Question

Let \( \lambda \) be an integer. If the shortest distance between the lines

\( x-\lambda=2 y-1=-2 z \) and 

\[ x=y+2 \lambda=z-\lambda \] 

is \( \frac{\sqrt{7}}{2 \sqrt{2}} \) then the value of \( |\lambda| \) is

Answer 1

We know that distance between two show line

\(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\) and \(\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}\)

is d = \(\cfrac{\begin{vmatrix}x_2-x_1&y_2-y_1&z_2-z_1\\a_1&b_1&c_1\\a_2&b_2&c_2\end{vmatrix}}{\sqrt{(b_1c_2-b_2c_1)^2+(c_1a_2-c_2a_1)^2+(a_1b_2-a_2b_1)^2}}\)

Given lines are x - λ = 2y - 1 = -2z

⇒ \(\frac{x-\lambda}1=\cfrac{y-\frac12}{\frac12}=\cfrac{z}{\frac{-1}2}\)

and x = y + 2λ = z - λ

⇒ \(\frac{x-0}1 = \frac{y-(-2\lambda)}1 = \frac{z-\lambda}1\) 

\(\therefore\) x₁  = λ, y₁ = 1/2, z₁ = λ

a₁ = 1, b₁ = 1, c₂ = 1 and

a₂ = 1, b₂ = 1 c₂ = 1

\(\therefore\) d = \(\cfrac{\begin{vmatrix}0-\lambda&-2\lambda-1/2&\lambda-0\\1&1/2&-1/2\\1&1&1\end{vmatrix}}{\sqrt{(1/2+1/2)^2+(-1/2-1)^2+(1-1/2)^2}}\)

Given that distance between them is \(\frac{\sqrt7}{2\sqrt2}\)

\(\therefore\) \(\cfrac{\begin{vmatrix}-\lambda&-2\lambda-1/2&\lambda\\1&1/2&-1/2\\1&1&1\end{vmatrix}}{\sqrt{1+9/4+1/4}}\) \(=\frac{\sqrt7}{2\sqrt2}\) 

⇒ \(\cfrac{-\lambda\begin{vmatrix}1/2&-1/2\\1&1\end{vmatrix}+(2\lambda+1/2)\begin{vmatrix}1&-1/2\\1&1\end{vmatrix}+\lambda\begin{vmatrix}1&1/2\\1&1\end{vmatrix}}{\sqrt{\frac72}}\) = \(\frac{\sqrt7}{2\sqrt2}\)

⇒ -λ(1/2 + 1/2) + (2λ + 1/2)(1 + 1/2) + λ(1 - 1/2)

 = \(\frac{\sqrt7}{2\sqrt2}\times\frac{\sqrt7}{\sqrt2}\)

⇒ -λ + 3λ + 3/4 + λ/2 = 7/4

⇒ \(\frac{2\lambda+6\lambda+\lambda}2\) = \(\frac74-\frac34 = \frac44=1\) 

⇒ 5λ = 2

⇒ λ = 2/5