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Given |Ā1| = 2 , |Ā2|=3 and |Ā1+Ā2| = 3 . The magnitude of (Ā1+2Ā2).(3Ā1 - 4Ā2) is ?
Solution
|Ā1 + Ā2|2 = |Ā1|2 + |Ā2|2 + 2|Ā1||Ā2| cos 𝛼 32 = 22 + 32 + 2(2)(3) cos 𝛼
22 1 cos𝛼 = −2(2)(3) = −3.
(Ā1 + 2Ā2) ∙ (3Ā1 − 4Ā2) = 3|Ā1|2 − 4|Ā1||Ā2| cos 𝛼 + 6|Ā1||Ā2| cos 𝛼 − 8|Ā2|2
= 3(2)2 + 2(2)(3) (− 1) − 8(3)2 = −64. 3
The magnitude of (Ā1+2Ā2).(3Ā1 - 4Ā2) is 64.
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