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p=2a-3b,q=a-2b+c and r=-3a+b+2c, where a...

`p=2a-3b,q=a-2b+c` and `r=-3a+b+2c`, where `a,b,c` being non-coplanar vectors, then the vector `-2a+3b-c` is equal to

A

`p-4q`

B

`(-7q+r)/(5)`

C

`2p-3q+r`

D

`4p-2r`

Text Solution

Verified by Experts

The correct Answer is:
B
Let `-2a+3b-c=xp+yq+zr`
`implies-2a+3b-c=(2x+y-3z)a+(-3x-2y+z)b+(y+2z)c`
`therefore 2x+y-3z=-2,-3x-2y+z=3`
and `y+2z=-1`
On solving these we get `x=0,y=-(7)/(5),z=(1)/(5)`
`therefore-2a+3b-c=((-7q+r))/(5)`
trick check alternates one-by-ne
ie.., (a) `p-4q=-2a+5b-4c`
(b) `(-7q+r)/(5)=-2a+3b-c`
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