If the shortest distance between the line joining the points ( 1 , 2 , 3 ) and ( 2 , 3 , 4 ) , and the line x - 1 2 = y + 1 - 1 = z - 2 0 is , then 28 2 is equal to ______ . [2023]

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Solution

Q.
If the shortest distance between the line joining the points $(1, 2, 3)$ and $(2, 3, 4)$ , and the line $\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z - 2}{0}$ is $α$ , then $28 α^{2}$ is equal to ______ . [2023]

Ans.
(18)

$The equation of the line passing through (1, 2, 3) and (2, 3, 4) is$

$\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} + \hat{j} + \hat{k}) (∵ \vec{r} = \vec{a} + λ \vec{p})$

and vector form of equation of given second line is

$\vec{r} = (\hat{i} - \hat{j} + 2 \hat{k}) + μ (2 \hat{i} - \hat{j}) (∵ \vec{r} = \vec{b} + μ \vec{q})$

$Now,$ $\vec{p} \times \vec{q} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & - 1 & 0 \end{matrix} |$ $= \hat{i} + 2 \hat{j} - 3 \hat{k}$

$So, the shortest distance =$ $| \frac{(\vec{b} - \vec{a}) \cdot (\vec{p} \times \vec{q})}{| \vec{p} \times \vec{q} |} |$

$=$ $| \frac{(- 3 \hat{j} - \hat{k}) \cdot (\hat{i} + 2 \hat{j} - 3 \hat{k})}{\sqrt{14}} |$ $=$ $| \frac{- 6 + 3}{\sqrt{14}} |$ $= \frac{3}{\sqrt{14}} = α = \frac{3}{\sqrt{14}}$

$Now, 28 α^{2} = 28 \times \frac{9}{14} = 18$

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