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Question

# Given: →a,→b and →c are coplanar. Vectors →a−2→b+3c,−−→2a+→3b−→4c and−→b+→2c are non-coplanar vectors.

A
True
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B
False
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Solution

## The correct option is B FalseWe will check if the given 3 vectors are linearly dependent or not. If yes, then they are coplanar. Else not. So Let α=→a−2→b+3c β=2→a+3→b−4→c γ=−→b+2→c So if α,β, and γ are linearly dependent then, for α=x→β+y→γ There should be values of x and y which are not simultaneously zero. So let’s find out. →a−2→b+3→c=x(−2→a+3→b−4→c) +y(−→b+2→c) Equating the coefficients of →a,→b and →c. We get -2x + 0.y = 1 (equating coefficient of →a) 3x – y = -2 (equating coefficient of →b) -4x + 2y = 3 (equating coefficient of →c) Solving the system of equations, we get x=−12 y=12 These values satisfy all 3 equations and thus the system of equations is consistent and has a non-zero solution. So α,β and γ are linearly dependent and hence coplanar.

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