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Question : 7 of 120
Marks:
+1,
-0
Solution:
Here,
|a|=1,|b|=1,|c|=2 ×(×)−=0 (.)−(.)−=0 (.)=(.)+ (|a||c|cos‌θ)=(|a||a|cos‌0) + 2‌cos‌θ‌=+ 2‌cos‌θ‌−= Taking magnitude both sides, we get
4cos2θ|a|2+|c|2−2 ×2‌cos‌θ‌.=|b|2 4cos2θ+4−2×2‌cos‌θ×|a||c| cos‌θ=1 4cos2θ+4−8cos2θ=1 4cos2θ=3 cos2θ= cos‌θ= ∴θ=π∕6 Hence, option (2) is correct.
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