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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. The equation of the plane ABC is |
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Answer» `x+y+z-3=0`  `""|{:(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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