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Question : 12 of 120
Marks:
+1,
-0
Solution:
Given:
||=α,||=α and
|+|=α As we know,
|+|=√a2+b2+2ab‌cos‌θ ⇒α=√α2+α2+2(α)(α)‌cos‌θ ⇒α2=2α2+2α2‌cos‌θ ⇒−1=2‌cos‌θ ⇒cos‌θ=− Now,
|−|=√a2+b2−2ab‌cos‌θ ⇒|−|=√α2+α2−2(α)(α)‌cos‌θ ∵cos‌θ=− ⇒|−|=√2α2−2α2() ⇒|−|=√2α2+α2 ⇒|−|=√3α
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