The vectors a^i+2a^j−3a^k,(2a+1)^i+(2a+3)^j+(a+1)^k and (3a+5)^i+(a+5)^j+(a+2)^k are non-coplanar for the range of a in
A
{0}
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B
(0,∞)
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C
(−∞,1)
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D
(1,∞)
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Solution
The correct options are B(0,∞) D(1,∞) ∣∣
∣∣a2a−3a2a+12a+3a+13a+5a+5a+2∣∣
∣∣\
Taking a common from R1, we get
=a∣∣
∣∣12−32a+12a+3a+13a+5a+5a+2∣∣
∣∣
Applying R2→R2−(2a+1)R1;R3→R3−(3a+5)R1, we get =a∣∣
∣∣12−30−2a+17a+40−5(a+1)10a+17∣∣
∣∣=a(15a2+31a+37)≠0 for a≠0 Hence three vectors are coplanar only for a=0