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Question

If ¯a=^i+^j2^k,¯b=2^i^j+^k and ¯c=3^i^k then, find the scalars m and n such that ¯c=m¯a+n¯b.

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Solution

3¯i¯k=m(¯i+¯j2¯k)+n(2¯i¯j+¯k))
So we get an comparison,
m+2n=3mn=0
and 2m+n=1
on using first two and solving, we get m=n=1
and since this satisfies 2m+n=1
we can say that m=n=1.

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