Question

Verify that the two planes are parallel, and find the distance between the planes. $$ \begin{array}{l} -3 x+6 y+7 z=1 \\ 6 x-12 y-14 z=25 \end{array} $$

Short answer

Yes, the two planes are parallel, and the distance between the planes is approximated as 2.78 units.

Step-by-step solution

Verification of the Parallelism

Check if the equation coefficients of the given planes are proportional. The normal vector for the first plane is given by the coefficients (-3, 6, 7), while for the second plane it is (6, -12, -14). The second set of coefficients is actually -2 times the first set, so the planes are indeed parallel.

Calculate the Distance

Find the distance between the two parallel planes by using the distance formula. Here, a=-3, b=6, c=7 for first plane and for the second plane a=6, b=-12, c=-14. c1 = -1 and c2 = -25. Substitute these into the formula, we get d = |-1 - (-25)| / sqrt((-3)^2 + 6^2 + 7^2) = 24 / sqrt(74) = 24/ sqrt(74) = 24 / sqrt(74).

Simplify the Result

The expression can be simplified, finishing the calculation provides the distance between the planes as d = 24 / sqrt(74) = 2.78 units (rounded to the nearest hundredths).

Key concepts

Parallel Planes

In geometry, parallel planes are surfaces that always maintain the same distance from each other and never intersect, much like parallel lines in a two-dimensional space. Charting this concept unto three-dimensional space, we have two planes that are parallel if their normal vectors are either identical or scalar multiples of one another.

This concept plays a vital role in understanding the spatial relationship between flat surfaces in three-dimensional geometry. When two planes are described by linear equations, we recognize their parallelism by examining the coefficients of x, y, and z. If the coefficients in one plane's equation are proportional to the coefficients in another plane's equation, as seen in the given exercise, the planes are indeed parallel.

Distance Formula for Planes

Once it is established that two planes are parallel, we can determine the distance between them using a special formula. This distance is the perpendicular line segment that joins the two planes. To calculate it, we use the formula:\[ d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2 + c^2}} \]Here, a, b, and c are the coefficients of the normal vector to either plane, and c_1 and c_2 are the scalar constants from the plane equations. Notably, this formula derives from the concept of the shortest distance from a point to a plane, adapted for two planes. The simplicity of this formula enables students to compute the distance efficiently, provided they have the correct coefficients and understand how to apply the formula.

Vector Normal to Plane

A vector normal to a plane is a three-dimensional vector that is perpendicular to every line lying on the plane. This vector characterizes the orientation of the plane in space and is essential in many geometric and physical calculations. For a plane described by the general equation \( Ax + By + Cz + D = 0 \), the normal vector is \( \vec{n} = (A, B, C) \).

In the context of our exercise, the normal vectors of both planes can be directly derived from the coefficients of x, y, and z in their equations. These vectors are pivotal in not only confirming the parallelism of the planes but also in calculating the distance between them using the aforementioned distance formula.

Proportional Coefficients Method

The proportional coefficients method is a straightforward way to verify the parallelism between two planes. When looking at the coefficients of the variables x, y, and z in the plane equations, if one plane's coefficients can be obtained by multiplying the other's by a single scalar factor, the planes are parallel. The method is efficient because it avoids the need for visualizing the planes or performing complex calculations.

For example, in the exercise, by comparing the coefficients of the two plane equations, we find that multiplying the coefficients of the first by -2 yields the coefficients of the second. This quick assessment efficiently confirms their parallelism and brings simplicity to the overall process, especially when working with multiple planes.