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← JEE Main 2022 · 27 Jul · Shift 1
MathematicsSingle correct+4 / -1Question 13 of 90

JEE Main 2022 · 27 Jul · Shift 1, Mathematics question 13

Let a⃗=2i^−j^+5k^\vec a=2\hat i-\hat j+5\hat k and b⃗=αi^+βj^+2k^\vec b=\alpha\hat i+\beta\hat j+2\hat k. If ((a⃗×b⃗)×i^)⋅k^=232\left(\left(\vec a\times\vec b\right)\times\hat i\right)\cdot\hat k=\dfrac{23}2, then ∣b⃗×2j^∣\left|\vec b\times2\hat j\right| is equal to

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Answer and worked solution

Official answer: B

Triple product. (v⃗×i^)⋅k^=−v2(\vec v\times\hat i)\cdot\hat k=-v_2, and v2=5α−4v_2=5\alpha-4 for v⃗=a⃗×b⃗\vec v=\vec a\times\vec b. So 4−5α=2324-5\alpha=\frac{23}2, giving α=−32\alpha=-\frac32.

Cross product. b⃗×2j^=2(−2,0,α)\vec b\times2\hat j=2(-2,0,\alpha), magnitude 24+94=52\sqrt{4+\frac94}=5.

(B)