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

Options

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

Answer

A

Explanation

In Delta's `PQR` and `PSQ`, we have
`vec(PQ)+vec(QR)=vec(PR)` and `vec(PS)+vec(SQ)=vec(PQ)`
`impliesvec(PQ)+vec(QR)=vec(PR)` and `vec(PQ)-vec(PS)=vec(SQ)`
`impliesvec(PQ)+vec(PS)=vec(PR)` and `vec(PQ)-vec(PS)=vec(SQ)`
`impliesvec(PQ)=(vec(PR)+vec(SQ))/2` and `vec(PS)=vec(PS)=(vec(PR)-vec(SQ))/2`
`impliesvec(PQ)=2hati-hatj-3hatk` and `vec(PS)=hati+2hatj+hatk`
Thus we obtain `vec(PQ)=2hati-hatj-3hatk, vec(PS)=hati+2hatj+hatk` and
`vec(PT)=hati+2hatj+3hatk`

`:.[(vec(PT), vec(PQ), vec(PS))]=|(1,2,3),(2,-1,-3),(1,2,1)|=5-10+15=10`
Hence volume of the parallelopiped determined by the vectors `vec(PT),vec(PQ)` and `vec(PS)` is 10 cubic units.