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← JEE Main 2022 · 29 Jun · Shift 1
MathematicsSingle correct+4 / -1Question 5 of 90

JEE Main 2022 · 29 Jun · Shift 1, Mathematics question 5

Let a⃗=αi^+3j^−k^\vec a=\alpha\hat i+3\hat j-\hat k, b⃗=3i^−βj^+4k^\vec b=3\hat i-\beta\hat j+4\hat k and c⃗=i^+2j^−2k^\vec c=\hat i+2\hat j-2\hat k where α,β∈R\alpha,\beta\in\mathbf R, be three vectors. If the projection of a⃗\vec a on c⃗\vec c is 103\dfrac{10}3 and b⃗×c⃗=−6i^+10j^+7k^\vec b\times\vec c=-6\hat i+10\hat j+7\hat k, then the value of α+β\alpha+\beta is equal to :

Choose one option
Answer and worked solution

Official answer: A

Projection. α+6+23=103\frac{\alpha+6+2}3=\frac{10}3, so α=2\alpha=2.

Cross product. (3,−β,4)×(1,2,−2)=(2β−8,10,6+β)=(−6,10,7)(3,-\beta,4)\times(1,2,-2)=(2\beta-8,10,6+\beta)=(-6,10,7), so β=1\beta=1.

α+β=3\alpha+\beta=3.

(A)

Topics:Cross productDot product
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