If the shortest distance between the lines (x-4)/1 = (y+1)/2 = (z)/-3 and (x-λ)/2 = (y+1)/4 = (z-2)/

Q: Wait, why do we use the absolute value in the shortest distance formula?

That is a fantastic question! The shortest distance is a physical length, and length must always be non-negative. The dot product in the numerator can result in a negative value depending on the order of vectors or the direction of the normal vector. The absolute value ensures that our final distance is always a positive, meaningful quantity.

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

Great observation. If the direction vectors $\vec{d_1}$ and $\vec{d_2}$ are proportional (i.e., $\vec{d_1} = k\vec{d_2}$), the lines are parallel. If they are not proportional, they are either intersecting or skew. In this problem, the direction vectors are $\langle 1, 2, -3 \rangle$ and $\langle 2, 4, -5 \rangle$. Since $\frac{1}{2} \neq \frac{2}{4}$, they are not parallel, confirming they are skew.

Q: Why do we get two values for $\lambda$?

When we solve the equation $|\lambda - 4| = 3$, we are looking for all numbers whose distance from $4$ on the number line is exactly $3$. This naturally leads to two solutions: $4 + 3 = 7$ and $4 - 3 = 1$. Both values of $\lambda$ satisfy the geometric condition of the shortest distance being $\frac{6}{\sqrt{5}}$.

Question

If the shortest distance between the lines $\frac{x-4}{1} = \frac{y+1}{2} = \frac{z}{-3}$ and $\frac{x-\lambda}{2} = \frac{y+1}{4} = \frac{z-2}{-5}$ is $\frac{6}{\sqrt{5}}$, then the sum of all possible values of $\lambda$ is :

Options:

(A) 5

(B) 8

(C) 7

(D) 10

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, three-dimensional grid. You see two airplanes flying in the sky, their paths represented by two lines, $L_1$ and $L_2$.

These lines are skew; they never meet, and they are not parallel. They are like two ships passing in the night, separated by a specific, minimum distance.

Our mission today is to find the value of $\lambda$ that maintains this precise separation of $\frac{6}{\sqrt{5}}$.

### Extracting the DNA of the Lines

Every line in 3D space has a unique signature: a point it passes through and a direction in which it travels.

For $L_1$, the symmetric equation $\frac{x-4}{1} = \frac{y+1}{2} = \frac{z}{-3}$ tells us it passes through point $A = (4, -1, 0)$ and moves along the direction vector $\vec{d_1} = \hat{i} + 2\hat{j} - 3\hat{k}$.

For $L_2$, the equation $\frac{x-\lambda}{2} = \frac{y+1}{4} = \frac{z-2}{-5}$ reveals point $B = (\lambda, -1, 2)$ and direction vector $\vec{d_2} = 2\hat{i} + 4\hat{j} - 5\hat{k}$. These are the building blocks of our universe.

### The Common Normal: The Bridge Between Lines

To find the shortest distance, we need a bridge—a vector that is perpendicular to both lines. This is the common normal, found by the cross product $\vec{n} = \vec{d_1} 	imes \vec{d_2}$.

Setting up the determinant, we have:

$$
\vec{n} = 
\begin{vmatrix} 
\hat{i} & \hat{j} & \hat{k} \ 
1 & 2 & -3 \ 
2 & 4 & -5 
\end{vmatrix}
$$

Expanding this, we get $\hat{i}(-10 - (-12)) - \hat{j}(-5 - (-6)) + \hat{k}(4 - 4)$, which simplifies to $2\hat{i} - \hat{j} + 0\hat{k}$.

The magnitude of this normal vector is $|\vec{n}| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{5}$. This $\sqrt{5}$ is a gift; it will simplify our final equation significantly.

### The Projection: Measuring the Gap

The shortest distance is the projection of the vector $\vec{AB}$ onto the normal vector $\vec{n}$. First, $\vec{AB} = \vec{b} - \vec{a} = (\lambda - 4)\hat{i} + 0\hat{j} + 2\hat{k}$.

The scalar triple product, which is the numerator of our distance formula, is:

$$
|\vec{AB} \cdot \vec{n}| = |(\lambda - 4)(2) + (0)(-1) + (2)(0)| = |2(\lambda - 4)|
$$

Now, we combine everything into the shortest distance formula:

$$
S.D. = \frac{|2(\lambda - 4)|}{\sqrt{5}}
$$

We are told this distance is $\frac{6}{\sqrt{5}}$. Equating them, we get $\frac{|2(\lambda - 4)|}{\sqrt{5}} = \frac{6}{\sqrt{5}}$. The $\sqrt{5}$ cancels out, leaving us with $|2(\lambda - 4)| = 6$, or simply $|\lambda - 4| = 3$.

### The Final Resolution

We have arrived at the final hurdle. The absolute value equation $|\lambda - 4| = 3$ splits into two paths:

1. $\lambda - 4 = 3 \Rightarrow \lambda = 7$
2. $\lambda - 4 = -3 \Rightarrow \lambda = 1$

Both are valid. The question asks for the sum of all possible values of $\lambda$.

Thus, the final result is $7 + 1 = \mathbf{8}$. You have successfully navigated the 3D landscape, mastered the cross product, and solved for the unknown parameter.

Animated Solution for Mathematics - Three Dimensional Geometry: 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 Skew Lines

Parameters of Lines

Shortest Distance Formula

Vector $A B$

Cross Product Setup

Calculating Cross Product

Magnitude of Cross Product

Scalar Triple Product

Equating Shortest Distance

Simplifying Equation

Solving for $λ$ (Case 1)

Solving for $λ$ (Case 2)

Sum of $λ$ Values

The Sigma Insight: Shortest Distance Between Two Skew Lines

Analyzing the Setup

Extracting the DNA of the Lines

The Common Normal

The Projection

The Final Resolution

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