Question

Consider a square $ABCD$ of diagonal length $2a$. The square is folded along the diagonal $AC$ so that the plane of $\Delta ABC$ is perpendicular to the plane of $\Delta ADC$. In this case the shortest distance between $AB$ and $CD$ is:

• A. $\frac{2a}{\sqrt{3}}$
• B. $\frac{a}{2\sqrt{3}}$
• C. $\frac{a}{\sqrt{3}}$
• D. $\frac{\sqrt{3}a}{2}$

Hint:

When folding conics or squares into perpendicular 3D vector spaces, always anchor the fold boundary directly onto a primary coordinate axis like the $x$-axis. It forces the remaining vertices to land safely on the remaining perpendicular axes, reducing the vector cross products to simple 1 and 0 calculations!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We have a 3D geometry problem. After folding, \( AB \) and \( CD \) are skew lines. The shortest distance between skew lines is the common perpendicular.
Step 2: Key Formula or Approach:
1. Set up a coordinate system with the origin at the center of the diagonal \( AC \).
2. Find the equations of lines \( AB \) and \( CD \).
3. Use the shortest distance formula: \( d = \frac{|(\vec{b} - \vec{a}) \cdot (\vec{u} \times \vec{v})|}{|\vec{u} \times \vec{v}|} \).
Step 3: Detailed Explanation:
Let the midpoint of diagonal \( AC \) be the origin \( (0, 0, 0) \).
Since diagonal length is \( 2a \), let \( A = (-a, 0, 0) \) and \( C = (a, 0, 0) \).
Before folding, \( B \) and \( D \) were at \( (0, a, 0) \) and \( (0, -a, 0) \).
After folding along \( AC \) so planes are perpendicular:
Let \( B = (0, a, 0) \) and \( D = (0, 0, a) \).
Coordinates: \( A(-a, 0, 0), B(0, a, 0), C(a, 0, 0), D(0, 0, a) \).
Line \( AB \): Passes through \( A(-a, 0, 0) \) with direction vector \( \vec{u} = B - A = (a, a, 0) \).
Line \( CD \): Passes through \( C(a, 0, 0) \) with direction vector \( \vec{v} = D - C = (-a, 0, a) \).
Normal vector \( \vec{n} = \vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
a & a & 0
-a & 0 & a \end{vmatrix} = a^2\hat{i} - a^2\hat{j} + a^2\hat{k} \).
Unit normal \( \hat{n} = \frac{(1, -1, 1)}{\sqrt{3}} \).
Shortest distance \( d = |(C - A) \cdot \hat{n}| = |(2a, 0, 0) \cdot \frac{(1, -1, 1)}{\sqrt{3}}| = \frac{2a}{\sqrt{3}} \).
Step 4: Final Answer:
The shortest distance is \( \frac{2a}{\sqrt{3}} \).