If a→ and b→ are unit vectors, then the vector (a→+b→)×(a→×b→) is parallel to the vector

|a|=1=|b| (a+b)×(a×b) =a×(a×b)+b×(a×b) =(a·b) a -b+a-(a·b) b =(1+a·b) (a-b)

If a→ and b→ are unit vectors, then the vector (a→+b→)×(a→×b→) is parallel to the vector

|a|=1=|b| (a+b)×(a×b) =a×(a×b)+b×(a×b) =(a·b) a -b+a-(a·b) b =(1+a·b) (a-b)

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If $\vec{a}$ and $\vec{b}$ are unit vectors, then the vector $(\vec{a} + \vec{b}) \times (\vec{a} \times \vec{b})$ is parallel to the vector

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$a + b$

$2 a + b$

$2 a - b$

$a - b$

answer is A.

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Detailed Solution

$\begin{array}{rcl} | a | & = & 1 = | b | \\ (a + b) \times (a \times b) \\ = & a \times (a \times b) + b \times (a \times b) \\ = & (a \cdot b) a - b + a - (a \cdot b) b \\ = & (1 + a \cdot b) (a - b) \end{array}$

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