1.

Given, p=3hati+2hatj+4hatk,a=hati+hatj,b=hatj+hatk,c=hati+hatk and P = xa + yb + zc, then x, y, z are respectively

Answer»

`(3)/(2), (1)/(2), (5)/(2)`
`(1)/(2),(3)/(2),(5)/(2)`
`(5)/(2),(3)/(2),(1)/(2)`
`(1)/(2),(5)/(2),(3)/(2)`

Solution :`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)`


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