If the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda}$ 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 $\lambda$ is :

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Three Dimensional Geometry

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}

\frac{1}{3}

, then the sum of all possible values of

λ

is :

Answer: (A)

For lines

\frac{x - x _{1}}{a _{1}} = \frac{y - y _{1}}{b _{1}} = \frac{z - z _{1}}{c _{1}}

and

\frac{x - x _{2}}{a _{2}} = \frac{y - y _{2}}{b _{2}} = \frac{z - z _{2}}{c _{2}}

, the shortest distance is

\frac{∣ ( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} ) ∣}{∣ b _{1} \times b _{2} ∣}

. Here

a_{1} = (1, 2, 3)

a_{2} = (2, 4, 5)

b_{1} = (2, 3, λ)

b_{2} = (1, 4, 5)

. Compute

b_{1} \times b_{2} = (15 - 4 λ, λ - 10, 5)

. The numerator is

∣ (1, 2, 2) \cdot (15 - 4 λ, λ - 10, 5) ∣ = ∣15 - 4 λ + 2 λ - 20 + 10∣ = ∣5 - 2 λ ∣

. Setting

\frac{∣5 - 2 λ ∣}{( 15 - 4 λ ) ^{2} + ( λ - 10 ) ^{2} + 25} = \frac{1}{3}

and squaring:

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

. Simplifying gives

5 λ^{2} - 80 λ + 275 = 0 \Rightarrow λ^{2} - 16 λ + 55 = 0 \Rightarrow λ = 5, 11

. Sum = 16.

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Single correct option