JEE Main 2022 · 29 Jul · Shift 1 · Mathematics Q14

← JEE Main 2022 · 29 Jul · Shift 1

MathematicsSingle correct+4 / -1Question 14 of 90

Let $\hat a$ and $\hat b$ be two unit vectors such that the angle between them is $\dfrac\pi4$ . If θ is the angle between the vectors $(\hat a+\hat b)$ and $(\hat a+2\hat b+2(\hat a\times\hat b))$ , then the value of 164 cos $^2$ θ is equal to :

90+27\sqrt2

45+18\sqrt2

90+3\sqrt2

54+90\sqrt2

Answer and worked solution

Official answer: A

Dot products with $c=\hat a\cdot\hat b=\frac1{\sqrt2}$ (the cross product is perpendicular to both): $u\cdot v=3+3c$ , $|u|^2=2+2c$ , $|v|^2=|\hat a+2\hat b|^2+4|\hat a\times\hat b|^2=7+4c$ .

So $\cos^2\theta=\dfrac{9(1+c)}{2(7+4c)}$ and $164\cos^2\theta=\dfrac{738(\sqrt2+1)}{7\sqrt2+4}=9(\sqrt2+1)(7\sqrt2-4)\cdot\frac{82}{82}=90+27\sqrt2$ .

(A)

Topics:Dot product Cross product

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JEE Main 2022 · 29 Jul · Shift 1, Mathematics question 14