If $\left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&{1 + {a^3}}\\b&{{b^2}}&{1 + {b^3}}\\c&{{c^2}}&{1 + {c^3}}\end{array}\,} \right| = 0$ and $a = (1,\,a,\,{a^2}),\,b = (1,\,b,\,{b^2}),$ and $c = (1,\,c,\,{c^2})$ are non-coplanar vectors, then $abc$ is equal to
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Answer
a
(a) Since $(1,\,\,a\,,\,{a^2}),\,\,(1,\,\,b,\,\,{b^2})$ and $(1,\,\,c,\,\,{c^2})$ are non-coplanar, therefore $\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2}}\\1&b&{{b^2}}\\1&c&{{c^2}}\end{array}} \right| \ne 0 = \Delta \,({\rm{say}})$
and $\left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&{1 + {a^3}}\\b&{{b^2}}&{1 + {b^3}}\\c&{{c^2}}&{1 + {c^3}}\end{array}\,} \right|\, = \,\left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&1\\b&{{b^2}}&1\\c&{{c^2}}&1\end{array}\,} \right| + \left| {\,\begin{array}{*{20}{c}}a&{{a^2}}&{{a^3}}\\b&{{b^2}}&{{b^3}}\\c&{{c^2}}&{{c^3}}\end{array}\,} \right| = 0$
$ \Rightarrow \Delta + abc\,\Delta = 0$
$ \Rightarrow \Delta (abc + 1) = 0$$ \Rightarrow abc = - 1.$
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