If the shortest distance between the straight lines \(3(x-1)=6(y-2)=2(z-1)\) and \(4(x-2)=2(y-\lambda)=(z-3), \lambda \in R\) is \(\frac{1}{\sqrt{38}}\), then the integral value of \(\lambda\) is equal to:

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Three Dimensional Geometry

If the shortest distance between the straight lines

3 (x - 1) = 6 (y - 2) = 2 (z - 1)

and

4 (x - 2) = 2 (y - λ) = (z - 3), λ \in R

\frac{1}{38}

, then the integral value of

λ

is equal to:

Answer: (A)

Writing each equation in symmetric form,

3 (x - 1) = 6 (y - 2) = 2 (z - 1)

becomes

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

and

4 (x - 2) = 2 (y - λ) = (z - 3)

becomes

\frac{x - 2}{1} = \frac{y - λ}{2} = \frac{z - 3}{4}

. The lines pass through

A (1, 2, 1)

and

B (2, λ, 3)

with directions

b_{1} = 2 \hat{i} + \hat{j} + 3 \hat{k}

and

b_{2} = \hat{i} + 2 \hat{j} + 4 \hat{k}

, and

b_{1} \times b_{2} = - 2 \hat{i} - 5 \hat{j} + 3 \hat{k}, b_{1} \times b_{2} = 38

With

A B = \hat{i} + (λ - 2) \hat{j} + 2 \hat{k}

, the shortest distance between the skew lines is

d = \frac{A B \cdot ( b _{1} \times b _{2} )}{b _{1} \times b _{2}} = \frac{∣ - 2 - 5 ( λ - 2 ) + 6∣}{38} = \frac{∣14 - 5 λ ∣}{38}

Equating this to

\frac{1}{38}

gives

∣14 - 5 λ ∣ = 1

, so

λ = \frac{13}{5}

λ = 3

. The integral value is

λ = 3

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Single correct option