1.

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`

Answer» 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))`
`=>OC^2 = a^2+b^2+2abcos(theta+alpha)->(3)`
Adding (2) and (3),
`c^2+OC^2 = 2(a^2+b^2)`
`=> OC^2 = 2(a^2+b^2)-c^2`
Now, in `Delta BOC`, using cosine law,
`a^2 = b^2+OC^2-2b|vecOC|costheta`
`=>a^2 = b^2+2a^2+2b^2-c^2-2b|vecOC|costheta`
`=>2b|vecOC|costheta = 3b^2+a^2-c^2`
`=>b|vecOC|costheta = (3b^2+a^2-c^2)/2`
From (1),
`vec(OC).vec(OB) = (a^2+3b^2-c^2)/2`


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