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 options are A |→u| C |→u|+|→u.→b|Let θ be the angle between →a and →b. Since, →a and →a and →a are non - collinear vectors, then θ≠0 and θ≠π. 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=|→a−cos θ →b|2⇒|→u|2=|→a|2+cos2θ|→b|2−2 cos θ (→a.→b)⇒|→u|2=1+cos2θ−2 cos2θ⇒|→u|2=1−cos2θ⇒|→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)2−cos2θ=1−cos2 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.→b−→a.→b|→b|2=→a.→b−→a.→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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