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d o t p r o d u c t o f t w o v e c t o r s \overset{⇀}{a} a n d \overset{⇀}{b} i s g i v e n b y, = \overset{⇀}{a} . \overset{⇀}{b} = |\overset{⇀}{a}| |\overset{⇀}{b}| C o s θ w e c o n s i d e r, |\overset{⇀}{a}| = 3 a n d |\overset{⇀}{b}| = 4 t h e n, t h e d o t p r o d u c t o f g i v e n v e c t o r s = 12 C o s θ . . . . . . (i) a c c o r d i n g t o t h e q u e s t i o n, r e s u l \tan t, R = 1 w e k n o w, R^{2} = a^{2} + b^{2} + 2 a b C o s θ \Rightarrow 1^{2} = 3^{2} + 4^{2} + 2 \times 3 \times 4 C o s θ \Rightarrow 1 = 9 + 16 + 24 C o s θ \Rightarrow 24 C o s θ = - 24 \Rightarrow C o s θ = \frac{- 24}{24} = - 1 p u t t i n g t h e v a l u e o f C o s θ i n e q u a t i o n (i), w e g e t t h e d o t p r o d u c t o f g i v e n v e c t o r s = - 12

option(A) is correct.

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