Let `vec(PR) = 3hati + hatj -2hatk and vec(SQ) =hati-3hatj - 4hatk` determine diagonals of a parallelogram PQRS. And `vec(PT) = hati + 2hatj + 3hatk ` be onther vector. Then the volume of the parallelepiped determined by the vectors `vec(PT), vec(PQ) and vec(PS)` is

Options

A. 5
B. 20
C. 10
D. 30

Answer

c

Explanation

Area of base (PQRS)
`=1/2|vec(PR)xx vec(SQ)|=1/2|{:(hati,hatj,hatk),(3,1,-2),(1,-3,-4):}|`
`= 1/2 |- 10hati + 10 hatj -10hatk|`
` 5|hati -hatj +hatk|= 5sqrt3`
Height= projection of PT on `hati -hatj +hatk`
`|(1-2+3)/sqrt3|=2/sqrt3`
volume = `(5 sqrt3) (2/sqrt3)` = 10 cu. unit