Consider the line L passing through the points (1, 2, 3) and (2, 3, 5). The distance of the point from the line L along the line is equal to [2024]

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Solution

Q.
Consider the line L passing through the points (1, 2, 3) and (2, 3, 5). The distance of the point $(\frac{11}{3}, \frac{11}{3}, \frac{19}{3})$ from the line L along the line $\frac{3 x - 11}{2} = \frac{3 y - 11}{1} = \frac{3 z - 19}{2}$ is equal to [2024]

1 4

2 3

3 5

4 6

Ans.
(2)

Equation of line L is given by

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

i.e., $\frac{x - 1}{1} = \frac{y - 2}{1} = \frac{z - 3}{2} = λ$

$∴$ Any point on L is given by $(λ + 1, λ + 2, 2 λ + 3)$

Also, any point of $L_{1} : \frac{3 x - 11}{2} = \frac{3 y - 11}{1} = \frac{3 z - 19}{2} = μ$

is given by $(\frac{2 μ + 11}{3}, \frac{μ + 11}{3}, \frac{2 μ + 19}{3})$

Now, $λ + 1 = \frac{2 μ + 11}{3}, λ + 2 = \frac{μ + 11}{3}, 2 λ + 3 = \frac{2 μ + 19}{3}$

$\Rightarrow 3 λ - 2 μ = 8 and 3 λ - μ = 5$

On solving these two equations, we get $λ = \frac{2}{3}$ and $μ = - 3$

Point on L is $(\frac{5}{3}, \frac{8}{3}, \frac{13}{3})$

Required distance = $\sqrt{{(\frac{11}{3} - \frac{5}{3})}^{2} + {(\frac{11}{3} - \frac{8}{3})}^{2} + {(\frac{19}{3} - \frac{13}{3})}^{2}}$

$= \sqrt{4 + 1 + 1} = 3$ units.

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