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Question

Let a and b be two non-collinear unit vectors. If u=a(a.b)b and v=a×b, then |v| is

A
|u|
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B
|u|+|u.a|
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C
|u|+|u.b|
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D
|u|+u.(a+b)
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Solution

The correct option is C |u|+|u.b|
Let θ be the angle between a and b. Since, a and a and a are non - collinear vectors, then θ0 and θπ.
We have. a.b=|a||a| cos θ
=cos θ [because|a|=1,|b|=1,given]
\text{Now}, u=a(a.b)b|u|=|a(a.b)b)||u|2=|a(a. b)b|2|u|2=|acos θ b|2|u|2=|a|2+cos2θ|b|22 cos θ (a.b)|u|2=1+cos2θ2 cos2θ|u|2=1cos2θ|u|2=sin2 θAlso,v=a×b [given]|v|2=|a×b|2|v|2=|a|2|b|2.sin2 θ|v|2=sin2θ |u|2=|v|2Now,u.a=[a(a.b)b].a=a.a(a.b)(b.a)=(a)2cos2θ=1cos2 θ=sin2 θ|u|+|u.a|=sin θ sin2 θ |v|
u.b=[a(a.b)b].b=a.b(a.b)(b.b)=a.ba.b|b|2=a.ba.b=0|u|+|u.b|=|u|+0=|u|=|v|Also,u.(a+b)=u.a+u.b=u.a|u|+u.(a+b)=|u|+u.a|v|

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