Let a→,b→,c→ be unit vectors such that a→+b→+c→=x→, a→.x→=1, b→.x→=32,x→=2. Then angle between c→ and x→ is

a+b+c=x, a=b=c=1 a·x+b·x+c·x=x2 1+32+c·x=4 ⇒ c.x=3-32=32 ⇒ 1×2×cos θ=32 ⇒ θ=cos-134

Let a→,b→,c→ be unit vectors such that a→+b→+c→=x→, a→.x→=1, b→.x→=32,x→=2. Then angle between c→ and x→ is

a+b+c=x, a=b=c=1 a·x+b·x+c·x=x2 1+32+c·x=4 ⇒ c.x=3-32=32 ⇒ 1×2×cos θ=32 ⇒ θ=cos-134

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Let $\vec{a}, \vec{b}, \vec{c}$ be unit vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{x}, \vec{a} . \vec{x} = 1, \vec{b} . \vec{x} = \frac{3}{2}, |\vec{x}| = 2 .$ Then angle between $\vec{c}$ and $\vec{x}$ is

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$\cos^{- 1} (\frac{3}{4})$

$\cos^{- 1} (\frac{3}{8})$

$\cos^{- 1} (\frac{5}{8})$

$\cos^{- 1} (\frac{1}{4})$

answer is B.

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Detailed Solution

$a + b + c = x, |a| = |b| = |c| = 1 a \cdot x + b \cdot x + c \cdot x = {|x|}^{2} 1 + \frac{3}{2} + c \cdot x = 4 \Rightarrow c . x = 3 - \frac{3}{2} = \frac{3}{2} \Rightarrow 1 \times 2 \times \cos θ = \frac{3}{2} \Rightarrow θ = \cos^{- 1} (\frac{3}{4})$