If the shortest distance between the lines x –3 = y – 2 – 1 = z – 1 1 and x + 2 – 3 = y + 5 2 = z – 4 4 is 44 30 , then the largest possible value of is equal to [2024]

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Solution

Q.
If the shortest distance between the lines $\frac{x - λ}{3} = \frac{y - 2}{- 1} = \frac{z - 1}{1}$ and $\frac{x + 2}{- 3} = \frac{y + 5}{2} = \frac{z - 4}{4}$ is $\frac{44}{\sqrt{30}}$ , then the largest possible value of $| λ |$ is equal to __________. [2024]

Ans.
(43)

We have, $\vec{a_{1}} = λ \hat{i} + 2 \hat{j} + \hat{k}, \vec{a_{2}} = - 2 \hat{i} - 5 \hat{j} + 4 \hat{k}$

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

Now, $\vec{n_{1}} \times \vec{n_{2}} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & - 1 & 1 \\ - 3 & 2 & 4 \end{matrix} |$ $= - 6 \hat{i} - 15 \hat{j} + 3 \hat{k}$

Shortest distance between lines = $| \frac{(\vec{a_{2}} - \vec{a_{1}}) \cdot (\vec{n_{1}} \times \vec{n_{2}})}{| \vec{n_{1}} \times \vec{n_{2}} |} |$

$\Rightarrow \frac{44}{\sqrt{30}} =$ $| \frac{(- 2 - λ) \vec{i} - 7 \vec{j} + 3 \vec{k}) \cdot (- 6 \vec{i} - 15 \vec{j} + 3 \vec{k})}{\sqrt{36 + 225 + 9}} |$

$\Rightarrow \frac{44}{\sqrt{30}} =$ $| \frac{- 6 (- 2 - λ) + 105 + 9}{\sqrt{270}} |$ $\Rightarrow$ $| \frac{6 λ + 126}{\sqrt{270}} |$ $= \frac{44}{\sqrt{30}}$

$\Rightarrow | 6 λ + 126 | = 132 \Rightarrow | λ + 21 | = 22$

$\Rightarrow λ + 21 = \pm 22 \Rightarrow | λ |_{\max} = 43$ .

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