1.

ABCD is a regular tetrahedron P & Q are the mid -points of the edges AC and AB respectively, G is the cenroid of the face BCD and theta is the angle between the vectors vec(PG) and vec(DQ), then

Answer»

The angle between `vec(AB)` and `vec(CD)` is `90^(@)`
The angle `theta` is `pi-cos^(-1)(5/(3sqrt(3)))`
The angle `theta` is `pi-cos^(-1)(5/(6sqrt(3)))`
The angle between `vec(AB)` and `vec(CD)` is `120^(@)`

Solution :`vec(DA)=veca,vec(DB),VECB, vec(DC)=vecc`
`vec(AD).vec(BC)=(-veca).(vecc-vecb)`
`=-veca.vecc+veca.vecb`
`=0`
Hence `vec(AD)_|_^(r) vec(BC)impliesvec(AB)_|_vec(CD)`
Now, `vec(PG)=-1/6(3veca-2vecb+vecc)` & `vec(DQ)=1/2(veca+vecb)`
Let `|veca|=|vecb|=|vecc|=K`
`implies|vec(PG)|=K/2` & `|vec(DQ)|=(sqrt(3))/2K`
`impliesvec(PG).vec(DQ)=-1/6(3veca-2vecb+vecc).1/2(veca+vecb)`
`costheta=-5/(6sqrt(3))impliestheta-pi-cos^(-1)(5/(6sqrt(3)))`


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