JEE Mathematics 2023 Question with Solution

Given: The lines are

$$\frac{x-5}{1} = \frac{y-2}{2} = \frac{z-4}{-3}$$

and

$$\frac{x+3}{1} = \frac{y+5}{4} = \frac{z-1}{-5}$$

Find: The shortest distance between these two lines.

From the solution, the conclusion stated is that the correct answer is option B, and the computed distance is $6\sqrt{3}$. This disagrees with the listed options, where $6\sqrt{3}$ is option C.

Using the determinant formula shown in the solution for shortest distance between two lines,

$$d = \frac{\begin{vmatrix} x_1-x_2 & y_1-y_2 & z_1-z_2\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{vmatrix}}{\sqrt{(a_1b_3-a_3b_2)^2+(a_1b_3-a_3b_1)^2+(a_1b_2-a_2b_1)^2}}$$

Substituting the values shown in the solution,

$$d = \frac{\begin{vmatrix} 5-(3) & 2-(-5) & 4-1\\ 1 & 2 & -3\\ 1 & 4 & -5 \end{vmatrix}}{\sqrt{(-10+12)^2+(-5+3)^2+(4-2)^2}}$$ $$= \frac{\begin{vmatrix} 8 & 7 & 3\\ 1 & 2 & -3\\ 1 & 4 & -5 \end{vmatrix}}{\sqrt{(2)^2+(-2)^2+(2)^2}}$$ $$= \frac{\begin{vmatrix} 8(-10+12)-7(-5+3)+3(4-2)\end{vmatrix}}{\sqrt{4+4+4}}$$ $$= \frac{\begin{vmatrix} 16+14+6\end{vmatrix}}{\sqrt{12}}$$ $$= \frac{36}{\sqrt{12}} = \frac{36}{2\sqrt{3}}$$ $$= \frac{18}{\sqrt{3}} = 6\sqrt{3}$$

Therefore, the shortest distance is $6\sqrt{3}$. The solution labels this as option B, so the answer is recorded as B based on the solution, despite the option-value mismatch.

Given: The same two lines are provided.

Find: The shortest distance.

The second approach shown in the solution states:

Step 1: Shortest distance formula between skew lines:

$$d = \frac{|(a_1 b_2 - a_2 b_1) + (a_2 b_3 - a_3 b_2) + (a_3 b_1 - a_1 b_3)|} {\sqrt{(a_1 - a_2)^2 + (b_1 - b_2)^2 + (c_1 - c_2)^2}}$$

Step 2: Substituting values,

$$d = \frac{8(-10 + 12) - 7(-5 + 3) + 3(4 - 2)}{\sqrt{4 + 4 + 4}}$$ $$= \frac{16 + 14 + 6}{\sqrt{12}} = \frac{36}{\sqrt{12}} = \frac{36}{2\sqrt{3}}$$ $$= \frac{18}{\sqrt{3}} = 6\sqrt{3}$$

Hence, the shortest distance is $6\sqrt{3}$.