If →a,→b,→c are three mutually perpendicular unit vectors and d is a unit vector making equal angles with →a,→b,→c, then ∣∣→a+→b+→c+→d∣∣2 is
A
4
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B
4±√3
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C
4±2√3
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D
None of these
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Solution
The correct option is B4±2√3 We have
|¯¯¯a+¯¯b+¯¯c+¯¯¯d|2=4+2[¯¯¯a.¯¯b+¯¯b.¯¯c+¯¯c.¯¯¯d.+¯¯¯d.¯¯¯a+¯¯¯a.¯¯c+¯¯b.¯¯¯d] =4+2(0+0+0+¯¯¯d(¯¯¯a+¯¯b+¯¯c)) Let ¯¯¯d=λ¯¯¯a+μ¯¯b+v¯¯c ¯¯¯d.¯¯¯a=λ,¯¯¯d.¯¯b=μ,¯¯¯d.¯¯c=v But ¯¯¯d.¯¯¯a=¯¯¯d.¯¯b=¯¯¯d.¯¯c=cosθ But λ2+μ2+v2=1 3cos2θ=1