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Question : 31 of 160
Marks:
+1,
-0
Solution:
Given
|a|=2,|b|=3,|c|=5‌(a⋅b)=‌‌a⋅b=3‌ and ‌b⋅c=‌Given,
|a+b+c|=√69On squaring both sides, we get
‌|a|2+|b|2+|c|2+2(a⋅b+b⋅c+c⋅a)=69‌⇒‌‌4+9+25+2(3+‌+c⋅a)=69‌⇒2(‌+c⋅a)=31‌⇒‌‌21+2⋅c=31‌⇒‌‌2(c⋅a)=10‌⇒‌‌c⋅a=5‌⇒‌‌|c||a|cos‌θ=5‌⇒‌‌5×2‌cos‌θ=5‌⇒‌‌cos‌θ=‌⇒θ=‌
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