The vectors →a=x^i+(x+1)^j+(x+2)^k,→b=(x+3)^i+(x+4)^j+(x+5)^k and

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The vectors

\to a = x^i + (x + 1)^j + (x + 2)^k, \to b = (x + 3)^i + (x + 4)^j + (x + 5)^k

and

\to 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

\to a, \to b, \to c

are coplanar, iff

[\to a \to b \to c] = 0

We have,

[\to a \to b \to c] = ∣ ∣∣ ∣ \begin{matrix} x & x + 1 & x + 2 x + 3 & x + 4 & x + 5 x + 6 & x + 7 & x + 8 \end{matrix} ∣ ∣∣ ∣

= ∣ ∣∣ ∣ \begin{matrix} x & x + 1 & x + 2 3 & 3 & 3 6 & 6 & 6 \end{matrix} ∣ ∣∣ ∣ [\begin{matrix} A p p l y i n g R_{2} \to & R_{2} - & R_{1}, R_{3} \to & R_{3} - & R_{1} \end{matrix}]

= 0 for all x[

∵ R_{1}

and

R_{2}

are proportional]

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