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Question

If a=^i+^j+^k,b=2^i^j+3^k and c=^i2^j+^k, find a unit vector parallel to the vector 2ab+3c .

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Solution

We have,
a=^i+^j+^k,b=2^i^j+3^k and c=^i2^j+^k
2ab+3c=2(^i+^j+^k)(2^i^j+3^k+3(^i2^j+^k
=2^i+2^j+2^k2^i+^j3^k+3^i6^j+3^k
=3^i^j+2^k
2ab+3c=32+(3)2+22=9+9+4=22
Hence, the unit vector along 2ab+3c is
2ab+3c|2ab+3c|=3^i3^j+2^k22=322^i322^j+222^k.

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