The shortest distance between the lines (x-3)/2 = (y-2)/3 = (z-1)/-1 and (x+3)/2 = (y-6)/1 = (z-5)/3

Q: Wait, why can't I just calculate the distance between the two points $(3, 2, 1)$ and $(-3, 6, 5)$ given in the equations?

That is a classic trap! The points $(3, 2, 1)$ and $(-3, 6, 5)$ are simply arbitrary points on the lines. The shortest distance connects two specific points on the lines, which are generally not the ones you see in the equation. You must use the common perpendicular to find the true minimum distance.

Q: What happens if the lines are parallel? Does this formula still work?

Great question! If the lines are parallel, the direction vectors $\vec{b}_1$ and $\vec{b}_2$ are proportional, meaning their cross product $\vec{b}_1 \times \vec{b}_2$ becomes the zero vector. The formula would involve division by zero, so it fails. For parallel lines, you would use a different approach: the distance from a point on one line to the other line.

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

Distance must always be a positive scalar. The dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$ can be negative depending on the order of your vectors or the orientation of the cross product, but the physical distance between two lines is a magnitude. The absolute value ensures our result is physically meaningful.

Question

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:

Options:

(A) $\frac{18}{\sqrt{5}}$

(B) $\frac{22}{3\sqrt{5}}$

(C) $\frac{46}{3\sqrt{5}}$

(D) $6\sqrt{3}$

Sigma Percentile · Accepted Answer

### The Geometry of the Gap

When two lines are skew, they exist in parallel planes. The shortest path between them is the unique segment that is perfectly perpendicular to both lines simultaneously, known as the **common perpendicular**.

If you were to place a ruler between these two lines, the shortest distance is the length of that ruler when it is held at a 90-degree angle to both flight paths. To find this, we rely on the power of vectors.

The distance $d$ is given by the projection of the vector connecting the two lines onto the common normal vector:

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

### Step 1: The Extraction

We are given two lines in symmetric form:
$$\frac{x-3}{2} = \frac{y-2}{3} = \frac{z-1}{-1} \quad 	ext{and} \quad \frac{x+3}{2} = \frac{y-6}{1} = \frac{z-5}{3}$$

From the first line, we extract the point $\vec{a}_1 = (3, 2, 1)$ and the direction vector $\vec{b}_1 = (2, 3, -1)$. From the second line, we extract the point $\vec{a}_2 = (-3, 6, 5)$ and the direction vector $\vec{b}_2 = (2, 1, 3)$.

Always watch your signs! The equation $\frac{x+3}{2}$ implies $x - (-3)$, so the coordinate is $-3$. Missing this sign is the most common way to lose marks in the exam.

### Step 2: The Connecting Vector

Now, we find the vector that bridges our two starting points. We calculate $\vec{a}_2 - \vec{a}_1 = (-3-3, 6-2, 5-1)$, which gives us the vector $(-6, 4, 4)$.

This vector represents one possible path between the lines, but it is likely slanted. We need to isolate the component of this vector that is perpendicular to both lines.

### Step 3: The Common Perpendicular

To find a direction perpendicular to both lines, we use the cross product $\vec{n} = \vec{b}_1 	imes \vec{b}_2$. We set up the determinant:

$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & -1 \ 2 & 1 & 3 \end{vmatrix}$$

Calculating this, we get $\vec{n} = \hat{i}(9 - (-1)) - \hat{j}(6 - (-2)) + \hat{k}(2 - 6)$, which simplifies to $\vec{n} = 10\hat{i} - 8\hat{j} - 4\hat{k}$. This vector $(10, -8, -4)$ is our common normal.

### Step 4: The Final Calculation

We are in the home stretch. First, we find the magnitude of our normal vector:

$$|\vec{n}| = \sqrt{10^2 + (-8)^2 + (-4)^2} = \sqrt{100 + 64 + 16} = \sqrt{180} = 6\sqrt{5}$$

Next, we find the dot product of our connecting vector and the normal vector:

$$(\vec{a}_2 - \vec{a}_1) \cdot \vec{n} = (-6)(10) + (4)(-8) + (4)(-4) = -60 - 32 - 16 = -108$$

Finally, we plug these into our distance formula:

$$d = \frac{|-108|}{6\sqrt{5}} = \frac{108}{6\sqrt{5}} = \frac{18}{\sqrt{5}}$$

There you have it! By systematically breaking down the 3D geometry into vector components, we have conquered the problem. The final shortest distance is **$\frac{18}{\sqrt{5}}$**.

Animated Solution for Mathematics - Three Dimensional Geometry: 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:

Select Answer:

Visualized Solution

Visualizing Skew Lines

The Shortest Distance Formula

Extracting Data from Line 1

Extracting Data from Line 2

Finding the Connecting Vector

Setting up the Cross Product

Computing the Normal Vector

Magnitude of the Normal Vector

Calculating the Dot Product

Final Distance Calculation

The Sigma Insight: Shortest Distance Between Two Skew Lines

The Geometry of the Gap

Step 1

Step 2

Step 3

Step 4

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