Problem 23 - Calculus an Applied Approach

Question

Determine whether the planes \(a_{1} x+b_{1} y+c_{1} z=d_{1}\) and \(a_{2} x+b_{2} y+c_{2} z=d_{2}\) are parallel, perpendicular, or neither. The planes are parallel if there exists a nonzero constant \(k\) such that \(a_{1}=k a_{2}, b_{1}=k b_{2}\), and \(c_{1}=k c_{2},\) and are perpendicular if \(a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0\). $$ x-5 y-z=1,5 x-25 y-5 z=-3 $$

Step-by-Step Solution

Answer

The given planes are parallel.

Step 1: Identify coefficients

For the first plane \(x-5y-z=1\), the coefficients are \(a_{1}=1\), \(b_{1}=-5\), \(c_{1}=-1\). For the second plane \(5x-25y-5z=-3\), the coefficients are \(a_{2}=5\), \(b_{2}=-25\), \(c_{2}=-5\).

Step 2: Check if planes are parallel

For the planes to be parallel, there must exist a nonzero constant \(k\) such that \(a_{1}=k a_{2}\), \(b_{1}=k b_{2}\), and \(c_{1}=k c_{2}\). Here, as \(5 = 5*1\), \(-25 = 5*(-5)\), and \(-5 = 5*(-1)\), it can be concluded that \(k=5\), hence the planes are parallel.

Step 3: Check if planes are perpendicular

The planes are perpendicular if \(a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0\). This step is already fully covered in Step 2, as the planes were found to be parallel. Thus, the planes are not perpendicular.

Key Concepts

Parallel Planes

In three-dimensional geometry, two planes are considered parallel if they do not intersect. This means they are always a fixed distance apart from each other. To determine if two planes are parallel, we analyze their normal vectors.A plane can be expressed in the general form \(ax + by + cz = d\), where \(a\), \(b\), and \(c\) are the coefficients representing the normal vector of the plane. For two planes given by equations: \(a_1x + b_1y + c_1z = d_1\) and \(a_2x + b_2y + c_2z = d_2\), they are parallel if the normal vectors are scalar multiples of each other. This can be checked by assessing if there is a nonzero constant \(k\) such that:

• \(a_1 = k \, a_2\)
• \(b_1 = k \, b_2\)
• \(c_1 = k \, c_2\)

Therefore, if you can find a consistent \(k\) for all three comparisons, the planes are parallel. For example, with planes \(x - 5y - z = 1\) and \(5x - 25y - 5z = -3\), calculations show \(k = 5\). Hence, they are parallel.

Perpendicular Planes

Perpendicular planes intersect at a right angle. This distinct relationship can be identified by analyzing their normal vectors. If two planes, \(a_1x + b_1y + c_1z = d_1\) and \(a_2x + b_2y + c_2z = d_2\), are perpendicular, the dot product of their normal vectors is zero.The dot product is calculated by combining the products of the corresponding coefficients:\[a_1a_2 + b_1b_2 + c_1c_2 = 0\]In the equations provided, the normal vectors do not satisfy this condition because their dot product yields a non-zero result. Specifically, when calculated for the given example, \(1 \times 5 + (-5) \times (-25) + (-1) \times (-5)\) results in a nonzero value. Therefore, these planes are not perpendicular.

Plane Equations

Plane equations in three-dimensional space describe a flat surface using a linear equation. The general form of a plane equation is \(ax + by + cz = d\). This format incorporates:

• Coefficients \(a\), \(b\), and \(c\), which determine the plane's orientation through the normal vector \(\langle a, b, c \rangle\).
• The variable \(d\), which adjusts the plane's position relative to the origin.

Each plane's equation uniquely defines its positioning and inclination in space. For instance, the equation \(x - 5y - z = 1\) outlines a specific plane determined by the normal vector \(\langle 1, -5, -1 \rangle\) and its specific offset represented by \(1\). Understanding these equations aids in visualizing spatial relationships and interactions such as parallelism and perpendicularity, as seen in the original exercise.