Given a parallelogram `OACB`. The lengths of the vectors `vec (OA)`, `vec (OB)` & `vec (AB)` are `a`, `b` & `c` respectively. The scalar product of the vectors `vec (OC)` & `vec (OB)` is
(A) `(a^2-3b^2+c^2)/2`
(B) `(3a^2+b^2-c^2)/2`
(C) `(3a^2-b^2+c^2)/2`
(D) `(a^2+3b^2-c^2)/2`

Explanation

With the given details, we can create a diagram.

Please refer to video for the diagram.

Here,` vec(OC).vec(OB) =|vec(OB)||vec(OC)|costheta`

`=> vec(OC).vec(OB) =b|vec(OC)|costheta ->(1)`

In `Delta AOB`, using cosine law,

`c^2 = a^2+b^2-2abcos(theta+alpha)->(2)`

In `Delta AOC`, using cosine law,

`OC^2 = a^2+b^2-2abcos(pi-(theta+alpha))`

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