lf a=b=1 and a×b=a.b, then a+b2 =

|a×b|=|a·b| ⇒ 1×1×sinθ=1×1×cosθ ⇒ tanθ=1 ⇒ θ=π4|a+b|2=1+1+2 a·b =1+1+2·1×1×12 =2+2

lf a=b=1 and a×b=a.b, then a+b2 =

|a×b|=|a·b| ⇒ 1×1×sinθ=1×1×cosθ ⇒ tanθ=1 ⇒ θ=π4|a+b|2=1+1+2 a·b =1+1+2·1×1×12 =2+2

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$lf |a| = |b| = 1 and |a \times b| = a . b, then {|a + b|}^{2} =$

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$\sqrt{2}$

$2 + \sqrt{2}$

answer is B.

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Detailed Solution

$\begin{array}{rcl} | a \times b | & = & | a \cdot b | \Rightarrow 1 \times 1 \times sinθ = 1 \times 1 \times cosθ \\ \Rightarrow & tanθ = 1 \Rightarrow θ = \frac{π}{4} \\ | a + b |^{2} & = & 1 + 1 + 2 a \cdot b \\ = & 1 + 1 + 2 \cdot 1 \times 1 \times \frac{1}{\sqrt{2}} \\ = & 2 + \sqrt{2} \end{array}$