If the shortest distance between the line joining the points (1, 2, 3) and (2, 3, 4) and the line (x

Q: Wait, why can't we just find the distance between the two points given in the problem?

That is a classic trap! The points $(1, 2, 3)$ and $(2, 3, 4)$ are just arbitrary points on the first line, and the point $(1, -1, 2)$ is just one point on the second line. The shortest distance is not necessarily the distance between these specific points. It is the distance between the lines themselves, which is measured along the common perpendicular.

Q: What if the lines were parallel? Does the formula still work?

Great question! If the lines were parallel, their direction vectors $\vec{b}_1$ and $\vec{b}_2$ would be proportional, making their cross product $\vec{b}_1 \times \vec{b}_2$ equal to the zero vector. Since we cannot divide by zero, the formula would fail. For parallel lines, you would use a different approach: the distance from any point on one line to the other line.

Q: Why do we take the absolute value in the numerator?

Distance is a physical quantity and must always be non-negative. The dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$ can result in a negative value depending on the order of the vectors or the direction of the common perpendicular. Taking the absolute value ensures our final distance $\alpha$ is a positive scalar.

Question

If the shortest distance between the line joining the points $(1, 2, 3)$ and $(2, 3, 4)$ and the line $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-2}{0}$ is $\alpha$, then $28\alpha^2$ is equal to ______.

Sigma Percentile · Accepted Answer

### The Geometry of Skew Lines: A 3D Odyssey

Welcome, fellow explorer of the mathematical universe. Today, we are diving into the elegant world of 3D geometry. We are tasked with finding the shortest distance between two skew lines.

Imagine two highways in the sky, passing over each other at different altitudes. They never touch, and they aren't parallel. How do we measure the gap between them? This is the essence of our problem.

### Phase 1: Deconstructing the Lines

First, let us look at our lines. We have line $L_1$ passing through $A(1, 2, 3)$ and $B(2, 3, 4)$.

To define a line in 3D, we need a point and a direction. Our point $\vec{a}_1$ is simply $\hat{i} + 2\hat{j} + 3\hat{k}$.

The direction vector $\vec{b}_1$ is the vector $\vec{AB}$, calculated as $(2-1)\hat{i} + (3-2)\hat{j} + (4-3)\hat{k}$, which simplifies to $\vec{b}_1 = \hat{i} + \hat{j} + \hat{k}$.

Now, consider $L_2$, given by the symmetric form:

$$\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-2}{0}$$

This symmetric form is a gift! The numerators reveal a point on the line: $\vec{a}_2 = \hat{i} - \hat{j} + 2\hat{k}$.

The denominators give us the direction vector: $\vec{b}_2 = 2\hat{i} - \hat{j} + 0\hat{k}$. We have our building blocks.

### Phase 2: The Common Perpendicular

Since these lines are skew, the shortest distance $\alpha$ is the length of the segment that is perpendicular to both. The direction of this segment is the cross product of the two direction vectors: $\vec{n} = \vec{b}_1 	imes \vec{b}_2$.

We calculate this using a determinant:

$$\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 2 & -1 & 0 \end{vmatrix}$$

Expanding this, we get $\hat{i}(0 - (-1)) - \hat{j}(0 - 2) + \hat{k}(-1 - 2) = \hat{i} + 2\hat{j} - 3\hat{k}$.

This vector $\vec{n} = \hat{i} + 2\hat{j} - 3\hat{k}$ is the backbone of our shortest distance.

### Phase 3: The Calculation

The formula for the shortest distance is:

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

We already have the denominator's vector. Its magnitude is:

$$|\vec{n}| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$$

Now, for the numerator, we need the vector connecting the two lines:

$$\vec{a}_2 - \vec{a}_1 = (1-1)\hat{i} + (-1-2)\hat{j} + (2-3)\hat{k} = -3\hat{j} - \hat{k}$$

Finally, the dot product:

$$(\vec{a}_2 - \vec{a}_1) \cdot \vec{n} = (0\hat{i} - 3\hat{j} - \hat{k}) \cdot (\hat{i} + 2\hat{j} - 3\hat{k}) = (0)(1) + (-3)(2) + (-1)(-3) = 0 - 6 + 3 = -3$$

Taking the absolute value, we get $3$.

