If the points `(2-x, 2, 2), (2, 2-y, 2) and (2, 2, 2-z)` are coplanar then prove `2/x+2/y+2/z=1`

Options

A. `1/x+1/y+1/z=1`
B. `x+y+z=1`
C. `1/(1-x)+1/(1-y)+1/(1-z)=1`
D. `1/(1-x)+1/(1-y)+1/(1-z)=2`

Answer

A

Explanation

`vec(AB)=vec(OB)-vec(OA)={2hati+(2-y)hatj+2hatk}-`
`{(2-x)hati+2hatj+2hatk}=xhati-yhatj`
`vec(AC)=vec(OC)-vec(OA)`
`={(2hati+2hatj+(2-z)hatk}-{(2-x)hati+2hatj+2hatk}`
`=xhati-zhatk`
`vec(AD)=vec(OD)-vec(OA)`
`=(hati+hatj+hatk)-{(2-x)hati+2hatj+2hatk}=(x-1)hati-hatj-hatk`
As these vectors are coplanar, `|(x, -y, 0),(x, 0, -z),(x-1, -1, -1)|=0`
On simplication we get `1/x+1/y+1/z=1`