Question

If the shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{λ}\; and\; \frac{x-2}{1 }= \frac{y-4}{4 }= \frac{z-5}{5 }\;is\; \frac{1}{\sqrt3}$, then the sum of all possible values of $λ$ is :

• A. 16
• B. 6
• C. 12
• D. 15

Answer 1

Answer: (A)

16

Answer 2

$\overrightarrow a_1 = \hat i+2\hat j+3\hat k$

$\overrightarrow a_2 = 2\hat i + 4\hat j + 5\hat k$

$\overrightarrow p = 2\hat i+3\hat j+λ\hat k,\overrightarrow q = \hat i+4\hat j+5\hat k$

$∴ \overrightarrow p × \overrightarrow q = (15-4λ)\hat i-(10-λ)\hat j+5\hat k$

∴ Shortest distance between the lines = $| \frac{(15-4λ)-2(10-λ)+10}{√(15-4λ)^2+(10-λ)^2+25}| = \frac{1}{√3}$

$⇒ 5λ^2-80λ+275=0$

the sum of all possible values of λ is : = $\frac{80}{5} = 16$

Hence, the correct option is (A): $16$