If a→ and b→ are two vectors such that |a→|$$=1$$,|b→|$$=4$$,a·→b→$$=2$$ and c→$$=(2a$$→×b→$$)$$−$$3b$$→ then the angle between b→ and c→ is

Question

If a→ and b→ are two vectors such that |a→|$$=1$$,|b→|$$=4$$,a·→b→$$=2$$ and c→$$=(2a$$→×b→$$)$$−$$3b$$→ then the  angle between b→ and c→ is

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Given  c→$$=(2a$$→×b→$$)$$−$$3b$$→b→⋅c→$$=b$$→$${(2a$$→×b→$$)$$−$$3b$$→$$}b$$→⋅c→$$=$$−$$3$$|b→|$$2$$ Now |a→×b→|$$2=a$$→$$2b$$→$$2$$−(a$$→⋅b→$$)2=16$$−$$4=12$$ And c→$$2=((2a$$→×b→$$)−$$3b$$→$$)2$$|c→|$$2=4$$|a→×b→|$$2+9b$$→$$2$$−$$6b$$→.(2a$$→×b→$$)⏟zero $$=48+144=192$$|c→|$$=83$$ Now, $$cos$$⁡θ$$=b$$→⋅c→|b→||c→|$$=$$−$$3$$|b→|$$2$$|b→||c→|$$=$$−$$3$$|b→||c→|$$=$$−$$3(4)83=$$−$$32$$ Hence  $$(b$$→,c→$$)=5$$$$π$$$$6$$

$If \vec{a} and \vec{b} are two vectors such that | \vec{a} | = 1, | \vec{b} | = 4, \vec{a \cdot} \vec{b} = 2 and \vec{c} = (2 \vec{a} \times \vec{b}) - 3 \vec{b} then the$ $angle between \vec{b} and \vec{c} is$

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If a→ and b→ are two vectors such that |a→|=1,|b→|=4,a·→b→=2 and c→=(2a→×b→)−3b→ then the angle between b→ and c→ is

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$If \vec{a} and \vec{b} are two vectors such that | \vec{a} | = 1, | \vec{b} | = 4, \vec{a \cdot} \vec{b} = 2 and \vec{c} = (2 \vec{a} \times \vec{b}) - 3 \vec{b} then the$ $angle between \vec{b} and \vec{c} is$