The vectors →a=x^i+(x+1)^j+(x+2)^k,→b=(x+3)^i+(x+4)^j+(x+5)^k and →c=(x+6)^i+(x+7)^j+(x+8)^k are coplanar for
Options:
A .  
all values of x
B .  
x < 0
C .  
x > 0
D .  
None of these
Answer: Option A : A →a,→b,→c are coplanar, iff [→a→b→c]=0 We have, [→a→b→c]=∣∣ ∣∣xx+1x+2x+3x+4x+5x+6x+7x+8∣∣ ∣∣ =∣∣ ∣∣xx+1x+2333666∣∣ ∣∣[ApplyingR2→R2−R1,R3→R3−R1] = 0 for all x [∵R1 and R2 are proportional]
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