1.Let α⃗\vec{\alpha}α, β⃗\vec{\beta}β, γ⃗\vec{\gamma}γ be three unit vectors such that α⃗⋅β⃗=α⃗⋅γ⃗=0\vec{\alpha} \cdot \vec{\beta} = \vec{\alpha} \cdot \vec{\gamma} = 0α⋅β=α⋅γ=0 and the angle between β⃗\vec{\beta}β and γ⃗\vec{\gamma}γ is 30∘30^{\circ}30∘. Then α⃗\vec{\alpha}α isa.2(β⃗×γ⃗)2(\vec{\beta} \times \vec{\gamma})2(β×γ)b.−2(β⃗×γ⃗)-2(\vec{\beta} \times \vec{\gamma})−2(β×γ)c.±2(β⃗×γ⃗)\pm 2(\vec{\beta} \times \vec{\gamma})±2(β×γ)d.(β⃗×γ⃗)(\vec{\beta} \times \vec{\gamma})(β×γ)Login to continueOnly logged in users canattempt or see the solution.
