Given `p=3hat(i)+2hat(j)+4hat(k), a=hat(i)+hat(j), b=hat(j)+hat(k), c=hat(i)+hat(k)` and `p=x a +y b +z c`, then x, y and 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=x a + yb +zc`
`rArr" "3hati+2hatj+4hatk=x(hati+hatj)+y(hatj+hatk)+z(hati+hatk)`
`rArr" "3hati+2hatj+4hatk=(x+z)hati+(x+y)hatj+(y+z)hatk`
`x+z=3" ...(i)"`
`x+y=2" ...(ii)"`
`x+z=4" ...(iii)"`
Now, subtracting Eq. (ii) from Eq. (i), we get
`z-y=1" ...(iv)"`
From Eqs. (iii) and (iv), we gt `z=5/2, y=3/2`
From Eq. (ii) we get `x=1/2`
Hence, `x=1/2, y=3/2, z=5/2`