1.

The vertices of a triangle ABC are A-=(2,0,2), B(-1,1,1) and C-=(1,-2,4). The points D and E divide the sides AB and CA in the ratio 1:2 respectively. Another point F is taken in space such that the perpendicular drawn from F to the plane containing DeltaABC, meets the plane at the point of intersection of the line segments CD and BE. If the distance of F from plane of triangle ABC issqrt(2), units, ten

Answer»

the volume of the tetrahdron `ABCD` is `7/3` CUBIC units
the volume of the tetrahedron `ABCF` is `7/6` cubic units
one of the equation of the line `AF` is `vecr=(2hati+2hatk)+LAMDA(2hatk-hati)(lamda in R)`
one of the equuation of the line `AF` is `vecr=(2hati+2hatk)+mu(hati+7hatk)`

Solution :`CD:vecr(hati-2hatj+4hatk)+(lamda)/3(7hatj-7hatk)`
`BE:vecr=(-hati+hatj+k)+(mu)/3(7hati-7hatj+7hatk)`
`P-=(hati-hatj+3hatj)`
Area of tetrahedron `ABCD`
`=1/3` (Area of base triangle) `XX` height `=7/3` cubic units
`vec(AB)xxvec(AC)=7hatj+7hatk,|vec(PF)|=PF=sqrt(2)` units
`vec(PF)=sqrt(2)((7hatj+7hatk)/(sqrt(49+49)))=hatj+hatk=P.V`. of `F-P.V.` of `P`
`P.V.` of `F=hati+4hatk`
Vector equation of `AF` is `vecr=2(hati+hatk)+ALPHA(-hati+2hatk)`


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