Skip to content
dronacharya.ai
← JEE Main 2022 · 27 Jul · Shift 2
MathematicsInteger+4 / -1Question 30 of 90

JEE Main 2022 · 27 Jul · Shift 2, Mathematics question 30

Let a⃗,b⃗,c⃗\vec a,\vec b,\vec c be three non-coplanar vectors such that a⃗×b⃗=4c⃗\vec a\times\vec b=4\vec c, b⃗×c⃗=9a⃗\vec b\times\vec c=9\vec a and c⃗×a⃗=αb⃗\vec c\times\vec a=\alpha\vec b, α>0\alpha>0. If ∣a⃗∣+∣b⃗∣+∣c⃗∣=136\left|\vec a\right|+\left|\vec b\right|+\left|\vec c\right|=\dfrac1{36}, then α\alpha is equal to ______.

Answer and worked solution

Official answer: Dropped: full marks to everyone

This question was dropped in NTA's final key.

Why. The vectors are mutually perpendicular. From ∣a∣∣b∣=4∣c∣|a||b|=4|c| and ∣b∣∣c∣=9∣a∣|b||c|=9|a| we get ∣b⃗∣=6|\vec b|=6, so ∣a⃗∣+∣b⃗∣+∣c⃗∣=136|\vec a|+|\vec b|+|\vec c|=\frac1{36} is impossible.