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The base vectors veca(1), veca(2) and ve...

The base vectors `veca_(1), veca_(2)` and `veca_(3)` are given in terms of base vectors `vecb_(1), vecb_(2)` and `vecb_(3)` as `veca_(1) = 2vecb_(1)+3vecb_(2)-vecb_(3)`, `veca-(2)=vecb_(1)-2vecb_(2)+2vecb_(3)` and `veca_(3) =-2vecb_(1) + vecb_(2)-2vecb_(3)`, if `vecF = 3vecb_(1)-vecb_(2)+2vecb_(3)`, then vector `vecF` in terms of `veca_(1), veca_(2)` and `veca_(3)` is

A

`vecF=3veca_(1) + 2veca_(2) + 5veca_(3)`

B

`vecF=3veca_(1) - 5veca_(2)-2veca_(3)`

C

`vecF=3veca_(1)+5veca_(2)+3veca_(3)`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
C
`veca_(1)=2vecb_(1)+3vecb_(2)-vecb_(3)`
`veca_(2)=vecb_(1)-2vecb_(2)+2vecb_(3)`
`veca_(3)=-2vecb_(1)+vecb_(2)-2vecb_(3)`
Let `vecF = lambda_(1)veca_(1)+lambda_(2)-2lambda_(3)vecb_(1)+(3lambda_(1)-2lambda_(2)+lambda_(3))vecb_(2)+(-lambda_(1)+2lambda_(2)-2lambda_(3))vecb_(3)=3vecb_(1)-vecb_(2)+2vecb_(3)`
Comparing coefficients of `b_(1),b_(2)` and `b_(3)`
(`therefore vecb_(1),vecb_(2)` and `vecc_(3)` are base vectors hence non-coplanar)
`2lambda_(1)+lambda_(2)-2lambda_(3)=-1`
`-lambda_(1)+2lambda_(2)-2lambda_(2)=2`
Solving above equations, we get
`lambda_(1)=2`
`lambda_(2)=5`
`lambda_(3)=3`
`therefore vecF=2veca_(1)+5veca_(2)+3veca_(3)`
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