Let L₁: (x-1)/1 = (y-2)/-1 = (z-1)/2 and L₂: (x+1)/-1 = (y-2)/2 = z/1 be two lines. Let L₃ be a line

Q: Wait, why do we reject the negative root here? Or do we need to solve for $t$ and $s$?

Great question! You don't actually need to solve for the specific values of $t$ and $s$. Because the expression $5\alpha - 11\beta - 8\gamma$ is designed to be invariant for any point on the line $L_3$, the parameters $t$ and $s$ will cancel out algebraically. This is a common feature in JEE problems—look for the structure of the expression to see if it simplifies before you dive into solving for variables.

Q: How do I know the direction vector of $L_3$ is $\vec{d}_1 \times \vec{d}_2$?

By definition, the cross product $\vec{A} \times \vec{B}$ results in a vector that is perpendicular to both $\vec{A}$ and $\vec{B}$. Since $L_3$ is perpendicular to both $L_1$ and $L_2$, its direction vector must be parallel to the cross product of the direction vectors of $L_1$ and $L_2$. It is the most direct tool in your 3D geometry toolkit for this situation.

Q: Is it possible for the expression to depend on $t$ or $s$?

In general, yes, an expression could depend on the parameters. However, in this specific problem, the coefficients $(5, -11, -8)$ were carefully chosen such that when you substitute the parametric forms, the terms involving $t$ and $s$ vanish. If they didn't vanish, it would imply the expression isn't constant, which would contradict the question asking for a single numerical value.

Question

Let $L_{1}:\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}$ and $L_{2}:\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}$ be two lines. Let $L_{3}$ be a line passing through the point $(\alpha,\beta,\gamma)$ and be perpendicular to both $L_{1}$ and $L_{2}$. If $L_{3}$ intersects $L_{1}$, then $|5\alpha-11\beta-8\gamma|$ equals:

Options:

(A) 20

(B) 18

(C) 25

(D) 16

Sigma Percentile · Accepted Answer

### The Geometry of 3D Lines: A Journey into Orthogonality

Imagine standing in a vast, three-dimensional coordinate system. You see two infinite highways, $L_1$ and $L_2$, stretching out into the distance.

They don't necessarily intersect, and they certainly aren't parallel. Your mission is to find a third path, $L_3$, that cuts through space, perfectly perpendicular to both of these highways.

This is not just a math problem; it is an exercise in visualizing the rigid, beautiful structure of 3D space.

### Phase 1: The Compass of Perpendicularity

To navigate this space, we first need to understand the orientation of our highways. The equations of our lines, 
$$L_1: \frac{x-1}{1} = \frac{y-2}{-1} = \frac{z-1}{2}$$ 
and 
$$L_2: \frac{x+1}{-1} = \frac{y-2}{2} = \frac{z}{1}$$ 
give us their direction vectors directly from their denominators.

We identify $\vec{d}_1 = (1, -1, 2)$ and $\vec{d}_2 = (-1, 2, 1)$.

Now, we need a direction for $L_3$. Since $L_3$ must be perpendicular to both $L_1$ and $L_2$, its direction vector $\vec{d}_3$ must be orthogonal to both $\vec{d}_1$ and $\vec{d}_2$.

This is the classic role of the cross product. We calculate $\vec{d}_3 = \vec{d}_1 	imes \vec{d}_2$:

$$\vec{d}_3 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 2 \ -1 & 2 & 1 \end{vmatrix} = \hat{i}(-1 - 4) - \hat{j}(1 - (-2)) + \hat{k}(2 - 1) = -5\hat{i} - 3\hat{j} + \hat{k}$$

Thus, our direction vector is $\vec{d}_3 = (-5, -3, 1)$. This vector is our compass; it defines the orientation of $L_3$ in the void.

### Phase 2: The Dance of Parameters

We know $L_3$ intersects $L_1$ at some point $P$. Since $P$ lies on $L_1$, we can describe it using a parameter $t$.

Any point on $L_1$ takes the form $P = (t+1, -t+2, 2t+1)$. This point $P$ is the anchor for our line $L_3$.

Since $L_3$ passes through $P$ and follows the direction $\vec{d}_3$, any point $(\alpha, \beta, \gamma)$ on $L_3$ can be reached by starting at $P$ and moving along $\vec{d}_3$ by some scalar $s$.

Mathematically, this is $(\alpha, \beta, \gamma) = P + s\vec{d}_3$. Breaking this down into components, we get:

$$\alpha = (t+1) - 5s$$
$$\beta = (-t+2) - 3s$$
$$\gamma = (2t+1) + s$$

We have successfully captured the entire line $L_3$ in terms of two parameters, $t$ and $s$. We are now ready for the final act.

### Phase 3: The Algebraic Miracle

The problem asks for the value of $|5\alpha - 11\beta - 8\gamma|$. Let's define this expression as $E$.

We substitute our parametric expressions for $\alpha, \beta, \gamma$ into $E$:

$$E = 5(t+1-5s) - 11(-t+2-3s) - 8(2t+1+s)$$

Now, we expand with caution. Precision is our best friend here:

$$E = (5t + 5 - 25s) + (11t - 22 + 33s) + (-16t - 8 - 8s)$$

Watch closely as we group the terms. The $t$ terms are $5t + 11t - 16t = 0t$.

The $s$ terms are $-25s + 33s - 8s = 0s$. The variables have vanished!

We are left only with the constants: $5 - 22 - 8 = -25$.

### Conclusion: The Elegance of Invariance

We find that $E = -25$. The question asks for the absolute value, so $|E| = |-25| = 25$.

It is truly satisfying to see how the complex geometry of 3D lines collapses into a simple, constant value.

This is the beauty of mathematics—the underlying structure often simplifies the most daunting problems if you trust the process. You have navigated the 3D space, mastered the cross product, and witnessed the algebraic cancellation. **The final answer is 25.**

Select Answer:

Visualized Solution

Identifying the Given Lines

Direction of Line $L_{3}$

Calculating $d_{3}$

Intersection Point $P$

General Point on $L_{3}$

Expressing $α, β, γ$

Substituting into the Target Expression

Expanding the Expression

Grouping and Simplifying

Final Absolute Value

The Sigma Insight: Equation of a Line in Space

The Geometry of 3D Lines

Phase 1

Phase 2

Phase 3

Conclusion

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