Vectors & 3D GeometryeasyPYQ · JEE Main · 4 Apr 2026 · Shift 2 (Afternoon)Free

Shortest Distance Between Two Skew Lines = 3 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
The shortest distance between the lines r⃗=(13i^+2j^+83k^)+λ(2i^−5j^+6k^)\vec r=\left(\dfrac13\hat i+2\hat j+\dfrac83\hat k\right)+\lambda\left(2\hat i-5\hat j+6\hat k\right) and r⃗=(−23i^−13k^)+μ(j^−k^)\vec r=\left(-\dfrac23\hat i-\dfrac13\hat k\right)+\mu\left(\hat j-\hat k\right), λ,μ∈R\lambda,\mu\in\mathbb{R}, is
A5\sqrt5
B33correct
C232\sqrt3
D15\sqrt{15}
Solution
Step 1: Direction vectors b⃗1=2i^−5j^+6k^\vec b_1=2\hat i-5\hat j+6\hat k, b⃗2=j^−k^\vec b_2=\hat j-\hat k. Cross product:
b⃗1×b⃗2=∣i^j^k^2−5601−1∣=−i^+2j^+2k^,∣b⃗1×b⃗2∣=1+4+4=3.\vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\2&-5&6\\0&1&-1\end{vmatrix}=-\hat i+2\hat j+2\hat k,\qquad |\vec b_1\times\vec b_2|=\sqrt{1+4+4}=3.
 Step 2: a⃗1−a⃗2=(13+23)i^+2j^+(83+13)k^=i^+2j^+3k^\vec a_1-\vec a_2=\left(\dfrac13+\dfrac23\right)\hat i+2\hat j+\left(\dfrac83+\dfrac13\right)\hat k=\hat i+2\hat j+3\hat k. Step 3: Shortest distance:
S.D.=∣(i^+2j^+3k^)⋅(−i^+2j^+2k^)∣3=∣−1+4+6∣3=93=3.\text{S.D.}=\frac{\left|(\hat i+2\hat j+3\hat k)\cdot(-\hat i+2\hat j+2\hat k)\right|}{3}=\frac{|-1+4+6|}{3}=\frac{9}{3}=3.
 Correct answer: (2)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Vectors & 3D Geometry · medium
If (2α+1, α2−3α, α−12)\left(2\alpha+1,\ \alpha^2-3\alpha,\ \dfrac{\alpha-1}{2}\right) is the image of (α, 2α, 1)(\alpha,\ 2\alpha,\ 1) in the line x−23=y−12=z1\dfrac{x-2}{3}=\dfrac{y-1}{2}=\dfrac{z}{1}, then the possible value(s) of α\alpha is(are)
Vectors & 3D Geometry · medium
Let u^\hat u and v^\hat v be unit vectors inclined at an acute angle such that ∣u^×v^∣=32|\hat u\times\hat v|=\dfrac{\sqrt3}{2}. If A⃗=λu^+v^+(u^×v^)\vec A=\lambda\hat u+\hat v+(\hat u\times\hat v), then λ\lambda is equal to
Vectors & 3D Geometry · medium
A line with direction ratios 1,−1,21,-1,2 intersects the lines x2=y3=z+13\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z+1}{3} and x+1−1=y−21=z4\dfrac{x+1}{-1}=\dfrac{y-2}{1}=\dfrac{z}{4} at the points PP and QQ respectively. If the length of the line segment PQPQ is α\alpha, then 225α2225\alpha^2 is equal to
Vectors & 3D Geometry · easy
The square of the distance of the point (−2,−8,6)(-2,-8,6) from the line x−11=y−12=z−1\dfrac{x-1}{1}=\dfrac{y-1}{2}=\dfrac{z}{-1} along the line x+51=y+5−1=z2\dfrac{x+5}{1}=\dfrac{y+5}{-1}=\dfrac{z}{2} is equal to

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath