Let the line L intersect the lines x - 2 = -y = z - 1, 2(x + 1) = 2(y - 1) = z + 1 and be parallel t

Q: Why do we use parameters $\lambda$ and $\mu$ instead of just solving the equations directly?

Great question! In 3D geometry, lines are infinite. We cannot 'solve' for a single point because there isn't one. By using parameters, we are essentially saying, 'I don't know where the intersection points are yet, but I know they must exist somewhere on these lines.' It turns a geometric problem into a manageable algebraic one.

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

In this specific problem, it doesn't matter! Whether the lines are skew or intersecting, the logic of finding a transversal line parallel to a given vector remains identical. The parametric approach is robust—it works regardless of the relative orientation of the lines.

Q: Why can't we just use the cross product of the direction vectors?

The cross product of the direction vectors of $L_1$ and $L_2$ would give you a vector perpendicular to both, which is useful for finding the shortest distance. But here, we need a line *parallel* to $L_3$, not perpendicular to $L_1$ and $L_2$. Always keep your objective in mind: are you finding the shortest distance or a specific transversal?

Question

Let the line $L$ intersect the lines $x - 2 = -y = z - 1, 2(x + 1) = 2(y - 1) = z + 1$ and be parallel to the line $\frac{x-2}{3} = \frac{y-1}{1} = \frac{z-2}{2}$. Then which of the following points lies on L?

Options:

(A) $(-\frac{1}{3}, 1, -1)$

(B) $(-\frac{1}{3}, -1, 1)$

(C) $(-\frac{1}{3}, 1, 1)$

(D) $(-\frac{1}{3}, -1, -1)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional space. You see two lines, $L_1$ and $L_2$, stretching infinitely into the void. Your mission is to find a third line, $L$, that acts as a bridge—it must touch both $L_1$ and $L_2$.

There is a specific constraint: this bridge must be perfectly parallel to a third line, $L_3$. This is the essence of 3D geometry in JEE Advanced: it is not just about formulas; it is about visualizing constraints.

### The Parametric Bridge

To find this line $L$, we need to identify two points: one on $L_1$ and one on $L_2$. Let us call them $M$ and $N$. If we can find these points, the line passing through them is our target.

For $L_1$, given by $x - 2 = -y = z - 1$, we introduce a parameter $\lambda$. This allows us to express any point $M$ on $L_1$ as:
$$M(\lambda + 2, -\lambda, \lambda + 1)$$

Similarly, for $L_2$, given by $2(x + 1) = 2(y - 1) = z + 1$, we introduce a parameter $\mu$. This gives us the general point $N$ as:
$$N\left(\frac{\mu}{2} - 1, \frac{\mu}{2} + 1, \mu - 1\right)$$

### The Vector of Direction

Now, consider the vector $\vec{MN}$ connecting these two points. This vector represents the direction of our line $L$. By subtracting the coordinates of $M$ from $N$, we obtain:
$$\vec{MN} = \left(\left(\frac{\mu}{2}-1\right) - (\lambda+2), \left(\frac{\mu}{2}+1\right) - (-\lambda), (\mu-1) - (\lambda+1)\right)$$

Simplifying this, we find the direction vector of our line $L$:
$$\vec{MN} = \left(\frac{\mu}{2} - \lambda - 3, \frac{\mu}{2} + \lambda + 1, \mu - \lambda - 2\right)$$

### The Parallelism Constraint

The problem states that $L$ is parallel to $L_3$, which has direction ratios $(3, 1, 2)$. In the language of vectors, this means $\vec{MN}$ must be proportional to $(3, 1, 2)$. We set up the proportionality:

$$\frac{\frac{\mu}{2} - \lambda - 3}{3} = \frac{\frac{\mu}{2} + \lambda + 1}{1} = \frac{\mu - \lambda - 2}{2}$$

By equating the first two ratios, we obtain the linear equation:
$$\mu + 4\lambda = -6$$

By equating the last two ratios, we find:
$$\mu - 2\lambda = 4$$

Solving this system of equations, we find $\lambda = -\frac{4}{3}$ and $\mu = -\frac{2}{3}$.

### Final Resolution

With $\lambda = -\frac{4}{3}$, we pinpoint the exact location of $M$ on $L_1$. Substituting $\lambda$ into our expression for $M$, we get:
$$M\left(\frac{2}{3}, \frac{4}{3}, -\frac{1}{3}\right)$$

Now, we have a fixed point on $L$ and a direction vector $(3, 1, 2)$. The equation of our line $L$ is:
$$\frac{x - 2/3}{3} = \frac{y - 4/3}{1} = \frac{z + 1/3}{2} = k$$

By setting $k = -\frac{1}{3}$, we find the point $(-\frac{1}{3}, 1, -1)$. This point satisfies the equation perfectly, confirming the **correct path through the 3D landscape**.

Animated Solution for Mathematics - Three Dimensional Geometry: Let the line $L$ intersect the lines $x - 2 = - y = z - 1, 2 (x + 1) = 2 (y - 1) = z + 1$ and be parallel to the line $\frac{x - 2}{3} = \frac{y - 1}{1} = \frac{z - 2}{2}$ . Then which of the following points lies on L?

Select Answer:

Visualized Solution

Visualizing the Geometry

Parametric Form of $L_{1}$

Parametric Form of $L_{2}$

Finding Vector $M N$

The Parallelism Condition

Solving for $λ$ and $μ$ : Part 1

Solving for $λ$ and $μ$ : Part 2

Calculating Point $M$

Equation of Line $L$

Checking the Options

Final Conclusion

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Parametric Bridge

The Vector of Direction

The Parallelism Constraint

Final Resolution

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