1.

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

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=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`


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