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← JEE Main 2022 · 26 Jul · Shift 1
MathematicsSingle correct+4 / -1Question 15 of 90

JEE Main 2022 · 26 Jul · Shift 1, Mathematics question 15

Let a⃗=αi^+j^−k^\vec a=\alpha\hat i+\hat j-\hat k and b⃗=2i^+j^−αk^\vec b=2\hat i+\hat j-\alpha\hat k, α>0\alpha>0. If the projection of a⃗×b⃗\vec a\times\vec b on the vector −i^+2j^−2k^-\hat i+2\hat j-2\hat k is 30, then α\alpha is equal to :

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Answer and worked solution

Official answer: D

Cross product a⃗×b⃗=(1−α, α2−2, α−2)\vec a\times\vec b=(1-\alpha,\,\alpha^2-2,\,\alpha-2).

Projection −(1−α)+2(α2−2)−2(α−2)3=2α2−α−13=30\dfrac{-(1-\alpha)+2(\alpha^2-2)-2(\alpha-2)}3=\dfrac{2\alpha^2-\alpha-1}3=30: α=7\alpha=7.

(D)

Topics:Cross productDot product
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