JEE Mathematics 2025 Question with Solution

Given: The lines are $\frac{x-2}{1} = \frac{y-1}{2} = \frac{z+3}{-3}$ and $\frac{x+1}{2} = \frac{y+3}{4} = \frac{z+5}{-5}$.

Find: The value of $m+n$ if the square of the shortest distance is $\frac{m}{n}$.

Write the lines in vector form using points and direction vectors:

$$\vec{r}_1 = (2,1,-3) + t(1,2,-3)$$ $$\vec{r}_2 = (-1,-3,-5) + s(2,4,-5)$$

So, take

$$\vec{a}_1 = (1,2,-3), \quad \vec{a}_2 = (2,4,-5)$$

and points

$$\vec{b}_1 = (2,1,-3), \quad \vec{b}_2 = (-1,-3,-5)$$

The shortest distance between two skew lines is

$$d = \frac{|(\vec{b}_2-\vec{b}_1) \cdot (\vec{a}_1 \times \vec{a}_2)|}{|\vec{a}_1 \times \vec{a}_2|}$$

Compute the cross product:

$$\vec{a}_1 \times \vec{a}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix} = (2, -1, 0)$$

Hence,

$$|\vec{a}_1 \times \vec{a}_2| = \sqrt{2^2 + (-1)^2} = \sqrt{5}$$

Now,

$$\vec{b}_2 - \vec{b}_1 = (-1,-3,-5) - (2,1,-3) = (-3,-4,-2)$$

Then,

$$(\vec{b}_2-\vec{b}_1) \cdot (\vec{a}_1 \times \vec{a}_2) = (-3,-4,-2) \cdot (2,-1,0) = -6 + 4 = -2$$

So,

$$d = \frac{|-2|}{\sqrt{5}} = \frac{2}{\sqrt{5}}$$

Therefore,

$$d^2 = \frac{4}{5}$$

Thus, $m = 4$ and $n = 5$, so

$$m+n = 9$$

Therefore, the correct option is B.

The first provided approach contains a sign error in the cross product. It states

$$\vec{a}_1 \times \vec{a}_2 = (2,1,0)$$

but the correct computation is

$$\vec{a}_1 \times \vec{a}_2 = \hat{i}(2 \cdot -5 - (-3) \cdot 4) - \hat{j}(1 \cdot -5 - (-3) \cdot 2) + \hat{k}(1 \cdot 4 - 2 \cdot 2)$$ $$= 2\hat{i} - \hat{j} + 0\hat{k} = (2,-1,0)$$

Using the incorrect vector gives a wrong dot product and hence the wrong value $21$. The second provided approach correctly concludes that the shortest distance squared is $\frac{4}{5}$, which gives $m+n = 9$.