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Question : 100 of 125
Marks:
+1,
-0
Solution:
Since,
||=√11,||=√22Now,
||=|×|=||||sin‌θ⇒√11=√50√22sin‌θ‌⇒sin‌θ=‌‌∵|+|2=||2+||2+2b⋅‌=||2+||2+2||||cos‌θ‌=50+22+2×√50×√22×‌=72+66‌⇒|72−|+|2|=66
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