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Question : 10 of 90
Marks:
+1,
-0
Solution:
‌a=+2−3‌a=2+−‌a×r=a×b;a⋅r=0‌⇒a×(r−b)=0‌⇒a=λ(r−b)‌a⋅a=λ(a⋅r−a⋅b)‌14=−7λ⇒λ=−2‌‌=r−b⇒r=b−‌‌=‌=‌Statement (I) is incorrect
‌cos‌2‌A+cos‌2‌B+cos‌2‌c≥−‌‌2A+2B+2C=2π‌cos‌2‌A+cos‌2‌B+cos‌2‌C‌=−1−4‌cos‌A⋅cos‌B⋅cos‌C≥−1−4×‌×‌×‌‌=−‌Statement (II) is correct.
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