CBSE 2025, Set 3

Question

**Case Study - 2**
An engineer is designing a new metro rail network in a city.

![Diagram Metro Rail](https://www.edvaya.com/images/cbse/2025/mathematics/set3/q37-metro-rail.webp)

Initially, two metro lines, Line A and Line B, each consisting of multiple stations are designed. The track for Line A is represented by
$l_1: \frac{x-2}{3} = \frac{y+1}{-2} = \frac{z-3}{4}$ while the track for Line B is represented by
$l_2: \frac{x-1}{2} = \frac{y-3}{1} = \frac{z+2}{-3}$.

Based on the above information, answer the following question:
(iii) (b) Find the shortest distance between Line A and Line B.

Edvaya · Accepted Answer

The correct answer is Subjective. **Step 1:** Extract the position vectors and direction vectors of both lines. From the cartesian equations, the lines pass through points $\vec{a_1}$ and $\vec{a_2}$ with directions $\vec{b_1}$ and $\vec{b_2}$: Line A ($l_1$): $\vec{a_1} = 2\hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b_1} = 3\hat{i} - 2\hat{j} + 4\hat{k}$ Line B ($l_2$): $\vec{a_2} = \hat{i} + 3\hat{j} - 2\hat{k}$ and $\vec{b_2} = 2\hat{i} + \hat{j} - 3\hat{k}$ **Step 2:** State the shortest distance formula. The shortest distance $d$ between two skew lines $\vec{r} = \vec{a_1} + \lambda\vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu\vec{b_2}$ is: $$d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2})}{|\vec{b_1} 	imes \vec{b_2}|} ight|$$ **Step 3:** Calculate the necessary vector components. First, compute the vector difference $(\vec{a_2} - \vec{a_1})$: $$\vec{a_2} - \vec{a_1} = (1 - 2)\hat{i} + (3 - (-1))\hat{j} + (-2 - 3)\hat{k}$$ $$\vec{a_2} - \vec{a_1} = -\hat{i} + 4\hat{j} - 5\hat{k}$$ Next, compute the cross product $\vec{b_1} 	imes \vec{b_2}$ (as calculated in the previous part): $$\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -2 & 4 \ 2 & 1 & -3 \end{vmatrix} = 2\hat{i} + 17\hat{j} + 7\hat{k}$$ **Step 4:** Calculate the dot product and magnitude. Calculate the dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2})$: $$= (-1)(2) + (4)(17) + (-5)(7)$$ $$= -2 + 68 - 35 = 31$$ Calculate the magnitude of the cross product $|\vec{b_1} 	imes \vec{b_2}|$: $$|\vec{b_1} 	imes \vec{b_2}| = \sqrt{2^2 + 17^2 + 7^2}$$ $$= \sqrt{4 + 289 + 49} = \sqrt{342}$$ **Step 5:** Calculate the shortest distance. Substitute the values back into the distance formula: $$d = \left| \frac{31}{\sqrt{342}} ight|$$ $$d = \frac{31}{\sqrt{342}} 	ext{ units}$$ Therefore, the shortest distance between Line A and Line B is $\frac{31}{\sqrt{342}} 	ext{ units}$.

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