Let a^ and b^ be mutually perpendicular unit vectors. Then for any arbitrary r→
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a
r→=(r→⋅a^)a^+(r→⋅b^)b^+(r→⋅(a^×b^))(a^×b^)
b
r→=(r→⋅a^)−(r→⋅b^)b^−(r→⋅(a^×b^))(a^×b^)
c
r→=(r→⋅a^)a^−(r→⋅b^)b^+(r→⋅(a^×b^))(a^×b^)
d
none of these
answer is A.
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Detailed Solution
Let r→=x1a^+x2b^+x3(a^×b^)⇒ r→⋅a^=x1+x2a^⋅b^+x3a^⋅(a^×b^)=x1 Also, r→⋅b^=x1a^⋅b^+x2+x3b^⋅(a^+b^)=x2 and r→⋅(a^×b^)=x1a^⋅(a^×b^)+x2b^⋅(a^×b^)+x3(a^×b^)⋅(a^×b^)=x3 ⇒ r→=(r→⋅a^)a^+(r→⋅b^)b^+(r→⋅a^×b^))(a^×b^)