Let the values of p , for which the shortest distance between the lines (x+1)/(3)=(y)/(4)=(z)/(5) and r =(p i +2 j + k )+λ(2 i +3 j +4 k ) is 1 6 , be a, b , (a b) . Then the length of the latus rectum of the ellipse

Correct answer: C. (2)/(3) shortest distance = |( a - b )| ·( p × q ) | p × q | where aligned & a =- i +0 j +0 k & b =p i +2 j + k & p =3 i +4 j +5 k - b =(-1-p) i -2 j - k & q =2 i +3 j +4 k & (1)/(16)= |-1-p+4-1| 6 & |-p+2|=1 & p=3 & x^2 1^2 + y^2 3^3 =1 & L · R= 2 a^2 b =(2 × 1)/(3)=(2)/(3) aligned option (3)

JEE Main 20254 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 4 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Three Dimensional Geometry

JEE Main 2025 · 4 April · Evening Shift · Mathematics · Three Dimensional Geometry

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

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

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

and

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

\frac{1}{\sqrt{6}}

, be

\mathrm{a}, \mathrm{b}

(a \lt b)

. Then the length of the latus rectum of the ellipse

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

is :-

Options

A $9$
B $\frac{3}{2}$
C $\frac{2}{3}$
D $18$

Correct answer

C. $\frac{2}{3}$

Step-by-step solution

\text{ shortest distance }=\frac{|(\overline{\mathrm{a}}-\overline{\mathrm{b}})| \cdot(\overline{\mathrm{p}} \times \overline{\mathrm{q}})}{|\overline{\mathrm{p}} \times \overline{\mathrm{q}}|}

where

\begin{aligned}& \overline{\mathrm{a}}=-\hat{\mathrm{i}}+0 \hat{\mathrm{j}}+0 \hat{\mathrm{k}} \\ & \overline{\mathrm{ b}}=\mathrm{p} \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & \overline{\mathrm{p}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}-\overline{\mathrm{b}}=(-1-\mathrm{p}) \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-\hat{\mathrm{k}} \\ & \overline{\mathrm{q}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\ & \frac{1}{16}=\frac{|-1-\mathrm{p}+4-1|}{\sqrt{6}} \\ & |-\mathrm{p}+2|=1 \\ & \mathrm{p}=3 \\ & \frac{\mathrm{x}^2}{1^2}+\frac{\mathrm{y}^2}{3^3}=1 \\ & \mathrm{ L} \cdot \mathrm{R}=\frac{2 \mathrm{a}^2}{\mathrm{ b}}=\frac{2 \times 1}{3}=\frac{2}{3}\end{aligned}

option (3)

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