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← JEE Advanced 2018 · Paper 1
MathematicsNumerical+3Question 48 of 54

JEE Advanced 2018 · Paper 1, Mathematics question 12

Let a⃗\vec{a} and b⃗\vec{b} be two unit vectors such that a⃗⋅b⃗=0\vec{a} \cdot \vec{b} = 0. For some x,y∈Rx, y \in \mathbb{R}, let c⃗=xa⃗+yb⃗+(a⃗×b⃗)\vec{c} = x\vec{a} + y\vec{b} + (\vec{a} \times \vec{b}). If ∣c⃗∣=2|\vec{c}| = 2 and the vector c⃗\vec{c} is inclined at the same angle α\alpha to both a⃗\vec{a} and b⃗\vec{b}, then the value of 8cos⁡2α8\cos^2\alpha is _____ .

Answer and worked solution

Official answer: 3

a⃗\vec a, b⃗\vec b and a⃗×b⃗\vec a\times\vec b are mutually perpendicular unit vectors.

Magnitude. ∣c⃗∣2=x2+y2+1=4|\vec c|^2=x^2+y^2+1=4.

Equal angles. c⃗⋅a⃗=x=2cos⁡α\vec c\cdot\vec a=x=2\cos\alpha and c⃗⋅b⃗=y=2cos⁡α\vec c\cdot\vec b=y=2\cos\alpha.

Solve.
8cos⁡2α+1=4  ⟹  8cos⁡2α=3.8\cos^2\alpha+1=4\implies8\cos^2\alpha=3.

Answer: 3