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← JEE Main 2022 · 28 Jun · Shift 2
MathematicsSingle correct+4 / -1Question 19 of 90

JEE Main 2022 · 28 Jun · Shift 2, Mathematics question 19

Let a⃗\vec a be a vector which is perpendicular to the vector 3i^+12j^+2k^3\hat i+\dfrac12\hat j+2\hat k. If a⃗×(2i^+k^)=2i^−13j^−4k^\vec a\times\left(2\hat i+\hat k\right)=2\hat i-13\hat j-4\hat k, then the projection of the vector a⃗\vec a on the vector 2i^+2j^+k^2\hat i+2\hat j+\hat k is :

Choose one option
Answer and worked solution

Official answer: C

Unknowns. a⃗×(2,0,1)=(y, 2z−x, −2y)=(2,−13,−4)\vec a\times(2,0,1)=(y,\,2z-x,\,-2y)=(2,-13,-4) gives y=2y=2, 2z−x=−132z-x=-13; perpendicularity gives 3x+1+2z=03x+1+2z=0. So a⃗=(3,2,−5)\vec a=(3,2,-5).

Projection. 6+4−53=53\dfrac{6+4-5}3=\dfrac53.

(C)

Topics:Cross productDot product
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