3 and 4 .Determinants and Matrices
hard

જો $\alpha+\beta+\gamma=2 \pi$ તો સમીકરણ સંહતિ  $x+(\cos \gamma) y+(\cos \beta) z=0$  ;  $(\cos \gamma) x+y+(\cos \alpha) z=0$  ; $(\cos \beta) x+(\cos \alpha) y+z=0$ નો ઉકેલગણ . . .  ..

A

ખાલીગણ

B

અનંત ઉકેલ ધરાવે

C

માત્ર બેજ ઉકેલ ધરાવે

D

એક્જ ઉકેલ ધરાવે

(JEE MAIN-2021)

Solution

$\alpha+\beta+\gamma=2 \pi$

$\left|\begin{array}{ccc}1 & \cos \gamma & \cos \beta \\ \cos \gamma & 1 & \cos \alpha \\ \cos \beta & \cos \alpha & 1\end{array}\right|$

$=1+2 \cos \alpha \cdot \cos \beta \cdot \cos \gamma-\cos ^{2} \alpha-\cos ^{2} \beta-\cos ^{2} \gamma$

$=\sin ^{2} \gamma-\cos ^{2} \alpha-\cos ^{2} \beta+(\cos (\alpha+\beta)+\cos (\alpha-\beta)) \cos \gamma$

$=\sin ^{2} \gamma-\cos ^{2} \alpha-\cos ^{2} \beta+\cos ^{2} \gamma+\cos (\alpha-\beta) \cos \gamma$

$=\sin ^{2} \alpha-\cos ^{2} \beta+\cos (\alpha-\beta) \cos (\alpha+\beta)$

$=\sin ^{2} \alpha-\cos ^{2} \beta+\cos ^{2} \alpha-\sin ^{2} \beta=0$

Standard 12
Mathematics

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