Cube Geometry: Finding The Length Of Segment MN

Hey guys! Let's dive into a fascinating problem involving cube geometry. We're going to figure out how to find the length of a specific segment within a cube. The problem might seem a bit tricky at first, but don't worry, we'll break it down step by step so it's super clear. Grab your thinking caps, and let's get started!

Problem Statement

Imagine a cube, labeled $ABCDA_1B_1C_1D_1$. Inside this cube, we have a point $M$ on the segment $AC_1$ such that $AM$ is one-fourth of the length of $AC_1$. We also have another point $N$ on the segment $DD_1$, and $DN$ is three-fourths of the length of $DD_1$. If the cube's edge length is $\sqrt{56}$, our mission is to find the length of the segment $MN$. Sounds like a fun challenge, right?

Understanding the Cube and Key Points

Before we jump into calculations, let's make sure we have a solid picture of what's going on. A cube, as we know, has all its edges equal in length, and all its faces are squares. In our case, each edge has a length of $\sqrt{56}$.

Now, let's pinpoint the key points. $AC_1$ is a diagonal that runs across the cube, connecting vertex $A$ to the opposite vertex $C_1$. Point $M$ sits on this diagonal, closer to $A$ than to $C_1$. On the other hand, $DD_1$ is an edge of the cube, and point $N$ is located on this edge, closer to $D_1$ than to $D$. Visualizing these points within the cube is crucial for solving the problem. Think of it like setting the stage for our geometric drama!

Setting Up a Coordinate System

To make things easier, we'll use a coordinate system. This is a neat trick in geometry that turns spatial problems into algebraic ones. Let's place our cube in a 3D coordinate system with vertex $A$ at the origin $(0, 0, 0)$. We'll align the edges $AB$, $AD$, and $AA_1$ with the x, y, and z axes, respectively. This setup makes it straightforward to define the coordinates of the cube's vertices.

Since the edge length is $\sqrt{56}$, the coordinates of the vertices become:

• $A(0, 0, 0)$
• $B(\sqrt{56}, 0, 0)$
• $C(\sqrt{56}, \sqrt{56}, 0)$
• $D(0, \sqrt{56}, 0)$
• $A_1(0, 0, \sqrt{56})$
• $B_1(\sqrt{56}, 0, \sqrt{56})$
• $C_1(\sqrt{56}, \sqrt{56}, \sqrt{56})$
• $D_1(0, \sqrt{56}, \sqrt{56})$

With these coordinates in hand, we're ready to find the coordinates of points $M$ and $N$.

Finding the Coordinates of Points M and N

Point M Coordinates

Remember, point $M$ is on the segment $AC_1$, and AM = rac{1}{4}AC_1. This means we can use the section formula to find the coordinates of $M$. The section formula is a handy tool that helps us find the coordinates of a point dividing a line segment in a given ratio. In our case, $M$ divides $AC_1$ in the ratio $1:3$ (since $AM$ is one-fourth of the total length $AC_1$).

The coordinates of $M$ are given by:

M = A + rac{1}{4}(C_1 - A)

Substituting the coordinates of $A$ and $C_1$, we get:

M = (0, 0, 0) + rac{1}{4}((\sqrt{56}, \sqrt{56}, \sqrt{56}) - (0, 0, 0))

M = rac{1}{4}(\sqrt{56}, \sqrt{56}, \sqrt{56})

$M = (\frac{\sqrt{56}}{4}, \frac{\sqrt{56}}{4}, \frac{\sqrt{56}}{4})$

Point N Coordinates

Point $N$ is on the segment $DD_1$, and DN = rac{3}{4}DD_1. Again, we can use the section formula, but this time, $N$ divides $DD_1$ in the ratio $3:1$.

The coordinates of $N$ are given by:

N = D + rac{3}{4}(D_1 - D)

Substituting the coordinates of $D$ and $D_1$, we get:

N = (0, \sqrt{56}, 0) + rac{3}{4}((0, \sqrt{56}, \sqrt{56}) - (0, \sqrt{56}, 0))

N = (0, \sqrt{56}, 0) + rac{3}{4}(0, 0, \sqrt{56})

$N = (0, \sqrt{56}, \frac{3\sqrt{56}}{4})$

Now that we have the coordinates of both $M$ and $N$, we're one step closer to finding the length of segment $MN$!

Calculating the Length of Segment MN

With the coordinates of $M$ and $N$ in hand, we can finally calculate the length of the segment $MN$. We'll use the distance formula in 3D space, which is a direct extension of the Pythagorean theorem. This formula tells us the distance between two points in 3D space given their coordinates.

If we have two points, $M(x_1, y_1, z_1)$ and $N(x_2, y_2, z_2)$, the distance between them is:

$MN = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Plugging in the coordinates of $M(\frac{\sqrt{56}}{4}, \frac{\sqrt{56}}{4}, \frac{\sqrt{56}}{4})$ and $N(0, \sqrt{56}, \frac{3\sqrt{56}}{4})$, we get:

$MN = \sqrt{ (0 - \frac{\sqrt{56}}{4})^2 + (\sqrt{56} - \frac{\sqrt{56}}{4})^2 + (\frac{3\sqrt{56}}{4} - \frac{\sqrt{56}}{4})^2 }$

Let's simplify this step by step:

$MN = \sqrt{ (-\frac{\sqrt{56}}{4})^2 + (\frac{3\sqrt{56}}{4})^2 + (\frac{2\sqrt{56}}{4})^2 }$

$MN = \sqrt{ \frac{56}{16} + \frac{9 * 56}{16} + \frac{4 * 56}{16} }$

$MN = \sqrt{ \frac{56(1 + 9 + 4)}{16} }$

$MN = \sqrt{ \frac{56 * 14}{16} }$

$MN = \sqrt{ \frac{784}{16} }$

$MN = \sqrt{49}$

$MN = 7$

So, the length of the segment $MN$ is 7. Awesome!

Solution and Conclusion

Therefore, the length of the segment $MN$ is 7. We solved it! By carefully setting up a coordinate system, finding the coordinates of points $M$ and $N$ using the section formula, and then applying the distance formula, we cracked this cube geometry problem. Remember, the key to tackling these problems is breaking them down into smaller, manageable steps. Geometry can be super fun when you approach it systematically!

This problem showcases how geometry and algebra can work together to solve complex spatial challenges. Next time you encounter a geometry problem, think about how you can use coordinates and formulas to simplify it. Keep practicing, and you'll become a geometry whiz in no time! Keep exploring, guys!