Given ` vecP=3 hati+2hatj+5hatk, veca=hati+hatj, vecb=hatj+hatk, vecc=hati+hatk and vecP=xveca+yvecb+zvecc`, then x,y,z are respectively

Options

A. `(3)/(2), (1)/(2), (5)/(2)`
B. `(1)/(2),(3)/(2),(5)/(2)`
C. `(5)/(2),(3)/(2),(1)/(2)`
D. `(1)/(2),(5)/(2),(3)/(2)`

Answer

B

Explanation

`p=xa+yb+zc`
`implies3hati+2hatj+4hatk=x(hati+hatj)+y(hatj+hatk)+z(hati+hatk)`
`implies 3hati+2hatj+4hatk=(x+z)hati+(x+y)hatj+(y+z)hatk`
On comparing both sides, the coefficients of `hati, hatj, hatk`, we get
`x+z=3" ... (i)"`
`x+y=2" ... (ii)"`
and `y+z=4" ... (iii)"`
On solving Eqs. (i), (ii) and (iii), we get
`x=(1)/(2), y=(3)/(2), z=(5)/(2)`