Line L1 passes through the point (1, 2, 3) and is parallel to Z -axis. Line L2 passes through the po

Q: Wait, why do we reject the negative root here? Or do we?

We actually don't reject it! The equation $|\lambda - 1| = 3$ yields two solutions: $\lambda - 1 = 3$ (so $\lambda = 4$) and $\lambda - 1 = -3$ (so $\lambda = -2$). The problem gives us the condition $\lambda_2 < \lambda_1$, which helps us assign the values correctly. Always keep both roots until the problem explicitly restricts them!

Q: Why can't we just use the standard formula for the shortest distance between two skew lines?

You absolutely could use the formula $\frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$. However, in JEE Advanced, time is your greatest asset. Recognizing that the lines are parallel to the coordinate axes allows you to bypass the cross-product calculation entirely and solve it using simple coordinate geometry. Efficiency is key!

Q: How do I find the projection of a point on a line parallel to an axis?

Think of it as dropping a perpendicular. If a line is parallel to the $Z$-axis, its $x$ and $y$ coordinates are fixed. To project a point $P(x_0, y_0, z_0)$ onto this line, you simply keep the $x$ and $y$ coordinates of the line and use the $z$-coordinate of the point $P$. It's just like finding the shadow of the point on the line!

Question

Line $L_1$ passes through the point $(1, 2, 3)$ and is parallel to $Z$ -axis. Line $L_2$ passes through the point $(\lambda, 5, 6)$ and is parallel to $y$-axis. Let for $\lambda = \lambda_1, \lambda_2, \lambda_2 < \lambda_1$, the shortest distance between the two lines be $3$. Then the square of the distance of the point $(\lambda_1, \lambda_2, 7)$ from the line $L_1$ is

Options:

(A) $40$

(B) $32$

(C) $25$

(D) $37$

Sigma Percentile · Accepted Answer

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

Welcome, fellow explorer of the mathematical universe! Today, we are diving into the elegant world of 3D geometry.

Often, when students see lines in 3D space, they immediately reach for complex vector formulas. But wait—before you do that, let's pause and look at the geometry. There is a hidden simplicity in this problem that, once unlocked, turns a daunting task into a beautiful, intuitive exercise.

### Phase 1: Visualizing the Pillars

Imagine you are standing in a room. Line $L_1$ passes through $(1, 2, 3)$ and runs perfectly parallel to the $Z$-axis.

Think of it as a vertical pillar extending from the floor to the ceiling. Because it is parallel to the $Z$-axis, its $x$ and $y$ coordinates are locked in stone: $x=1$ and $y=2$. No matter how high or low you go on this line, those coordinates never change.

Now, consider line $L_2$. It passes through $(\lambda, 5, 6)$ and runs parallel to the $Y$-axis. This is like a horizontal beam crossing the room.

For this line, the $x$ and $z$ coordinates are fixed: $x=\lambda$ and $z=6$. By simply visualizing these constraints, we have already defined our lines without needing a single complex vector equation.

### Phase 2: The Shortest Distance

We are looking for the shortest distance between these two skew lines. In 3D space, the shortest distance between two lines is always measured along their common perpendicular.

Let's look at their directions. $L_1$ is directed along the $Z$-axis (the $\hat{k}$ vector), and $L_2$ is directed along the $Y$-axis (the $\hat{j}$ vector). What is the common perpendicular to both the $Y$ and $Z$ axes? It is the $X$-axis!

This means the shortest distance segment must be parallel to the $X$-axis. To find the length of this segment, we need the points where this perpendicular intersects both lines.

On $L_1$, the $z$-coordinate must match $L_2$, so $z=6$. On $L_2$, the $y$-coordinate must match $L_1$, so $y=2$. The intersection points are $(1, 2, 6)$ on $L_1$ and $(\lambda, 2, 6)$ on $L_2$.

The distance between these points is simply the difference in their $x$-coordinates: $|\lambda - 1|$. We are told this distance is $3$, so we have the equation $|\lambda - 1| = 3$.

### Phase 3: The Algebra of Possibilities

Solving $|\lambda - 1| = 3$ gives us two potential values: $\lambda = 4$ or $\lambda = -2$. The problem provides a crucial hint: $\lambda_2 < \lambda_1$.

This is our key to unlocking the final configuration. Clearly, $\lambda_1 = 4$ and $\lambda_2 = -2$. With these values, we have fully defined our system.

We are now ready to tackle the final challenge: finding the distance of point $P(\lambda_1, \lambda_2, 7)$ from line $L_1$.

### Phase 4: The Final Leap

Substituting our values, point $P$ is at $(4, -2, 7)$. We need the perpendicular distance from $P$ to $L_1$.

Recall that $L_1$ is defined by $x=1$ and $y=2$. To drop a perpendicular from $P$ to $L_1$, we simply project $P$ onto the line. The projection point will have the same $z$-coordinate as $P$, which is $7$.

Thus, the foot of the perpendicular is $(1, 2, 7)$. Now, we apply the 3D distance formula to find the distance $d$ between $P(4, -2, 7)$ and the foot $(1, 2, 7)$:

$$d = \sqrt{(4 - 1)^2 + (-2 - 2)^2 + (7 - 7)^2}$$

$$d = \sqrt{(3)^2 + (-4)^2 + 0^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

The problem asks for the square of this distance, $d^2$. Therefore, **$d^2 = 25$**.

### Conclusion

We have arrived at our destination! The answer is **$25$**.

Notice how we didn't need to perform heavy vector algebra or calculate complex cross products. By visualizing the lines as simple geometric entities, we reduced a 3D problem into a 1D distance calculation.

This is the power of spatial intuition in JEE Advanced. Keep practicing this visualization, and you will find that even the most intimidating problems have a simple, elegant core waiting to be discovered.

Select Answer:

Visualized Solution

Visualizing Line $L_{1}$

Visualizing Line $L_{2}$

Shortest Distance Between $L_{1}$ and $L_{2}$

Calculating the Shortest Distance

Solving for $λ$

Identifying $λ_{1}$ and $λ_{2}$

Locating the Target Point $P$

Distance from $P$ to $L_{1}$

Calculating the Square of the Distance

Final Conclusion

The Sigma Insight: Shortest Distance Between Two Skew Lines

The Geometry of Skew Lines

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

Similar Questions

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 $\frac{x - 3}{2} = \frac{y + 15}{- 7} = \frac{z - 9}{5}$ and $\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{- 3}$ is

Let the shortest distance between the lines $\frac{x - 3}{3} = \frac{y - α}{- 1} = \frac{z - 3}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - β}{4}$ be $330$ . Then the positive value of $5 α + β$ is

The shortest distance between lines $L_{1}$ and $L_{2}$ , where $L_{1} : \frac{x - 1}{2} = \frac{y + 1}{- 3} = \frac{z + 4}{2}$ and $L_{2}$ is the line passing through the points $A (- 4, 4, 3), B (- 1, 6, 3)$ and perpendicular to the line $\frac{x - 3}{- 2} = \frac{y}{3} = \frac{z - 1}{1}$ , is

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

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 $\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 8}{1}$ 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 :

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 :

Select Answer:

Visualized Solution

Visualizing Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

Visualizing Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Shortest Distance Between .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Calculating the Shortest Distance

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Identifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ2​

Locating the Target Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Distance from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​