If the shortest distance between the lines and is , where gcd(m, n) = 1, then the value of m + n equals [2024]

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Solution

Q.
If the shortest distance between the lines

$L_{1} : \vec{r} = (2 + λ) \hat{i} + (1 - 3 λ) \hat{j} + (3 + 4 λ) \hat{k}, λ \in R$

$L_{2} : \vec{r} = 2 (1 + μ) \hat{i} + 3 (1 + μ) \hat{j} + (5 + μ) \hat{k}, μ \in R$

is $\frac{m}{\sqrt{n}}$ , where gcd(m, n) = 1, then the value of m + n equals [2024]

1 384

2 387

3 390

4 377

Ans.
(2)

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

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

$∴ a_{1} = 2 \hat{i} + \hat{j} + 3 \hat{k}, a_{2} = 2 \hat{i} + 3 \hat{j} + 5 \hat{k}$

$n_{1} = \hat{i} - 3 \hat{j} + 4 \hat{k}, n_{2} = 2 \hat{i} + 3 \hat{j} + \hat{k}$

Shortest distance = $\frac{| ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{n}}_{1} \times {\vec{n}}_{2}) |}{| {\vec{n}}_{1} \times {\vec{n}}_{2} |}$

$= \frac{| (2 \hat{j} + 2 \hat{k}) \cdot | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 3 & 4 \\ 2 & 3 & 1 \end{matrix} | |}{| {\vec{n}}_{1} \times {\vec{n}}_{2} |}$

$=$ $| \frac{(2 \hat{j} + 2 \hat{k}) \cdot (- 15 \hat{i} + 7 \hat{j} + 9 \hat{k})}{\sqrt{{(15)}^{2} + 7^{2} + 9^{2}}} |$ $= \frac{32}{\sqrt{355}} = \frac{m}{\sqrt{n}}$

On comparing, we get m = 32 and n = 355.

So, m + n = 32 + 355 = 387.

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