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
A
all values of x
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B
x < 0
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C
x > 0
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D
None of these
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Solution
The correct option is A all values of x →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]