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Given four points A(2, 1, 0), B(1, 0, 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 a line L orthogonal to the plane determined by the points A, B and C.
The equation of the plane ABC is

A

`x+y+z-3=0`

B

`y+z-1=0`

C

`x+z-1=0`

D

`2y+z-1=0`

Text Solution

Verified by Experts

The correct Answer is:
b

`" "|{:(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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