Substitute the point \((1,3,5)\) into the equation \(Ax + By + Cz = D\) to find \(D\). Thus, \(0 \times 1 + 2 \times 3 + 4 \times 5 = D\). Calculating these gives \(0 + 6 + 20 = D\), so \(D = 26\).

Step 3: Write Plane Equation in Standard Form

With \(A = 0\), \(B = 2\), \(C = 4\), and \(D = 26\), substitute these values into the standard form to get the plane equation: \(0x + 2y + 4z = 26\).

Step 4: Convert to General Form

The general form of a plane equation is usually written as \(Ax + By + Cz - D = 0\). Using the values obtained earlier, write: \(0x + 2y + 4z - 26 = 0\). This simplifies to \(2y + 4z - 26 = 0\) as the general form.

Key Concepts

Normal Vector

A normal vector is essential in defining the orientation of a plane in 3D space. It is a vector that is perpendicular to the surface of the plane. For any plane, given by the equation \(Ax + By + Cz = D\), the corresponding normal vector is \(\vec{n} = \langle A, B, C \rangle\).
In our exercise, the normal vector provided is \(\vec{n} = \langle 0, 2, 4 \rangle\). This vector indicates the direction in which the plane extends across space and is perpendicular to every line lying on the plane. So any plane defined by this normal vector will have the same directionality.
It is important to understand the normal vector's role as it ultimately influences the plane's alignment and its equations.

Standard Form

The standard form of a plane equation is expressed as \(Ax + By + Cz = D\). This equation requires a normal vector \(\vec{n} = \langle A, B, C \rangle\) and a specific point \((x_0, y_0, z_0)\) through which the plane passes. To find \(D\), we substitute the coordinates of this point into the plane equation:

• \(A \times x_0 + B \times y_0 + C \times z_0 = D\)

In the given exercise, the point is \((1, 3, 5)\), and the normal vector provides \(A = 0, B = 2, C = 4\). Using the coordinates, we calculate \(D\) as follows:

• \(0 \times 1 + 2 \times 3 + 4 \times 5 = D\)
• \(0 + 6 + 20 = 26\)

Thus, the standard form of the plane becomes \(0x + 2y + 4z = 26\). This format is quite straightforward and helpful in identifying the scale factors of each variable that affect the plane's geometry.

General Form

The general form of a plane equation is a variant of the standard form but is written as \(Ax + By + Cz - D = 0\). This presentation is a rearrangement that aligns more closely with algebraic expressions by moving all terms to one side of the equation.
In our solved exercise, the values of \(A, B, C,\) and \(D\) remain the same, and from \(0x + 2y + 4z = 26\), it is rearranged to form the general equation:

• \(0x + 2y + 4z - 26 = 0\)

Which simplifies to \(2y + 4z - 26 = 0\). This form is highly useful in algebraic computations and when determining intersections or parallelism between planes.

Points in 3D Space

Points in 3D space are defined by three coordinates, usually written as \((x, y, z)\). These coordinates indicate a specific location within a three-dimensional environment. Each coordinate represents a distance along one of the three axes: the \(x\)-axis, \(y\)-axis, and \(z\)-axis.
In the context of the plane equation, a point \((x_0, y_0, z_0)\) is used to help define the plane's position by locating where the plane intercepts space. Thus a point greatly influences both the standard and general forms by dictating the value of \(D\) in the equations.
For the given exercise, the point \((1, 3, 5)\) helps establish exactly where the plane exists within 3D space, ensuring that the plane equation accurately represents all points lying on that plane. Understanding how points in 3D space interact with plane equations is crucial for visualizing and solving geometric problems.