If a, b are two unit vectors inclined at an angle π3 then a×(b+a×b)·b=

|a|=|b|=1 a·b=1×1×12=12 a×b+a+a×b·b =[a b b]+a·b a-|a|2 b·b =0+1212-1=-34

If a, b are two unit vectors inclined at an angle π3 then a×(b+a×b)·b=

|a|=|b|=1 a·b=1×1×12=12 a×b+a+a×b·b =[a b b]+a·b a-|a|2 b·b =0+1212-1=-34

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If $a, b$ are two unit vectors inclined at an angle $\frac{π}{3}$ then $\{a \times (b + a \times b)\} \cdot b =$

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$\frac{3}{4}$

$\frac{- 3}{4}$

$\frac{1}{2}$

$\frac{1}{4}$

answer is A.

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Detailed Solution

$\begin{array}{rcl} | a | & = & | b | = 1 a \cdot b = 1 \times 1 \times \frac{1}{2} = \frac{1}{2} \\ (a \times b + a + (a \times b)) \cdot b \\ = & [a b b] + ((a \cdot b) a - | a |^{2} b) \cdot b \\ = & 0 + \frac{1}{2} (\frac{1}{2}) - 1 = \frac{- 3}{4} \end{array}$