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Consider DeltaABC with A=(veca);B=(vecb)...

Consider `DeltaABC` with `A=(veca);B=(vecb) and C=(vecc).` If `vecb.(veca+vecc)=vecb.vecb+veca.vecc;|vecb-veca|=3;|vecc-vecb|=4` then the angle between the medians `AvecM and BvecD` is

A

`pi-cos^(-1)((1)/(5sqrt(13)))`

B

`pi-cos^(-1)((1)/(13sqrt(5)))`

C

`cos^(-1)((1)/(5sqrt(13)))`

D

`cos^(-1)((1)/(13sqrt(5)))`

Text Solution

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The correct Answer is:
A
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ARIHANT MATHS-PRODUCT OF VECTORS-Exercise (Single Option Correct Type Questions)
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