Let a¯ and b¯ be two non-collinear unit vectors . If u¯=a¯-a¯.b¯b¯ and v¯=a¯×b¯, then v¯=

|a|=1=|b|u=a-(a·b) b|u|2=|a|2+(a·b)2 |b|2-2(a·b) (a·b) =1+(a·b)2-2(a·b)2=1-(a·b)2 =|a|2 |b|2-(a·b)2=|a×b|2=|v|2 ⇒ |u|=|v|

Let a¯ and b¯ be two non-collinear unit vectors . If u¯=a¯-a¯.b¯b¯ and v¯=a¯×b¯, then v¯=

|a|=1=|b|u=a-(a·b) b|u|2=|a|2+(a·b)2 |b|2-2(a·b) (a·b) =1+(a·b)2-2(a·b)2=1-(a·b)2 =|a|2 |b|2-(a·b)2=|a×b|2=|v|2 ⇒ |u|=|v|

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$Let \bar{a} and \bar{b} be two non - collinear unit vectors . If \bar{u} = \bar{a} - (\bar{a} . \bar{b}) \bar{b} and \bar{v} = \bar{a} \times \bar{b}, then |\bar{v}| =$

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$|\bar{a}| |\bar{b}|$

$|\bar{u}|$

$|\bar{a}|$

$|\bar{b}|$

answer is A.

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Detailed Solution

$\begin{array}{rcl} | a | & = & 1 = | b | \\ u & = & a - (a \cdot b) b \\ | u |^{2} & = & | a |^{2} + {(a \cdot b)}^{2} | b |^{2} - 2 (a \cdot b) (a \cdot b) \\ = & 1 + {(a \cdot b)}^{2} - 2 {(a \cdot b)}^{2} = 1 - {(a \cdot b)}^{2} \\ = | a |^{2} | b |^{2} - {(a \cdot b)}^{2} = | a \times b |^{2} = | v |^{2} \\ \Rightarrow & | u | = | v | \end{array}$