### Phase 4: The Final Victory

Putting it all together, we find:

$$\alpha = \frac{3}{\sqrt{14}}$$

The problem asks for $28\alpha^2$. Squaring $\alpha$, we get:

$$\alpha^2 = \frac{9}{14}$$

Thus, the final result is:

$$28\alpha^2 = 28 	imes \frac{9}{14} = 2 	imes 9 = \mathbf{18}$$

We have arrived at the destination. The beauty of this problem lies in how the 3D complexity collapses into a single, elegant integer. Keep practicing, and these vectors will soon become second nature to you.

Animated Solution for Mathematics - Three Dimensional Geometry: If the shortest distance between the line joining the points $(1, 2, 3)$ and $(2, 3, 4)$ and the line $\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z - 2}{0}$ is $α$ , then $28 α^{2}$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Extracting Vectors for Line 1

Extracting Vectors for Line 2

Shortest Distance Formula

Vector Connecting the Lines

Common Perpendicular Direction

Evaluating the Cross Product

Magnitude of the Normal Vector

Projection Numerator

Calculating Shortest Distance $α$

Final Evaluation

The Sigma Insight: Shortest Distance Between Two Skew Lines

The Geometry of Skew Lines

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

The shortest distance between the lines $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 6}{2}$ and $\frac{x - 6}{3} = \frac{1 - y}{2} = \frac{z + 8}{0}$ is equal to ______.

If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x}{1} = \frac{y}{α} = \frac{z - 5}{1}$ is $\frac{5}{6}$ , then the sum of all possible values of $α$ is

Let the point $(- 1, α, β)$ lie on the line of the shortest distance between the lines $\frac{x + 2}{- 3} = \frac{y - 2}{4} = \frac{z - 5}{2}$ and $\frac{x + 2}{- 1} = \frac{y + 6}{2} = \frac{z - 1}{0}$ . Then $(α - β)^{2}$ is equal to___________

The shortest distance between the lines $\frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 1}{- 1}$ and $\frac{x + 3}{2} = \frac{y - 6}{1} = \frac{z - 5}{3}$ is:

The sum of all values of $α$ , for which the shortest distance between the lines $\frac{x + 1}{α} = \frac{y - 2}{- 1} = \frac{z - 4}{- α}$ and $\frac{x}{α} = \frac{y - 1}{2} = \frac{z - 1}{2 α}$ is $2$ , is

If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{λ}$ and $\frac{x - 2}{1} = \frac{y - 4}{4} = \frac{z - 5}{5}$ is $\frac{1}{3}$ , then the sum of all possible values of $λ$ is :

The shortest distance between the lines $\frac{x - 5}{1} = \frac{y - 2}{2} = \frac{z - 4}{- 3}$ and $\frac{x + 3}{1} = \frac{y + 5}{4} = \frac{z - 1}{- 5}$ is

If the shortest distance between the lines $\frac{x - 4}{1} = \frac{y + 1}{2} = \frac{z}{- 3}$ and $\frac{x - λ}{2} = \frac{y + 1}{4} = \frac{z - 2}{- 5}$ is $\frac{6}{5}$ , then the sum of all possible values of $λ$ is :

If the shortest distance between the lines $\frac{x - λ}{3} = \frac{y - 2}{- 1} = \frac{z - 1}{1}$ and $\frac{x + 2}{- 3} = \frac{y + 5}{2} = \frac{z - 4}{4}$ is $\frac{44}{30}$ , then the largest possible value of $∣ λ ∣$ is equal to _________