If the shortest distance between the lines (x-1)/2 = (y-2)/3 = (z-3)/λ and (x-2)/1 = (y-4)/4 = (z-5)

Q: Wait, why do we need the absolute value in the numerator of the distance formula?

Distance is a physical quantity—it must be non-negative. The dot product can result in a negative scalar depending on the orientation of the vectors, but the absolute value ensures we get the magnitude of the projection, which is the physical distance we seek.

Q: Is there a way to solve this without the formula? Maybe using calculus?

Absolutely! You could parameterize points $P$ on $L_1$ and $Q$ on $L_2$, define the distance vector $\vec{PQ}$, and minimize its magnitude. However, that leads to a much more complex optimization problem. The formula is a derived shortcut—use it to save time for the tougher questions!

Q: Do I need to solve for the individual values of $\lambda$?

Not at all! The question asks for the sum of all possible values. Once you arrive at the quadratic equation $\lambda^2 - 16\lambda + 55 = 0$, simply use the property of roots: the sum of roots is $-b/a$. It's a classic JEE trick to save you from unnecessary calculation.

Question

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

Options:

(A) 16

(B) 6

(C) 12

(D) 15

Sigma Percentile · Accepted Answer

### Analyzing the Architecture of Skew Lines

We are navigating the geometry of two skew lines, $L_1$ and $L_2$, in 3D space. Our objective is to determine the parameter $\lambda$ such that the shortest distance between these lines is exactly $\frac{1}{\sqrt{3}}$.

### Phase 1: Extracting the DNA

Every line in 3D space is defined by a point it passes through and a direction vector. For $L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{\lambda}$, we identify:
Position vector $\vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}$
Direction vector $\vec{b}_1 = 2\hat{i} + 3\hat{j} + \lambda\hat{k}$

For $L_2: \frac{x-2}{1} = \frac{y-4}{4} = \frac{z-5}{5}$, we identify:
Position vector $\vec{a}_2 = 2\hat{i} + 4\hat{j} + 5\hat{k}$
Direction vector $\vec{b}_2 = \hat{i} + 4\hat{j} + 5\hat{k}$

### Phase 2: The Common Perpendicular

The shortest distance $d$ between two skew lines is given by the formula:
$$d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2)|}{|\vec{b}_1 	imes \vec{b}_2|}$$

First, we calculate the vector connecting the lines:
$$\vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (4-2)\hat{j} + (5-3)\hat{k} = \hat{i} + 2\hat{j} + 2\hat{k}$$

Next, we compute the cross product $\vec{b}_1 	imes \vec{b}_2$ to find the direction of the common perpendicular:
$$\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & \lambda \ 1 & 4 & 5 \end{vmatrix} = (15 - 4\lambda)\hat{i} + (\lambda - 10)\hat{j} + 5\hat{k}$$

### Phase 3: The Algebraic Climax

Substituting these components into the distance formula, we obtain:
$$\frac{|(1)(15-4\lambda) + (2)(\lambda-10) + (2)(5)|}{\sqrt{(15-4\lambda)^2 + (\lambda-10)^2 + 5^2}} = \frac{1}{\sqrt{3}}$$

Simplifying the numerator, we get $|5 - 2\lambda|$. The denominator simplifies to $\sqrt{17\lambda^2 - 140\lambda + 350}$. Thus:
$$\frac{|5 - 2\lambda|}{\sqrt{17\lambda^2 - 140\lambda + 350}} = \frac{1}{\sqrt{3}}$$

Squaring both sides to eliminate the radical and absolute value yields:
$$3(5 - 2\lambda)^2 = 17\lambda^2 - 140\lambda + 350$$
$$3(25 - 20\lambda + 4\lambda^2) = 17\lambda^2 - 140\lambda + 350$$
$$12\lambda^2 - 60\lambda + 75 = 17\lambda^2 - 140\lambda + 350$$

Rearranging the terms results in the quadratic equation:
$$5\lambda^2 - 80\lambda + 275 = 0$$
$$\lambda^2 - 16\lambda + 55 = 0$$

### Phase 4: The Elegant Conclusion

To find the sum of all possible values of $\lambda$, we apply Vieta's formulas to the quadratic equation $\lambda^2 - 16\lambda + 55 = 0$. For a quadratic $a\lambda^2 + b\lambda + c = 0$, the sum of the roots is given by $-\frac{b}{a}$.

Sum of roots $= -\frac{-16}{1} = 16$.

The sum of all possible values of $\lambda$ is **16**.

Animated Solution for Mathematics - Three Dimensional Geometry: 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 :

Select Answer:

Visualized Solution

Visualizing the Skew Lines

The Shortest Distance Formula

Extracting Vectors $a$ and $b$

Calculating $a_{2} - a_{1}$

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

Expanding the Determinant

Calculating the Numerator

Calculating the Denominator

Setting up the Equation

Squaring Both Sides

Simplifying to a Quadratic

Final Step: Sum of Values

The Sigma Insight: Shortest Distance Between Two Skew Lines

Analyzing the Architecture of Skew Lines

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

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 :

If the shortest distance between the lines $\frac{x - λ}{- 2} = \frac{y - 2}{1} = \frac{z - 1}{1}$ and $\frac{x - 3}{1} = \frac{y - 1}{- 2} = \frac{z - 2}{1}$ is 1, then the sum of all possible values of $λ$ 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

The sum of all values of $α$ , for which the shortest distance between the lines $\frac{x + 1}{α} = \frac{y - 2}{- 1} = \frac{z - 4}{- α}$ and $\frac{x}{α} = \frac{y - 1}{2} = \frac{z - 1}{2 α}$ is $2$ , is

If the shortest distance between the lines $\frac{x - λ}{2} = \frac{y - 4}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{4} = \frac{y - 4}{6} = \frac{z - 7}{8}$ is $\frac{13}{29}$ , then a value of $λ$ is :

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

If the shortest distance between the lines $\frac{x - λ}{3} = \frac{y - 2}{- 1} = \frac{z - 1}{1}$ and $\frac{x + 2}{- 3} = \frac{y + 5}{2} = \frac{z - 4}{4}$ is $\frac{44}{30}$ , then the largest possible value of $∣ λ ∣$ is equal to _________

If the shortest distance between the straight lines $3 (x - 1) = 6 (y - 2) = 2 (z - 1)$ and $4 (x - 2) = 2 (y - λ) = (z - 3), λ \in R$ is $\frac{1}{38}$ , then the integral value of $λ$ 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