Problem 10 - APEX Calculus

Question

Give the equation of the described plane in standard and general forms. Passes through the points (5,3,8),(6,4,9) and (3,3,3) .

Step-by-Step Solution

Answer

The plane equation is \( 5x + 3y + 2z = 50 \).

Step 1: Determine Two Vectors from Points

To find the equation of the plane, we first need two vectors that lie on the plane. Let's designate the three given points as \( A(5,3,8) \), \( B(6,4,9) \), and \( C(3,3,3) \). Calculate two vectors using these points: \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \).\[ \overrightarrow{AB} = B - A = (6-5, 4-3, 9-8) = (1, 1, 1) \]\[ \overrightarrow{AC} = C - A = (3-5, 3-3, 3-8) = (-2, 0, -5) \]

Step 2: Find the Normal Vector by Cross Product

The normal vector (\( \mathbf{n} \)) to the plane can be found by calculating the cross product of the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \).\[ \mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC} \]\[ \mathbf{n} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \1 & 1 & 1 \-2 & 0 & -5 \\end{vmatrix} \]This determinant expands to:\[ \mathbf{n} = (1 \cdot 0 - 1 \cdot (-5)) \mathbf{i} - (1 \cdot -5 - 1 \cdot -2) \mathbf{j} + (1 \cdot 0 - 1 \cdot (-2)) \mathbf{k} \]\[ \mathbf{n} = (0 + 5) \mathbf{i} - (-5 + 2) \mathbf{j} + (0 + 2) \mathbf{k} \]\[ \mathbf{n} = (5, 3, 2) \]

Step 3: Write the Plane Equation in Standard Form

With the normal vector \( \mathbf{n} = (5, 3, 2) \), substitute it along with the point \( A(5, 3, 8) \) into the plane equation formula \( ax + by + cz = d \), where \( a = 5 \), \( b = 3 \), and \( c = 2 \). Plug in point \( A \) to find \( d \):\[ 5(5) + 3(3) + 2(8) = d \]\[ 25 + 9 + 16 = d \]\[ d = 50 \]Thus, the equation of the plane in standard form is \( 5x + 3y + 2z = 50 \) .

Step 4: Write the Plane Equation in General Form

The equation \( 5x + 3y + 2z = 50 \) is already in the general form of the plane, which is expressed as \( ax + by + cz = d \), where \( a, b, c \), and \( d \) are constants. There are no modifications needed for this form.

Key Concepts

Vectors

A vector is a fundamental mathematical object used to represent both direction and magnitude. In our problem, vectors help us describe positions and movements in space. We can think of a vector as an arrow pointing from one location to another. For instance, the vector \( \overrightarrow{AB} \) is the arrow from point \( A \) to point \( B \).

Here, given the points \( A(5,3,8) \) and \( B(6,4,9) \), we find the vector \( \overrightarrow{AB} \) as follows:

• Subtract the coordinates of \( A \) from \( B \)
• \( \overrightarrow{AB} = (6-5, 4-3, 9-8) = (1, 1, 1) \)

Vectors like \( \overrightarrow{AB} \) tell us about the relative position and direction from one point to another, and are crucial in defining planes in space.

Cross Product

The cross product is a powerful tool in vector mathematics used to find a vector that is perpendicular, or normal, to two given vectors. This is particularly useful when we need to derive the equation of a plane.
For two vectors, \( \overrightarrow{AB} = (1, 1, 1) \) and \( \overrightarrow{AC} = (-2, 0, -5) \), the cross product \( \overrightarrow{AB} \times \overrightarrow{AC} \) provides us with a normal vector to the plane containing these vectors:

• Use the formula \( \mathbf{n} = (b_1c_2 - b_2c_1, c_1a_2 - c_2a_1, a_1b_2 - a_2b_1) \)
• \( \mathbf{n} = (0 \times (-5) - 1 \times 0, 1 \times (-5) - 1 \times (-2), 1 \times 0 - (-2) \times 1) \)
• Simplified, this gives us \( \mathbf{n} = (5, 3, 2) \)

The cross product thus yields the normal vector \( \mathbf{n} \), essential to defining the plane.

Normal Vector

The normal vector of a plane is crucial because it provides the direction in which the plane tilts or inclines. It is perpendicular to any vector lying directly on the plane, and pivotal in defining the plane's orientation.
In our exercise, the normal vector was derived using the cross product and found to be \( (5, 3, 2) \).

• This means the plane is perpendicular to an invisible arrow pointing along the vector \( (5, 3, 2) \)
• The normal vector's components \( a, b, c \) become part of the plane's equation

Essentially, the normal vector anchors the concept of the plane into three-dimensional space, and is indispensable in creating the plane's equation.

Standard Form of a Plane

A plane's standard form equation is integral in representing its geometric properties in a simplified mathematical form. It is commonly expressed as \( ax + by + cz = d \), where \( a \), \( b \), and \( c \) come from the normal vector.
In our scenario, substituting the normal vector \( (5, 3, 2) \) and the specific point \( A(5, 3, 8) \) into the equation, we find:

• Calculate \( d \) by plugging \( A \) into the equation: \( 5(5) + 3(3) + 2(8) = d \)
• This solves to \( d = 50 \)

Thus, the standard form equation of the plane becomes \( 5x + 3y + 2z = 50 \). Those coefficients hold the keys to understanding the plane's inclination and position in space.

General Form of a Plane

The general form of a plane equation is quite similar to the standard form, often written as \( ax + by + cz = d \). In fact, the two terms are usually interchangeable, especially in elementary geometry.
For our calculated plane, the general form was already achieved as \( 5x + 3y + 2z = 50 \). There are no additional transformations or simplifications necessary:

• It remains as it is once derived from the normal vector and known point
• Serves as a clear method for specifying the plane in algebraic terms

Seeing the equation in this general form helps when solving related geometric problems, such as finding intersections with lines or other planes.