If the shortest distance between the lines x - 1 2 = y - 2 3 = z - 3 and x - 2 1 = y - 4 4 = z - 5 5 is 1 3 , then the sum of all possible values of is :

Question

If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{λ}$ and $\frac{x - 2}{1} = \frac{y - 4}{4} = \frac{z - 5}{5}$ is $\frac{1}{\sqrt{3}}$ , then the sum of all possible values of $λ$ is :

Smart Achievers · Accepted Answer

(1)

${\vec{a}}_{1} = (1, 2, 3),$ ${\vec{a}}_{2} = (2, 4, 5)$

${\vec{b}}_{1} = (2, 3, λ),$ ${\vec{b}}_{2} = (1, 4, 5)$

$Shortest distance:$

$\frac{| ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) |}{| {\vec{b}}_{1} \times {\vec{b}}_{2} |}$ $= \frac{1}{\sqrt{3}}$

${\vec{a}}_{2} - {\vec{a}}_{1} = (1, 2, 2)$

${\vec{b}}_{1} \times {\vec{b}}_{2} = (15 - 4 λ, λ - 10, 5)$

$So,$

$\frac{| 5 - 2 λ |}{\sqrt{17 λ^{2} - 140 λ + 350}} = \frac{1}{\sqrt{3}}$

$Squaring:$

$3 {(5 - 2 λ)}^{2} = 17 λ^{2} - 140 λ + 350$

$λ^{2} - 16 λ + 55 = 0$

$(λ - 5) (λ - 11) = 0$

$λ = 5, 11$

$Sum = 16$

Solution

Q.
If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{λ}$ and $\frac{x - 2}{1} = \frac{y - 4}{4} = \frac{z - 5}{5}$ is $\frac{1}{\sqrt{3}}$ , then the sum of all possible values of $λ$ is :

1 16

2 6

3 12

4 15

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