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MCQMediumJEE 2025Skew Lines & Shortest Distance

JEE Mathematics 2025 Question with Solution

Let the values of pp, for which the shortest distance between the lines x+13=y4=z5\frac{x + 1}{3} = \frac{y}{4} = \frac{z}{5} and r⃗=(pi^+2j^+k^)+λ(2i^+3j^+4k^)\vec{r} = (p \hat{i} + 2 \hat{j} + \hat{k}) + \lambda (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) is 16\frac{1}{\sqrt{6}}, be a,ba, b, where a<ba < b. Then the length of the latus rectum of the ellipse x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 is:

  • A

    99

  • B

    32\frac{3}{2}

  • C

    23\frac{2}{3}

  • D

    1818

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The lines are

x+13=y4=z5\frac{x+1}{3}=\frac{y}{4}=\frac{z}{5}

and

r⃗=(pi^+2j^+k^)+λ(2i^+3j^+4k^)\vec r=(p\hat i+2\hat j+\hat k)+\lambda(2\hat i+3\hat j+4\hat k)

The shortest distance between them is

16\frac{1}{\sqrt6}

Find: If the corresponding values of pp are a,ba,b with $$a

For the first line, a point on it is (−1,0,0)(-1,0,0) and its direction vector is

p⃗=3i^+4j^+5k^\vec p=3\hat i+4\hat j+5\hat k

For the second line, a point on it is (p,2,1)(p,2,1) and its direction vector is

q⃗=2i^+3j^+4k^\vec q=2\hat i+3\hat j+4\hat k

The shortest distance between two skew lines is

d=∣(a⃗−b⃗)⋅(p⃗×q⃗)∣∣p⃗×q⃗∣d=\frac{|(\vec a-\vec b)\cdot(\vec p\times \vec q)|}{|\vec p\times \vec q|}

where

a⃗=−i^,\vec a=-\hat i, b⃗=pi^+2j^+k^\vec b=p\hat i+2\hat j+\hat k

So,

a⃗−b⃗=(−1−p)i^−2j^−k^\vec a-\vec b=(-1-p)\hat i-2\hat j-\hat k

Now compute the cross product:

p⃗×q⃗=∣i^j^k^345234∣\vec p\times \vec q= \begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & 4 & 5 \\ 2 & 3 & 4 \end{vmatrix} =i^(16−15)−j^(12−10)+k^(9−8)=\hat i(16-15)-\hat j(12-10)+\hat k(9-8) =i^−2j^+k^=\hat i-2\hat j+\hat k

Hence,

∣p⃗×q⃗∣=12+(−2)2+12=6|\vec p\times \vec q|=\sqrt{1^2+(-2)^2+1^2}=\sqrt6

Now,

(a⃗−b⃗)⋅(p⃗×q⃗)=((−1−p),−2,−1)⋅(1,−2,1)(\vec a-\vec b)\cdot(\vec p\times \vec q)=((-1-p),-2,-1)\cdot(1,-2,1) =(−1−p)+4−1=(-1-p)+4-1 =2−p=2-p

Therefore,

d=∣2−p∣6d=\frac{|2-p|}{\sqrt6}

Given that

d=16d=\frac{1}{\sqrt6}

so

∣2−p∣=1|2-p|=1

Thus,

p=1 or 3p=1 \text{ or } 3

Hence,

a=1,b=3a=1,\quad b=3

For the ellipse

x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

with a=1a=1 and b=3b=3, the semi-major axis is 33 and the semi-minor axis is 11. The length of the latus rectum is

L=2(1)23=23L=\frac{2(1)^2}{3}=\frac{2}{3}

Therefore, the length of the latus rectum is 23\frac{2}{3} and the correct option is C.

Using the distance condition directly

Given: The distance between the two lines is 16\frac{1}{\sqrt6}. Find: The latus rectum after determining the two values of pp.

A vector joining the chosen points (−1,0,0)(-1,0,0) and (p,2,1)(p,2,1) is

(−1−p,−2,−1)(-1-p,-2,-1)

A vector perpendicular to both lines is

(3,4,5)×(2,3,4)=(1,−2,1)(3,4,5)\times(2,3,4)=(1,-2,1)

Project the joining vector on this perpendicular direction:

(−1−p,−2,−1)⋅(1,−2,1)=2−p(-1-p,-2,-1)\cdot(1,-2,1)=2-p

Hence the shortest distance is

∣2−p∣12+(−2)2+12=∣2−p∣6\frac{|2-p|}{\sqrt{1^2+(-2)^2+1^2}}=\frac{|2-p|}{\sqrt6}

Equating with the given distance,

∣2−p∣6=16\frac{|2-p|}{\sqrt6}=\frac{1}{\sqrt6} ∣2−p∣=1|2-p|=1

So the two values are

p=1,3p=1,3

Since aa

Common mistakes

  • Using the wrong point on the first line. The line x+13=y4=z5\frac{x+1}{3}=\frac{y}{4}=\frac{z}{5} passes through (−1,0,0)(-1,0,0), not (1,0,0)(1,0,0). A wrong point changes the distance equation and gives incorrect values of pp.

  • Making an error in the cross product of the direction vectors. The perpendicular vector must be p⃗×q⃗=(1,−2,1)\vec p\times\vec q=(1,-2,1). If this is computed incorrectly, both the numerator and denominator of the distance formula become wrong.

  • Confusing which axis is major in the ellipse. Here a=1a=1 and b=3b=3 with aa

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