JEE Main 2022 · 24 Jun · Shift 1 · Mathematics Q16

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MathematicsSingle correct+4 / -1Question 16 of 90

Let $\hat a,\hat b$ be unit vectors. If $\vec c$ be a vector such that the angle between $\hat a$ and $\vec c$ is $\dfrac\pi{12}$ , and $\hat b=\vec c+2\left(\vec c\times\hat a\right)$ , then $\left|6\vec c\right|^2$ is equal to :

6(3-\sqrt3)

3+\sqrt3

6(3+\sqrt3)

6(\sqrt3+1)

Answer and worked solution

Official answer: C

Square the relation. $\vec c$ and $\vec c\times\hat a$ are perpendicular, so
$1=|\hat b|^2=|\vec c|^2+4|\vec c\times\hat a|^2=|\vec c|^2\left(1+4\sin^2\tfrac\pi{12}\right).$

Angle. $4\sin^2 15^\circ=2(1-\cos30^\circ)=2-\sqrt3$ , so $|\vec c|^2=\dfrac1{3-\sqrt3}=\dfrac{3+\sqrt3}6$ .

Value. $|6\vec c|^2=36\cdot\dfrac{3+\sqrt3}6=6(3+\sqrt3)$ .

(C)

Topics:Cross product Dot product

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JEE Main 2022 · 24 Jun · Shift 1, Mathematics question 16