Vectors & 3D GeometrymediumPYQ · JEE Main · 5 Apr 2026 · Shift 1 (Morning)Free

|3r|² for a Vector Equation = 44 | JEE Main 2026

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

Question
Let a⃗=7 i^+j^−k^\vec a=\sqrt7\,\hat i+\hat j-\hat k and b⃗=j^+2k^\vec b=\hat j+2\hat k. If r⃗\vec r is a vector such that r⃗×a⃗+a⃗×b⃗=0⃗\vec r\times\vec a+\vec a\times\vec b=\vec 0 and r⃗⋅a⃗=0\vec r\cdot\vec a=0, then ∣3r⃗∣2|3\vec r|^2 is equal to
A4444correct
B5454
C8686
D132132
Solution
Step 1: a⃗×b⃗=−b⃗×a⃗\vec a\times\vec b=-\vec b\times\vec a, so r⃗×a⃗+a⃗×b⃗=0⃗⇒(r⃗−b⃗)×a⃗=0⃗⇒r⃗−b⃗=λa⃗\vec r\times\vec a+\vec a\times\vec b=\vec0\Rightarrow(\vec r-\vec b)\times\vec a=\vec0\Rightarrow\vec r-\vec b=\lambda\vec a:
r⃗=b⃗+λa⃗.\vec r=\vec b+\lambda\vec a.
 Step 2: r⃗⋅a⃗=0⇒b⃗⋅a⃗+λ∣a⃗∣2=0\vec r\cdot\vec a=0\Rightarrow\vec b\cdot\vec a+\lambda|\vec a|^2=0. a⃗⋅b⃗=0+1−2=−1\vec a\cdot\vec b=0+1-2=-1, ∣a⃗∣2=7+1+1=9|\vec a|^2=7+1+1=9:
−1+9λ=0⇒λ=19.-1+9\lambda=0\Rightarrow\lambda=\frac19.
 Step 3: ∣r⃗∣2=∣b⃗∣2+29(a⃗⋅b⃗)+181∣a⃗∣2|\vec r|^2=|\vec b|^2+\dfrac{2}{9}(\vec a\cdot\vec b)+\dfrac{1}{81}|\vec a|^2, with ∣b⃗∣2=1+4=5|\vec b|^2=1+4=5:
∣r⃗∣2=5+29(−1)+981=5−29+19=449.|\vec r|^2=5+\frac{2}{9}(-1)+\frac{9}{81}=5-\frac{2}{9}+\frac{1}{9}=\frac{44}{9}.
 Step 4: ∣3r⃗∣2=9⋅449=44|3\vec r|^2=9\cdot\dfrac{44}{9}=44. Correct answer: (1)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Vectors & 3D Geometry · easy
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
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

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