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

`vecP=xveca+y vecb+zvecc`
`implies 3 hati + 2hatj+ 4hatk=x(hati+hatj)+y(hatj+hatk)+z(hati+hatk)`
`implies 3hati+3hatj+4hatk=(x+z)hati+(x+y)hatj+(y+z)hatk `
On comparing both sides the coefficients of ` hatk, hatj, hatk` , we get
x+z=3 .................(i)
x+y=2 ..............(ii) brgt and x+y=4 ........(iii)
On solving Eqs. (i) , (ii) and (iii) , we get
`x=(1)/(2) , y=(3)/(2), z=(5)/(2)`