Let a→ and b→ be unit vectors such that a→+b→=3, then the value of 2a→+5b→.3a→+b→+a→×b→ is equal to

a=|b|=1 , a+b2=3 ⇒ 1+1+2a·b=3 ⇒ a·b=13 =2a+5b·(3a+b+a×b) =6+2x12+0+1512+5+0 ⇒12+152=392

Let a→ and b→ be unit vectors such that a→+b→=3, then the value of 2a→+5b→.3a→+b→+a→×b→ is equal to

a=|b|=1 , a+b2=3 ⇒ 1+1+2a·b=3 ⇒ a·b=13 =2a+5b·(3a+b+a×b) =6+2x12+0+1512+5+0 ⇒12+152=392

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Let $\vec{a}$ and $\vec{b}$ be unit vectors such that $|\vec{a} + \vec{b}| = \sqrt{3}$ , then the value of $(2 \vec{a} + 5 \vec{b}) . (3 \vec{a} + \vec{b} + \vec{a} \times \vec{b})$ is equal to

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$\frac{39}{2}$

$\frac{11}{2}$

$\frac{23}{2}$

$\frac{13}{2}$

answer is C.

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$|a| = | b | = 1, {|a + b|}^{2} = 3 \Rightarrow 1 + 1 + 2 a \cdot b = 3 \Rightarrow a \cdot b = \frac{1}{3} = (2 a + 5 b) \cdot (3 a + b + a \times b) = 6 + 2 x \frac{1}{2} + 0 + 15 (\frac{1}{2}) + 5 + 0 \Rightarrow 12 + \frac{15}{2} = \frac{39}{2}$

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Let a→ and b→ be unit vectors such that a→+b→=3, then the value of 2a→+5b→.3a→+b→+a→×b→ is equal to

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Let $\vec{a}$ and $\vec{b}$ be unit vectors such that $|\vec{a} + \vec{b}| = \sqrt{3}$ , then the value of $(2 \vec{a} + 5 \vec{b}) . (3 \vec{a} + \vec{b} + \vec{a} \times \vec{b})$ is equal to