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← JEE Main 2022 · 28 Jul · Shift 1
MathematicsSingle correct+4 / -1Question 6 of 90

JEE Main 2022 · 28 Jul · Shift 1, Mathematics question 6

Let a vector a⃗\vec a has magnitude 9. Let a vector b⃗\vec b be such that for every (x,y)∈R×R−{(0,0)}(x,y)\in\mathbf R\times\mathbf R-\{(0,0)\}, the vector (xa⃗+yb⃗)\left(x\vec a+y\vec b\right) is perpendicular to the vector (6ya⃗−18xb⃗)\left(6y\vec a-18x\vec b\right). Then the value of ∣a⃗×b⃗∣\left|\vec a\times\vec b\right| is equal to :

Choose one option
Answer and worked solution

Official answer: B

Expand (xa⃗+yb⃗)⋅(6ya⃗−18xb⃗)=0(x\vec a+y\vec b)\cdot(6y\vec a-18x\vec b)=0 for all x,yx,y: a⃗⋅b⃗=0\vec a\cdot\vec b=0 and 6∣a⃗∣2=18∣b⃗∣26|\vec a|^2=18|\vec b|^2, so ∣b⃗∣=33|\vec b|=3\sqrt3.

Cross product 9×33=2739\times3\sqrt3=27\sqrt3. (B)