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Question : 15 of 90
Marks:
+1,
-0
Solution:
∵++=0then
a+=−then
(+)×=−×b∴‌‌×+×=0‌‌...(i)For
(S1):|×+×|−||=6(2√2−1)|(+)×|−||=6(2√2−1)||=6−12√2( not possible
) For (S2) : from (i)
+=−⇒⋅+⋅=−⋅⇒‌‌12+12=−6√2⋅2√3‌cos(π−∠ACB)∴‌‌cos(∠ACB)=√‌∴‌‌∠ACB=cos−1√‌∴‌‌S(2) is correct.
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