Question

Find the given distances. The distance between the parallel planes \(x+y+z=0\) and $$ (x-2)+(y-3)+(z+4)=0 $$

Short answer

The distance between the planes is \(3\sqrt{3}\).

Step-by-step solution

Identify the Equations of the Planes

The equations of the parallel planes are given as \(x+y+z=0\) and \(x-2+y-3+z+4=0\). Simplify the second equation: it becomes \(x+y+z-9=0\). So, the equations can be written as \(x+y+z=0\) and \(x+y+z=9\).

Verify the Normal Vector

Both planes have the form \(x+y+z = d\), which indicates that the normal vector to both planes is \(\mathbf{n} = [1, 1, 1]\). This means the planes are indeed parallel since they share the same normal vector.

Formula for Distance Between Parallel Planes

The formula to find the distance \(D\) between two parallel planes \(ax+by+cz=d_1\) and \(ax+by+cz=d_2\) is given by:\[D = \frac{|d_2 - d_1|}{\sqrt{a^2 + b^2 + c^2}}\]Identify \(d_1 = 0\) and \(d_2 = 9\), and \(a = b = c = 1\).

Substitute Values into the Formula

Insert the known values into the distance formula:\[D = \frac{|9 - 0|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{9}{\sqrt{3}}\]

Simplify the Expression

Simplify the expression:\[D = \frac{9}{\sqrt{3}} = \frac{9\sqrt{3}}{3} = 3\sqrt{3}\]

Final Answer

The distance between the parallel planes \(x+y+z=0\) and \(x+y+z=9\) is \(3\sqrt{3}\).

Key concepts

Parallel Planes

Parallel planes are fascinating as they extend infinitely without ever intersecting. Imagine two perfectly flat sheets of paper positioned alongside one another, always equidistant and never meeting or diverging. In mathematics, parallel planes are represented by linear equations that share identical normal vectors. This means the coefficients in front of the variables are the same.

For example, looking at the equations of our given planes: \(x+y+z=0\) and \(x+y+z=9\). Both share the same structure, ensuring they are parallel. They have equal normal vectors despite having different constant terms. This difference in constant values simply shifts the planes apart along a direction perpendicular to their surface, but doesn’t alter their infinite span.

Normal Vector

A normal vector is crucial in understanding the orientation of a plane. It is a vector that points perpendicular to the surface of a plane. Essentially, it helps determine the plane's tilt in three-dimensional space.

For any plane expressed as \(ax+by+cz=d\), the normal vector is given by \[\mathbf{n} = [a, b, c]\]. In our exercise, the normal vector for both planes is \[\mathbf{n} = [1, 1, 1]\], meaning these vectors point in the same direction. This is what confirms that the planes are parallel. Understanding normal vectors can help visualize how planes relate to each other in space, whether they are parallel or at an angle.

Distance Formula for Planes

Calculating the distance between two parallel planes is straightforward once you know they have the same normal vector. The distance formula between two parallel planes \(ax+by+cz=d_1\) and \(ax+by+cz=d_2\) is:

• \[D = \frac{|d_2 - d_1|}{\sqrt{a^2 + b^2 + c^2}}\]

This formula tells us how far apart the planes are by considering the difference in their constant terms. In our scenario, \(d_1 = 0\) and \(d_2 = 9\). Thus, the distance \(D\) is \(\frac{|9 - 0|}{\sqrt{1^2 + 1^2 + 1^2}}\), simplifying to \(3\sqrt{3}\) units.

Using the distance formula provides a clear method to measure separation with ease, even in the complex space of three dimensions.

Simplifying Equations

Simplifying equations is a vital skill in algebra and geometry. It often makes challenging problems more approachable by reducing them to their core essence. In our problem, the second plane's equation started as \(x-2+y-3+z+4=0\).

Through simplification, we combined like terms to reformulate it as \(x+y+z=9\). By handling the arithmetic operations, we streamlined the equation, aligning it with the \(x+y+z=0\) plane's format. Consequently, the adjustments offer clarity and enable effortless parallel plane comparisons.

Simplifying is not just about making equations tidy; it's about enabling patterns to emerge, solving equations more efficiently, and understanding relationships among various mathematical elements.