One vertex of a rectangular parallelopiped is at the origin O and the lengths of its edges along x,

Q: Wait, why can't we just use the distance formula between two points? Why are these lines skew?

Great question! In 3D space, lines don't always intersect or stay parallel. Skew lines are lines that lie on different planes and never meet. Because they don't intersect, there is no single 'point' to measure distance from. Instead, we must find the length of the unique line segment that is perpendicular to both lines simultaneously. That is the 'shortest distance'.

Q: How do I know which edge to pick? The problem says 'an edge parallel to the z-axis'. Does it matter which one?

It does not matter! Due to the symmetry of the rectangular parallelepiped, any edge parallel to the $z$-axis that does not pass through $O$ or $P$ will yield the same shortest distance. You can choose the edge passing through $(3,0,0)$ or $(0,4,0)$; the math will lead you to the same elegant result.

Q: Why is the shortest distance formula defined as a dot product with a cross product?

Think of it geometrically. The cross product $\vec{d}_1 \times \vec{d}_2$ gives you a vector perpendicular to both lines. The vector $(\vec{a}_2 - \vec{a}_1)$ connects a point on the first line to a point on the second. By taking the dot product of this connecting vector with the unit normal vector (which is the cross product divided by its magnitude), you are essentially finding the 'projection' of the separation vector onto the direction of the shortest path. It is the most efficient way to collapse 3D complexity into a single scalar value.

Question

One vertex of a rectangular parallelopiped is at the origin $O$ and the lengths of its edges along $x, y$ and $z$ axes are $3, 4$ and $5$ units respectively. Let $P$ be the vertex $(3, 4, 5)$. Then the shortest distance between the diagonal $OP$ and an edge parallel to $z$ axis, not passing through $O$ or $P$ is

Options:

(A) $\frac{12}{\sqrt{5}}$

(B) $12\sqrt{5}$

(C) $\frac{12}{5\sqrt{5}}$

(D) $\frac{12}{5}$

Sigma Percentile · Accepted Answer

### The Geometry of Space: A Journey into 3D

Welcome, student. Today, we are not just solving a problem; we are stepping into the third dimension. Imagine you are standing in a room and observing a rectangular parallelepiped sitting in the corner.

Our goal is to find the shortest distance between two lines that seem to dance around each other without ever touching—the skew lines of our box.

### Phase 1: Visualizing the Box

First, let us ground ourselves. We have a box with dimensions $3, 4,$ and $5$ along the $x, y,$ and $z$ axes. The origin $O$ is at $(0,0,0)$, and the vertex $P$ is at $(3,4,5)$.

The diagonal $OP$ is the line segment cutting through the very heart of this box. We consider an edge parallel to the $z$-axis, specifically the one passing through $A(3,0,0)$. This edge is a vertical line segment from $(3,0,0)$ to $(3,0,5)$.

### Phase 2: The Skew Line Challenge

We have two lines: $L_1$ (the diagonal $OP$) and $L_2$ (the edge through $A$). These are skew lines, meaning they never intersect and are not parallel.

To find the shortest distance, we need the common perpendicular. The formula for the shortest distance between two skew lines $L_1: \vec{r} = \vec{a}_1 + \lambda \vec{d}_1$ and $L_2: \vec{r} = \vec{a}_2 + \mu \vec{d}_2$ is given by:

$$d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{d}_1 	imes \vec{d}_2)|}{|\vec{d}_1 	imes \vec{d}_2|}$$

This formula serves as our mathematical compass.

### Phase 3: The Vector Toolkit

Let us define our vectors with precision. For line $L_1$ (the diagonal $OP$), it passes through the origin, so $\vec{a}_1 = \vec{0}$. Its direction vector $\vec{d}_1$ is the vector from $O$ to $P$:

$$\vec{d}_1 = 3\hat{i} + 4\hat{j} + 5\hat{k}$$

For line $L_2$ (the edge through $A(3,0,0)$), it passes through $A$, so $\vec{a}_2 = 3\hat{i}$. Since it is parallel to the $z$-axis, its direction vector $\vec{d}_2$ is simply $\hat{k}$.

### Phase 4: The Calculation

Now, we calculate the cross product $\vec{n} = \vec{d}_1 	imes \vec{d}_2$, which represents the common normal. Substituting our values:

$$\vec{n} = (3\hat{i} + 4\hat{j} + 5\hat{k}) 	imes \hat{k}$$

Using the properties of cross products ($\hat{i} 	imes \hat{k} = -\hat{j}$, $\hat{j} 	imes \hat{k} = \hat{i}$, and $\hat{k} 	imes \hat{k} = 0$), we find:

$$\vec{n} = 3(-\hat{j}) + 4(\hat{i}) + 5(0) = 4\hat{i} - 3\hat{j}$$

The magnitude of this normal vector is:

$$|\vec{n}| = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = 5$$

Finally, we calculate the numerator of the distance formula:

$$(\vec{a}_2 - \vec{a}_1) \cdot \vec{n} = (3\hat{i} - \vec{0}) \cdot (4\hat{i} - 3\hat{j}) = 12$$

The shortest distance is therefore:

$$d = \frac{12}{5}$$

### Conclusion

Look at that! The complexity of 3D space collapses into a simple fraction: **$\frac{12}{5}$**.

This is the beauty of vector algebra. It allows us to navigate the most complex geometric configurations with nothing more than a few dot and cross products. You have successfully navigated the skew lines of the parallelepiped.

Select Answer:

Visualized Solution

Geometry of the Parallelepiped

Identifying Key Vertices

The Diagonal $O P$

The Target Edge

Shortest Distance Concept

Shortest Distance Formula

Vectors for Line $O P$

Vectors for Target Edge

Cross Product Setup

Evaluating the Cross Product

Magnitude of Normal Vector

Evaluating the Numerator

Final Calculation

The Sigma Insight: Shortest Distance Between Two Skew Lines

The Geometry of Space

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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