1.

Given four points A(2, 1, 0), B(1, 0, 1), C(3, 0, 1) and D(0, 0, 2). Point D lies on aline L orthogonal to the plane determined by the points A, B and C.

Answer»

`sqrt(2)`
`1//2`
2
`1//sqrt2`

SOLUTION :
`""|{:(x-2,,y-1,,z),(1-2,,0-1,,1-0),(3-2,,0-1,,1-0):}|=0`
`""(x-2)[(-1)-(-1)]-(y-1)[(-1)-1]+z[1+1]=0`
or `""2(y-1)+2z=0`
or `""y+z-1=0`
The vector NORMAL to the PLANE is `vecr= 0hati+hatj+ hatk`
The equation of the line through `(0, 0, 2)` and PARALLEL to `vecn` is `vecr = 2hatk+lamda(hatj+hatk)`
The perpendicular distance of `D(0, 0, 2)` from plane
`ABC` is `|(2-1)/(sqrt(1^(2)+1^(2))|=(1)/(sqrt2)|`


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