Problem 9 - APEX Calculus

Question

Give the equation of the described plane in standard and general forms. Passes through the points (1,2,3),(3,-1,4) and (1,0,1) .

Step-by-Step Solution

Answer

The plane equation is \(2x + y - z = 1\) in general form, \(8(x - 1) + 4(y - 2) - 4(z - 3) = 0\) in standard form.

Step 1: Find two vectors on the plane

To begin, we find two vectors that lie on the plane using the given points. The first vector can be found using two of the points, say (1,2,3) and (3,-1,4). The vector is \( \mathbf{v_1} = (3-1, -1-2, 4-3) = (2,-3,1) \). Similarly, the second vector can be found using the points (1,2,3) and (1,0,1), which gives \( \mathbf{v_2} = (1-1, 0-2, 1-3) = (0,-2,-2) \).

Step 2: Calculate the normal vector using the cross product

Next, we determine the normal vector to the plane by taking the cross product of \(- \mathbf{v_1} \) and \(- \mathbf{v_2} \).\[ \mathbf{n} = \mathbf{v_1} \times \mathbf{v_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & -3 & 1 \ 0 & -2 & -2 \end{vmatrix} \]This determinant results in \[ \mathbf{n} = \mathbf{i}( ( -3 \cdot -2) - (1 \cdot -2) ) - \mathbf{j}( (2 \cdot -2) - (1 \cdot 0) ) + \mathbf{k}( (2 \cdot -2) - (0 \cdot -3) ) \]\[ = \mathbf{i}(6 + 2) - \mathbf{j}(-4) + \mathbf{k}(-4) \]\[ = 8\mathbf{i} + 4\mathbf{j} - 4\mathbf{k} \].Therefore, the normal vector is \( \mathbf{n} = (8, 4, -4) \).

Step 3: Write the plane equation in standard form

With the normal vector \( \mathbf{n} = (8, 4, -4) \), the standard form of the plane can be written using the general point \( (x, y, z) \) and a point on the plane, say (1,2,3):\[ 8(x - 1) + 4(y - 2) - 4(z - 3) = 0 \].

Step 4: Expand to get the general form of the plane equation

Now, we simplify the standard form to obtain the general form:\[ 8x - 8 + 4y - 8 - 4z + 12 = 0 \]Combine like terms:\[ 8x + 4y - 4z - 4 = 0 \].This is the general form, but we can simplify further by dividing everything by 4:\[ 2x + y - z - 1 = 0 \].

Step 5: Verify the equation with the given points

To ensure correctness, we verify using the points. For example, using (1,2,3):\[ 2(1) + 2 - 3 - 1 = 2 + 2 - 3 - 1 = 0 \],which satisfies the equation. Similarly, check for (3,-1,4) and (1,0,1) by substituting into the equation \( 2x + y - z = 1 \) to confirm they also satisfy the equation.

Key Concepts

Cross Product

The cross product is a fundamental operation in vector mathematics. It helps us determine a vector that is perpendicular to two given vectors. This operation is especially useful when we need to find the normal vector of a plane, as is the case in the exercise above.
To compute the cross product of two vectors, say \( \mathbf{v_1} = (a, b, c) \) and \( \mathbf{v_2} = (d, e, f) \), we use the following determinant form:

• \( \mathbf{n} = \mathbf{v_1} \times \mathbf{v_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a & b & c \ d & e & f \end{vmatrix} \)
• This expands into \( \mathbf{n} = \mathbf{i}(bf-ce) - \mathbf{j}(af-cd) + \mathbf{k}(ae-bd) \).

The computed vector \( \mathbf{n} \) is orthogonal to both \( \mathbf{v_1} \) and \( \mathbf{v_2} \), making it ideal for describing planes.

Vector Mathematics

Vector mathematics is a branch of mathematics that deals with vectors, which are quantities defined by both magnitude and direction. In the context of the equation of a plane, vectors represent directions within the plane or orthogonal to it.
Vectors are used to calculate various products like the cross product which, as discussed earlier, helps find a perpendicular vector. Other operations in vector mathematics include addition, subtraction, and scalar multiplication.

• Addition combines vectors \( \mathbf{u} + \mathbf{v} = (u_1 + v_1, u_2 + v_2, u_3 + v_3) \).
• Subtraction involves \( \mathbf{u} - \mathbf{v} = (u_1 - v_1, u_2 - v_2, u_3 - v_3) \).
• Scalar multiplication scales the vector \( c\mathbf{v} = (cv_1, cv_2, cv_3) \).

Mastering vector mathematics is key to understanding how planes and other geometric objects are described in space.

Plane Representation

Representing a plane in three-dimensional space is essential for various applications in geometry and physics. A plane can be easily represented by an equation derived using a point on the plane and a normal vector.
For our exercise, the standard form of a plane equation is expressed with a normal vector \( \mathbf{n} = (a, b, c) \) as follows:

• \( a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 \), where \((x_0, y_0, z_0)\) is a point on the plane.

Expanding this leads to the general form \( ax + by + cz + d = 0 \), which simplifies plane calculations and checks if a point lies on it.
Using this approach ensures that the calculations are consistent with the spatial properties of vectors and planes.

Normal Vector

A normal vector is a vector that is perpendicular to a surface or a plane. It is crucial for defining planes, as it conveys the orientation of the plane.
In the exercise, the normal vector is calculated using the cross product of two vectors on the plane, resulting in \( \mathbf{n} = (8, 4, -4) \). This vector acts like an anchor-point for the plane’s orientation in space.
The normal vector \( \mathbf{n} = (a, b, c) \) becomes part of the plane equation, influencing its form \( ax + by + cz = d \).

• The coefficients \( a, b, \) and \( c \) directly inform the angle and direction the plane sits relative to the axes.
• Having a clear understanding of the normal vector helps tremendously in fields such as computer graphics, physics simulations, and more.

Overall, the normal vector is a pivot concept that aids in comprehensively describing the plane.