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 · medium
If the distance of the point (a,2,5)(a,2,5) from the image of (1,2,7)(1,2,7) in the line x1=y−11=z−22\dfrac{x}{1}=\dfrac{y-1}{1}=\dfrac{z-2}{2}is 4, then the sum of all possible values of aa is:
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 · medium
Let OO be the origin with OP→=a⃗\overrightarrow{OP}=\vec{a} and OQ→=b⃗\overrightarrow{OQ}=\vec{b}. If RR is on OP→\overrightarrow{OP} such that OP→=5OR→\overrightarrow{OP}=5\overrightarrow{OR}, and MM is such that OQ→=5RM→\overrightarrow{OQ}=5\overrightarrow{RM}, then PM→\overrightarrow{PM} equals:
Vectors & 3D Geometry · medium
Let △PQR\triangle PQR be such that PP and QQ lie on x+38=y−42=z+12\dfrac{x+3}{8}=\dfrac{y-4}{2}=\dfrac{z+1}{2} and are at distance 6 from R(1,2,3)R(1,2,3). If (α,β,γ)(\alpha,\beta,\gamma) is the centroid of △PQR\triangle PQR, then α+β+γ\alpha+\beta+\gamma equals:

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