If →a=(2,1,−1),→b=(1,−1,0),→c=(5,−1,1), then the unit vector parallel to →a+→b−→c in the opposite direction is
A
(−2^i+^j−2^k3)
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B
(2^i−^j+2^k3)
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C
(−^i+2^j−^k√6)
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D
(^i−2^j+^k√6)
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Solution
The correct option is B(2^i−^j+2^k3) Given →a=(2,1,−1),→b=(1,−1,0),→c=(5,−1,1) →a+→b−→c=(2^i+^j−^k)+(^i−^j+0^k)−(5^i−^j+^k) →a+→b−→c=−2^i+^j−2^k
Magnitude of the vector is √4+1+4=3
So the unit vector in the opposite direction =−(−2^i+^j−2^k3) ∴ Required unit vector =(2^i−^j+2^k3)