Let a→ and b→ be two vectors such that |a→|$$=1$$,|b→|$$=4$$ and a→⋅b→$$=2$$. If c→$$=(2a$$→×b→$$)$$−$$3b$$→ then find the angle between

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Let a→ and b→ be two vectors such that |a→|$$=1$$,|b→|$$=4$$ and a→⋅b→$$=2$$. If c→$$=(2a$$→×b→$$)$$−$$3b$$→ then find the angle between

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|a→|$$=1$$,|b→|$$=4$$,a→⋅b→$$=2c$$→$$=(2a$$→×b→$$)$$−$$3b$$→ or c→$$+3b$$→$$=2a$$→×b→∵ a→⋅b→$$=2$$⇒ |a→|⋅|b→|$$cos$$⁡θ$$=2or$$  $$cos$$⁡θ$$=2$$|a→|⋅|b→|$$=24=12or$$   θ$$=$$π$$3or$$  θ$$=$$π$$3$$⇒ |c→$$+3b$$→|$$2=$$|$$2a$$→×b→|$$2or$$  |c→|$$2+9$$|b→|$$2+2c$$→⋅$$3b$$→$$=4$$|a→|$$2$$|b→|$$2sin2$$⁡θor  |c→|$$2+144+6b$$→⋅c→$$=48or$$  |c→|$$2+96+6(b$$→⋅c→$$)=0------ior$$  c→$$=2a$$→×b→−$$3b$$→⇒b→⋅c→$$=0$$−$$3$$×$$16=$$−$$48Putting$$ value of b b→c→ in Eq. (i), we have|c→|$$2+96$$−$$6$$×$$48=0=48$$×$$4=192Again$$, putting the value of c→ in Eq. (i), we have $$192+96+6$$|b→|⋅|c→|$$cos$$⁡α$$=0or$$  $$6$$×$$4$$×$$83cos$$⁡α$$=$$−$$288or$$   $$cos$$⁡α$$=$$−$$2886$$×$$4$$×$$83=$$−$$323=$$−$$32or$$   α$$=5$$$$π$$$$6$$

Let a→ and b→ be two vectors such that |a→|=1,|b→|=4 and a→⋅b→=2. If c→=(2a→×b→)−3b→ then find the angle between

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