Let the values of p, for which the shortest distance between the lines $\frac{\mathrm{x}+1}{3}=\frac{\mathrm{y}}{4}=\frac{\mathrm{z}}{5}$ and $\vec{r}=(p \hat{i}+2 \hat{j}+\hat{k})+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})$ is $\frac{1}{\sqrt{6}}$, be $a, b$, $(\mathrm{a}<\mathrm{b})$. Then the length of the latus rectum of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is :-

JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Three Dimensional Geometry

Let the values of p, for which the shortest distance between the lines

\frac{x + 1}{3} = \frac{y}{4} = \frac{z}{5}

and

r = (p \hat{i} + 2 \hat{j} + \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 4 \hat{k})

\frac{1}{6}

, be

a, b

(a < b)

. Then the length of the latus rectum of the ellipse

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

is :-

Answer: (C)

With direction vectors

(3, 4, 5)

and

(2, 3, 4)

, their cross product is

(1, - 2, 1)

of length

6

. The shortest distance between the lines is

\frac{( p + 1 ) ( 1 ) + 2 ( - 2 ) + 1 ( 1 )}{6} = \frac{∣ p - 2∣}{6}

, and setting this equal to

\frac{1}{6}

gives

∣ p - 2∣ = 1

, so

a = 1, b = 3

. For

\frac{x ^{2}}{1} + \frac{y ^{2}}{9} = 1

the major axis lies along

y

, and the latus rectum is

\frac{2 ( 1 )}{3} = \frac{2}{3}

JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Single correct option