A) \[\alpha \]
B) p
C) d
D) a
Correct Answer: B
Solution :
Let\[\Delta =\left| \begin{matrix} 1 & \alpha & {{\alpha }^{2}} \\ \cos (p-d)a & \cos pa & \cos (p-d)a \\ \sin (p-d)a & \sin pa & \sin (p-d)a \\ \end{matrix} \right|\] Applying\[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}\], we get \[\Rightarrow \Delta =\left| \begin{matrix} 1 & \alpha & {{\alpha }^{2}}-1 \\ \cos (p-d)a & \cos pa & 0 \\ \sin (p-d)a & \sin pa & 0 \\ \end{matrix} \right|\] \[=({{\alpha }^{2}}-1)\{-\cos pa\sin (p-d)a\] \[+\sin pa\cos (p-d)a\}\] \[=({{\alpha }^{2}}-1)\sin \{-p(p-d)a+pa\}\] \[\Rightarrow \] \[\Delta =({{\alpha }^{2}}-1)\sin da\] which is independent of\[p\].
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