Exams Joint Entrance Examination Mathematics Vector Algebra Question 108 Question 108 of 379 Medium NUMERICAL
Let a ⃗ \vec{a} a and b ⃗ \vec{b} b be two vectors such that ∣ a ⃗ ∣ = 1 , ∣ b ⃗ ∣ = 4 |\vec{a}|=1,|\vec{b}|=4 ∣ a ∣ = 1 , ∣ b ∣ = 4 , and a ⃗ ⋅ b ⃗ = 2 \vec{a}\cdot \vec{b}=2 a ⋅ b = 2 . If c ⃗ = ( 2 a ⃗ × b ⃗ ) − 3 b ⃗ \vec{c}=(2\vec{a}\times \vec{b})-3\vec{b} c = ( 2 a × b ) − 3 b and the angle between b ⃗ \vec{b} b and c ⃗ \vec{c} c is α \alpha α , then 192 sin 2 α 192{\sin }^{2}\alpha 192 sin 2 α is equal to \_\_\_\_\_\_\_\_.
Answer: 48
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b → ⋅ c → = ( 2 a → × b → ) ⋅ b → − 3 ∣ b ∣ 2 ∣ b ∣ ∣ c ∣ cos α = − 3 ∣ b ∣ 2 ∣ c ∣ cos α = − 12 ∣ b ∣ = 4 a → ⋅ b → = 2 cos θ = 1 2 ⇒ θ = π 3 ∣ c ∣ 2 = ∣ ( 2 a → × b → ) − 3 b → ∣ 2 = 64 × 3 4 + 144 = 192 ∣ c ∣ 2 cos 2 α = 144 192 cos 2 α = 144 192 sin 2 α = 48 \begin{gathered} \overset{\to }{b}\cdot \overset{\to }{c}=(2\overset{\to }{a}\times \overset{\to }{b})\cdot \overset{\to }{b}-3|b{|}^{2} \\ |b||c|\cos \alpha =-3|b{|}^{2} \\ |c|\cos \alpha =-12|b|=4 \\ \overset{\to }{a}\cdot \overset{\to }{b}=2 \\ \cos \theta =\frac{1}{2}\Rightarrow \theta =\frac{\pi }{3} \\ |c{|}^{2}=|(2\overset{\to }{a}\times \overset{\to }{b})-3\overset{\to }{b}{|}^{2} \\ =64\times \frac{3}{4}+144=192 \\ |c{|}^{2}{\cos }^{2}\alpha =144 \\ 192{\cos }^{2}\alpha =144 \\ 192{\sin }^{2}\alpha =48 \end{gathered} b → ⋅ c → = ( 2 a → × b → ) ⋅ b → − 3∣ b ∣ 2 ∣ b ∣∣ c ∣ cos α = − 3∣ b ∣ 2 ∣ c ∣ cos α = − 12∣ b ∣ = 4 a → ⋅ b → = 2 cos θ = 2 1 ⇒ θ = 3 π ∣ c ∣ 2 = ∣ ( 2 a → × b → ) − 3 b → ∣ 2 = 64 × 4 3 + 144 = 192 ∣ c ∣ 2 cos 2 α = 144 192 cos 2 α = 144 192 sin 2 α = 48 Answer the question to unlock