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← JEE Main 2026 · 5 Apr · Shift 1
MathematicsSingle correct+4 / -1Question 14 of 75

JEE Main 2026 · 5 Apr · Shift 1, Mathematics question 14

Let a⃗=7i^+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=\vec0 and r⃗⋅a⃗=0\vec r\cdot\vec a=0, then ∣3r⃗∣2\left|3\vec r\right|^2 is equal to:

Choose one option
Answer and worked solution

Official answer: A

Form of r⃗\vec r. r⃗×a⃗=b⃗×a⃗\vec r\times\vec a=\vec b\times\vec a gives (r⃗−b⃗)×a⃗=0⃗(\vec r-\vec b)\times\vec a=\vec0, so r⃗=b⃗+λa⃗\vec r=\vec b+\lambda\vec a.

Using r⃗⋅a⃗=0\vec r\cdot\vec a=0. a⃗⋅b⃗=1−2=−1\vec a\cdot\vec b=1-2=-1, ∣a⃗∣2=9|\vec a|^2=9, so λ=19\lambda=\tfrac19.

Length.
∣r⃗∣2=∣b⃗∣2+2λ a⃗⋅b⃗+λ2∣a⃗∣2=5−29+19=449,|\vec r|^2=|\vec b|^2+2\lambda\,\vec a\cdot\vec b+\lambda^2|\vec a|^2=5-\tfrac29+\tfrac19=\tfrac{44}9,
and ∣3r⃗∣2=44|3\vec r|^2=44.

(A)