Question

Prove that the volume of a tetrahedron with coterminus edges `overlinea, overlineb` and `overlinec` is `1/6[(overlinea, overlineb, overlinec)]`.

Hence, find the volume of tetrahedron whose coterminus edges are `overlinea = hati + 2hatj + 3hatk, overlineb = -hati + hatj + 2hatk` and `overlinec = 2hati + hatj + 4hatk`.

Answer 1

Let `overline(OA) = overlinea, overline(OB) = overlineb` and `overline(OC) = overlinec` be coterminus edges of tetrahedron OABC.

Let AP be the height of the tetrahedron.

Volume of tetrahedron = `1/3`(Area of base ΔOCB) (Height AP)

But AP = Scalar Projection of `overlinea` on `(overlineb xx overlinec)`

= `((overlineb xx overlinec).overlinea)/|overlineb xx overlinec|`  ...`(∵  "scalar projection of"  overlinep  "on"  overlineq  "is"  (overlinep.overlineq)/overlineq)`

Area of ΔOBC = `1/2 |overlineb xx overlinec|`

Volume of tetrahedron = `1/3 xx 1/2 |overlineb xx overlinec| ((overlineb xx overlinec).overlinea)/|overlineb xx overlinec|`

= `1/6 [(overlineb xx overlinec).overlinea]`

= `1/6[(overlinea, overlineb, overlinec)]`

Hence proved.

Now `overlinea = hati + 2hatj + 3hatk, overlineb = -hati + hatj + 2hatk` and `overlinec = 2hati + hatj + 4hatk`

∴ Volume of tetrahedron = `1/6[(overlinea, overlineb, overlinec)]`

= `1/6|(1, 2, 3),(-1, 1, 2),(2, 1, 4)|`

= `1/6 [1(4 - 2) - 2(-4 - 4) + 3(-1 - 2)]`

= `1/6 [1(2) - 2(-8) + 3(-3)]`

= `1/6 [2 + 16 - 9]`

= `1/6 [18 - 9]`

= `1/6 xx 9`

= `3/2`

= 1.5

Thus the Volume of tetrahedron is 1.5 cubic units.