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 sum of possible values of $\lambda$ is

• A. 16
• B. 11
• C. 12
• D. 15

Hint:

Geometry Tip: The shortest distance between skew lines formula $d = \frac{|[\vec{a_2}-\vec{a_1}, \vec{b_1}, \vec{b_2}]|}{|\vec{b_1} \times \vec{b_2}|}$ is purely derived from projecting the displacement vector between the lines onto their mutually perpendicular normal vector.

Answer 1

The Correct Option is A

Concept: 3D Geometry - Shortest Distance Between Two Skew Lines.

Step 1: Identify the passing points and direction vectors of both lines. For line $L_1$, it passes through point $A(1, 2, 3)$ and its direction vector is $\vec{b_1} = 2\hat{i} + 3\hat{j} + \lambda\hat{k}$. For line $L_2$, it passes through point $B(2, 4, 5)$ and its direction vector is $\vec{b_2} = 1\hat{i} + 4\hat{j} + 5\hat{k}$. Calculate the vector connecting the points on the two lines: $\vec{AB} = B - A = (2-1)\hat{i} + (4-2)\hat{j} + (5-3)\hat{k} = \hat{i} + 2\hat{j} + 2\hat{k}$.

Step 2: Calculate the scalar triple product for the numerator. The shortest distance $d$ is given by $d = \frac{|(\vec{b_1} \times \vec{b_2}) \cdot \vec{AB}|}{|\vec{b_1} \times \vec{b_2}|}$. The numerator is the determinant formed by the components of $\vec{AB}, \vec{b_1},$ and $\vec{b_2}$: Numerator = $\begin{vmatrix} 1 & 2 & 2
2 & 3 & \lambda
1 & 4 & 5 \end{vmatrix}$.
Expanding along the first row: $1(15 - 4\lambda) - 2(10 - \lambda) + 2(8 - 3)$. Simplify: $15 - 4\lambda - 20 + 2\lambda + 10 = 5 - 2\lambda$.

Step 3: Calculate the magnitude of the cross product for the denominator. Find the cross product $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & 3 & \lambda
1 & 4 & 5 \end{vmatrix}$.
$\vec{b_1} \times \vec{b_2} = \hat{i}(15 - 4\lambda) - \hat{j}(10 - \lambda) + \hat{k}(8 - 3) = (15 - 4\lambda)\hat{i} + (\lambda - 10)\hat{j} + 5\hat{k}$.
The magnitude $|\vec{b_1} \times \vec{b_2}| = \sqrt{(15 - 4\lambda)^2 + (\lambda - 10)^2 + 5^2}$.
Expand the squares: $\sqrt{(225 - 120\lambda + 16\lambda^2) + (\lambda^2 - 20\lambda + 100) + 25}$. Combine like terms: $\sqrt{17\lambda^2 - 140\lambda + 350}$.

Step 4: Set up the equation using the given shortest distance. Substitute the numerator and denominator into the distance formula and equate it to the given value $\frac{1}{\sqrt{3}}$: $\frac{1}{\sqrt{3}} = \frac{|5 - 2\lambda|}{\sqrt{17\lambda^2 - 140\lambda + 350}}$.
To remove the absolute value and the square roots, square both sides of the equation: $\frac{1}{3} = \frac{(5 - 2\lambda)^2}{17\lambda^2 - 140\lambda + 350}$.
$\frac{1}{3} = \frac{25 - 20\lambda + 4\lambda^2}{17\lambda^2 - 140\lambda + 350}$.

Step 5: Solve the quadratic equation for $\lambda$. Cross-multiply the equation: $17\lambda^2 - 140\lambda + 350 = 3(25 - 20\lambda + 4\lambda^2)$. $17\lambda^2 - 140\lambda + 350 = 75 - 60\lambda + 12\lambda^2$.
Move all terms to the left side: $5\lambda^2 - 80\lambda + 275 = 0$.
Divide the entire equation by 5 to simplify: $\lambda^2 - 16\lambda + 55 = 0$.
Factor the quadratic: $(\lambda - 5)(\lambda - 11) = 0$. The possible values are $\lambda = 5$ or $\lambda = 11$. The sum of these possible values is $5 + 11 = 16$. $$ \therefore \text{The sum of possible values of } \lambda \text{ is } 16. $$