Question
किन्हीं भी दो सदिशों $a$ व $ b$ के लिये, ${(a \times b)^2}$ =
$ = {a^2}{b^2}{\sin ^2}\theta = {a^2}{b^2}(1 - {\cos ^2}\theta )$
$ = {a^2}{b^2} - {a^2}{b^2}{\cos ^2}\theta = {a^2}{b^2} - {(a\,.\,b)^2}$.
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$\left| {\begin{array}{*{20}{c}}a&{a + 1}&{a - 1}\\{ - b}&{b + 1}&{b - 1}\\c&{c - 1}&{c + 1}\end{array}} \right| + \left| {\begin{array}{*{20}{c}}{a + 1}&{b + 1}&{c - 1}\\{a - 1}&{b - 1}&{c + 1}\\{{{\left( { - 1} \right)}^{n + 2}} \cdot a}&{{{\left( { - 1} \right)}^{n + 1}} \cdot b}&{{{\left( { - 1} \right)}^n} \cdot c}\end{array}} \right| = 0$
तो $n$ का मान है