QJEEM24Maef6b: JEE Main 2024 Mathematics - Class 12 JEEM24Maef6b Solution: If the shortest distance between the lines and is 1 , then the…

Given the two lines:

$L_1: \frac{x-\lambda}{-2} = \frac{y-2}{1} = \frac{z-1}{1}$

$L_2: \frac{x-\sqrt{3}}{1} = \frac{y-1}{-2} = \frac{z-2}{1}$

We observe that these lines are not parallel as their directional vectors are not proportional. The directional vector for $L_1$ is $(-2,1,1)$ and for $L_2$ is $(1,-2,1)$ . The shortest distance between two skew (non-intersecting and non-parallel) lines in the three-dimensional space is along the line that is perpendicular to both lines. This implies we can find a vector that is perpendicular to both directional vectors by taking their cross product.

The directional vector for $L_1$ is $d_1 = \langle -2, 1, 1 \rangle$ , and for $L_2$ is $d_2 = \langle 1, -2, 1 \rangle$ . The cross product of $d_1$ and $d_2$ , which will be perpendicular to both lines, is given by:

$d = d_1 \times d_2 = \left| \begin{matrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\-2 & 1 & 1 \\1 & -2 & 1 \\\end{matrix} \right|$

Expanding the determinate gives:

$d = \mathbf{i}((1)(1) - (1)(-2)) - \mathbf{j}((-2)(1) - (1)(1)) + \mathbf{k}((-2)(-2) - (1)(1))$

$d = \mathbf{i}(1 + 2) - \mathbf{j}(-2 - 1) + \mathbf{k}(4 - 1)$

$d = 3\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}$

$d = \langle 3, 3, 3 \rangle$

The shortest distance $D$ between the two lines can then be given by the formula:

$D = \frac{\left| (\mathbf{a_2} - \mathbf{a_1}) \cdot d \right|}{\|d\|}$

Where $\mathbf{a_1}$ and $\mathbf{a_2}$ are position vectors to any points on line $L_1$ and line $L_2$ , respectively, and ' $\cdot$ ' denotes the dot product.

From the lines' equations, we can choose a point on each line (when the parameter is zero). Thus, for $L_1$ , let's choose the point $A(\lambda, 2, 1)$ , and for $L_2$ , let's choose the point $B(\sqrt{3}, 1, 2)$ . These points correspond to the vectors $\mathbf{a_1} = \langle \lambda, 2, 1 \rangle$ and $\mathbf{a_2} = \langle \sqrt{3}, 1, 2 \rangle$ , respectively.

The vector $\mathbf{a_2} - \mathbf{a_1}$ is:

$\mathbf{a_2} - \mathbf{a_1} = \langle \sqrt{3}, 1, 2 \rangle - \langle \lambda, 2, 1 \rangle$

$\mathbf{a_2} - \mathbf{a_1} = \langle \sqrt{3} - \lambda, 1 - 2, 2 - 1 \rangle$

$\mathbf{a_2} - \mathbf{a_1} = \langle \sqrt{3} - \lambda, -1, 1 \rangle$

We can now substitute this, along with $d$ , into the distance formula:

$D = \frac{\left| (\langle \sqrt{3} - \lambda, -1, 1 \rangle) \cdot \langle 3, 3, 3 \rangle \right|}{\|\langle 3, 3, 3 \rangle\|}$

$D = \frac{\left| 3(\sqrt{3} - \lambda) + 3(-1) + 3(1) \right|}{\sqrt{3^2 + 3^2 + 3^2}}$

$D = \frac{\left| 3\sqrt{3} - 3\lambda - 3 + 3 \right|}{\sqrt{27}}$

$D = \frac{\left| 3\sqrt{3} - 3\lambda \right|}{3\sqrt{3}}$

$D = \frac{\left| \sqrt{3} - \lambda \right|}{\sqrt{3}}$

Given that the shortest distance $D$ between the lines is 1, we can equate the above result to 1, and solve for $\lambda$ :

$\frac{\left| \sqrt{3} - \lambda \right|}{\sqrt{3}} = 1$

$\left| \sqrt{3} - \lambda \right| = \sqrt{3}$

This absolute value equation gives us two possible cases:

Case 1: $\sqrt{3} - \lambda = \sqrt{3}$ , which gives $\lambda = 0$ .

Case 2: $\sqrt{3} - \lambda = -\sqrt{3}$ , which gives $\lambda = 2\sqrt{3}$ .

Therefore, the sum of all possible values of $\lambda$ is:

$\lambda_{sum} = \lambda_1 + \lambda_2 = 0 + 2\sqrt{3} = 2\sqrt{3}$

Hence, option B ( $2 \sqrt{3}$ ) is the correct answer.