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

Concept: To find the shortest distance between two skew lines in a 3D coordinate system, we can set up vectors representing the lines and use the standard vector distance formula: \[ d = \frac{\left| (\vec{r}_2 - \vec{r}_1) \cdot (\vec{b}_1 \times \vec{b}_2) \right|}{|\vec{b}_1 \times \vec{b}_2|} \] where $\vec{b}_1$ and $\vec{b}_2$ are the directional vectors of the lines, and $\vec{r}_1, \vec{r}_2$ are position points located on each individual line. Step 1: Set up a 3D coordinate frame for the folded shape.
Let the folded diagonal line $AC$ lie along the $x$-axis, with the midpoint of the diagonal located at the origin $O(0,0,0)$. We are given that the total length of the diagonal is $2a$, so the coordinates of the vertices on this axis are: \[ A = (-a, 0, 0) \quad \text{and} \quad C = (a, 0, 0) \] Since the square was folded so that the two triangular planes are perpendicular to each other:
• Let the plane of $\Delta ABC$ lie entirely within the horizontal $xy$-plane. The vertex $B$ sits on the $y$-axis at a distance of $a$ from the origin: $B = (0, a, 0)$.
• Let the plane of $\Delta ADC$ lie entirely within the vertical $xz$-plane. The vertex $D$ sits on the $z$-axis at a distance of $a$ from the origin: $D = (0, 0, a)$.

Step 2: Write out the directional vectors for lines $AB$ and $CD$.
Using our vertex coordinates, calculate the directional vectors for each straight line:
• Directional vector for line $AB$: $\vec{b}_1 = \vec{B} - \vec{A} = (0 - (-a))\hat{i} + (a - 0)\hat{j} + 0\hat{k} = a\hat{i} + a\hat{j}$
• Directional vector for line $CD$: $\vec{b}_2 = \vec{D} - \vec{C} = (0 - a)\hat{i} + 0\hat{j} + (a - 0)\hat{k} = -a\hat{i} + a\hat{k}$ We can drop the common scalar scaling factor $a$ to get the simplified base direction vectors: \[ \vec{b}_1 = (1, 1, 0) \quad \text{and} \quad \vec{b}_2 = (-1, 0, 1) \]

Step 3: Compute the cross product of the direction vectors.
Find the normal vector $\vec{n} = \vec{b}_1 \times \vec{b}_2$ by evaluating the vector determinant: \[ \vec{b}_1 \times \vec{b}_2 = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} 1 & 1 & 0 -1 & 0 & 1 \end{matrix} \right| = \hat{i}(1 - 0) - \hat{j}(1 - 0) + \hat{k}(0 - (-1)) = \hat{i} - \hat{j} + \hat{k} \] Calculate the absolute magnitude of this cross product vector: \[ |\vec{b}_1 \times \vec{b}_2| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3} \]

Step 4: Calculate the shortest distance.
Pick a point on each line to form our connection position vector: select point $A(-a, 0, 0)$ on line $AB$ and point $C(a, 0, 0)$ on line $CD$: \[ \vec{r}_2 - \vec{r}_1 = \vec{C} - \vec{A} = (a - (-a), 0, 0) = (2a, 0, 0) \] Now take the dot product of this connection vector with our normal direction vector: \[ (\vec{r}_2 - \vec{r}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (2a, 0, 0) \cdot (1, -1, 1) = 2a \cdot 1 = 2a \] Substitute these values back into our core skew line distance formula: \[ d = \frac{2a}{\sqrt{3}} \] This matches option (A) perfectly.