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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`.
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Solution
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.
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