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Question

If AB=3^i2^j+^k, BC=^i+2^j^k, CD=2^i+^j+3^k, OA=^i+^j+^k, then the position vector of OD is

A
7^i2^j3^k
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B
7^i2^j+3^k
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C
7^i+2^j3^k
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D
7^i+2^j+4^k
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Solution

The correct option is D 7^i+2^j+4^k
OA+AB+BC+CD=OD
Adding all vectors we get
OD=7^i+2^j+4^k
So Position Vector of OD is
D=7^i+2^j+4^k

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