If d1 is the shortest distance between the lines x + 1 = 2y = -12z, x = y + 2 = 6z - 6 and d2 is the

Q: Wait, why do we have to divide the line equations by a constant? Can't I just use the coefficients as they are?

Great question! The standard form $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$ requires the coefficients of $x, y,$ and $z$ to be exactly $1$. If they aren't, your direction vector components will be wrong, leading to a cascade of errors. Always normalize the coefficients to $1$ first!

Q: How do I know if the lines are skew or intersecting?

If the shortest distance $d$ comes out to be $0$, the lines intersect. If $d > 0$, they are skew. In this problem, we are explicitly told they are skew, so expect a non-zero value!

Q: Is the order of $\vec{a}_2 - \vec{a}_1$ important in the numerator?

Not at all! Because we take the absolute value of the scalar triple product, the order of subtraction doesn't change the final magnitude. Just be consistent with your vectors.

Q: Why is the denominator the magnitude of the cross product?

Think of the cross product $\vec{b}_1 \times \vec{b}_2$ as a vector perpendicular to both lines. The shortest distance is essentially the projection of the vector connecting any two points on the lines onto this common perpendicular. Dividing by the magnitude normalizes this projection.

Q: I got a huge number for the cross product. Did I make a mistake?

Don't panic! You can always factor out common constants from the cross product vector before calculating its magnitude. It makes the arithmetic much cleaner and less prone to silly calculation errors.

Question

If $d_1$ is the shortest distance between the lines $x + 1 = 2y = -12z, x = y + 2 = 6z - 6$ and $d_2$ is the shortest distance between the lines $\frac{x-1}{2} = \frac{y+8}{-7} = \frac{z-4}{5}, \frac{x-1}{2} = \frac{y-2}{1} = \frac{z-6}{-3}$, then the value of $\frac{32\sqrt{3} d_1}{d_2}$ is :

Sigma Percentile · Accepted Answer

### The Geometry of Skew Lines: A Journey Through 3D Space

Imagine you are standing in a vast, empty room. You have two thin, straight wires suspended in the air. They don't touch, and they aren't parallel.

In the language of JEE mathematics, these are 'skew lines.' They exist in their own planes, separated by a unique, invisible bridge—the shortest distance between them. Today, we are going to find that bridge.

### Phase 1: Decoding the Lines

Before we can calculate anything, we must speak the language of the lines. The equations provided, such as $x + 1 = 2y = -12z$, are in a 'disguised' form.

To extract the soul of the line—its direction vector $\vec{b}$ and a point $\vec{a}$ on it—we must force them into the standard form:

$$\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$$

For the first line, we divide by $12$ to get $\frac{x+1}{12} = \frac{y}{6} = \frac{z}{-1}$. Now, the direction is clear: $\vec{b}_1 = 12\hat{i} + 6\hat{j} - \hat{k}$.

We repeat this for the second line, normalizing it to $\frac{x}{6} = \frac{y+2}{6} = \frac{z-1}{1}$. This gives us $\vec{b}_2 = 6\hat{i} + 6\hat{j} + \hat{k}$.

### Phase 2: The Common Perpendicular

To find the shortest distance, we need a vector that is perpendicular to both lines. This is the magic of the cross product. We compute $\vec{b}_1 	imes \vec{b}_2$ using the determinant method:

$$\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 12 & 6 & -1 \ 6 & 6 & 1 \end{vmatrix} = 12\hat{i} - 18\hat{j} + 36\hat{k}$$

Notice how we can factor out a $6$ to simplify our lives: $6(2\hat{i} - 3\hat{j} + 6\hat{k})$. This is the direction of our 'bridge.'

### Phase 3: Calculating $d_1$ and $d_2$

With the bridge defined, we use the scalar triple product formula:

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

For the first pair, the vector connecting the points is $\vec{a}_2 - \vec{a}_1 = (1, -2, 1)$. Plugging this into our formula, the numerator becomes $|1(12) - 2(-18) + 1(36)| = 84$.

The magnitude of our cross product is $\sqrt{12^2 + (-18)^2 + 36^2} = 42$. Thus, $d_1 = \frac{84}{42} = 2$.

We repeat this exact process for the second pair of lines. It is a test of patience and precision. After calculating the cross product $\vec{b}_3 	imes \vec{b}_4 = 16(\hat{i} + \hat{j} + \hat{k})$ and the point difference $\vec{a}_4 - \vec{a}_3 = (0, 10, 2)$, we find $d_2 = \frac{192}{16\sqrt{3}} = \frac{12}{\sqrt{3}}$.

### Phase 4: The Final Synthesis

We have arrived at the final hurdle. We need to evaluate $\frac{32\sqrt{3} d_1}{d_2}$. Substituting our values, we get:

$$\frac{32\sqrt{3} 	imes 2}{\frac{12}{\sqrt{3}}} = \frac{64\sqrt{3} 	imes \sqrt{3}}{12} = \frac{64 	imes 3}{12} = \frac{192}{12} = 16$$

Look at that! The complexity collapses into a clean, elegant integer. This is the beauty of JEE problems—they lead you through a labyrinth of calculations only to reward you with a moment of perfect clarity.

**The final answer is 16.** You didn't just solve a problem; you navigated the geometry of space itself. Keep that momentum going!

Enter Numerical Value:

Visualized Solution

$S h or t es t D i s t an ce F or m u l a$

$S t an d a r d F or m o f L_{1}$

$S t an d a r d F or m o f L_{2}$

$b_{1} \times b_{2}$

$S h or t es t D i s t an ce d_{1}$

$V ec t or s f or L_{3} an d L_{4}$

$b_{3} \times b_{4}$

$S h or t es t D i s t an ce d_{2}$

$E v a l u a t in g \frac{32 3 d _{1}}{d _{2}}$

$F ina l A n s w er$

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 $x + 1 = 2 y = - 12 z$ and $x = y + 2 = 6 z - 6$ is

Shortest distance between the lines $\frac{x - 1}{2} = \frac{y + 8}{- 7} = \frac{z - 4}{5}$ and $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{- 3}$ is

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 shortest distance between the lines $\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 8}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ is:

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 ______.

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

The shortest distance between the lines $\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ 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 $\frac{x - 3}{2} = \frac{y + 15}{- 7} = \frac{z - 9}{5}$ and $\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{- 3}$ is

The shortest distance between the lines $\frac{x - 1}{0} = \frac{y + 1}{- 1} = \frac{z}{1}$ and $x + y + z + 1 = 0, 2 x - y + z + 3 = 0$ is:

Enter Numerical Value:

Visualized Solution

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