If the square of the shortest distance between the lines and is , where m, n are coprime numbers, then m + n is equal to : [2025]

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Solution

Q.
If the square of the shortest distance between the lines $\frac{x - 2}{1} = \frac{y - 1}{2} = \frac{z + 3}{- 3}$ and $\frac{x + 1}{2} = \frac{y + 3}{4} = \frac{z + 5}{- 5}$ is $\frac{m}{n}$ , where m, n are coprime numbers, then m + n is equal to : [2025]

1 6

2 14

3 21

4 9

Ans.
(4)

We have, ${\vec{a}}_{1} = 2 \hat{i} + \hat{j} - 3 \hat{k}, {\vec{b}}_{1} = \hat{i} + 2 \hat{j} - 3 \hat{k}$

${\vec{a}}_{2} = - \hat{i} - 3 \hat{j} - 5 \hat{k}, {\vec{b}}_{2} = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}$

Now, ${\vec{b}}_{1} \times {\vec{b}}_{2} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & - 3 \\ 2 & 4 & - 5 \end{matrix} |$ $= 2 \hat{i} - \hat{j}$

and ${\vec{a}}_{2} - {\vec{a}}_{1} = - 3 \hat{i} - 4 \hat{j} - 2 \hat{k}$

$∴$ The shortest distance (d) between given lines

$= \frac{| ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) |}{| ({\vec{b}}_{1} \times {\vec{b}}_{2}) |} \Rightarrow d = \frac{2}{\sqrt{5}} \Rightarrow d^{2} = \frac{4}{5}$

$∴ m = 4, n = 5 \Rightarrow m + n = 9$

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