If `vecr.veca=vecr.vecb=vecr.vecc=1/2` for some non zero vector `vecr and veca,vecb,vecc` are non coplanar, then the area of the triangle whose vertices are `A(veca),B(vecb) and C(vecc0` is (A) `|[veca vecb vecc]|` (B) `|vecr|` (C) `|[veca vecb vecr]vecr|` (D) none of these

Options

A. `|[veca vecb vecc]|`
B. `|vecr|`
C. `|[veca vecb vecc]vecr|`
D. none of these

Answer

c

Explanation

Any vector, `vecr` can be represented in terms of three non- coplanar vectors, `veca, vecb and vecc` as
`vecr = x (veca xx vecb) + y (vecb xx vecc) +z (vecc xx veca) ` (i)
taking dot product with `veca , vecb and vecc` respectively, we have.
`x=(vecr.vecc)/([vecavecb vecc]),y = (vecr.veca)/([veca vecb vecc])and z= (vecr.vecb)/([veca vecb vecc])`
From (i) we have
`[veca vecb vecc] vecr= 1/2 (veca xx vecb + vecb+vecc +veccxxveca)`
Area of `triangleABC`
` 1/2 |veca xx vecb+vecb xx vecc +vecc +vecc xx veca|`
`|[veca vecb vecc] vecr|`