(i) Find the equation of the plane passing through the points`(2,1,0),(5,0,1)a n d(4,11)dot` (ii) If `P` s the point `(2,1,6),` then the find the point `Q` such that `P Q` is perpendicular to the plane in (i) and the midpoint of `P Q` lies on it.

Answer

(i) `x+y-2z=3" "(ii) Q(6,5,-2)`

Explanation

(i) Equation of plane passing throuth (2,1,0) is
`a(x-2)+b(y-1)+c(z-0)=0`
It also passes through (5,0,1) and (4,1,1).
`implies3a-b+c=0" and "2a-0b+c=0`
On solving, we get `(a)/(-1)=(b)/(-1)=( c )/(2)`
`:.` Equation of plane is
`-(x-2)-(y-1)+2(z-0)=0`
`-(x-2)-y+1+2z=0`
`implies" "x+y-2z=3`
(ii) Let the coordinates of Q be `(alpha,beta,gamma).`
Equation of line `PQimplies(x-2)/(1)=(y-1)/(1)=(z-6)/(-2)`
Since, mid-point of P and Q
`((alpha+2)/(2),(beta+1)/(2),(gamma+6)/(-2)),`
which lies in line PQ.
`implies" "((a-2)/(2)-2)/(2)=((beta+1)/(2)-1)/(1)=((gamma+6)/(2)-6)/(-2)`
`(1((alpha+2)/(2)-2)+1((beta+1)/(2)-1)-2((gamma+6)/(2)-6))/(1.1+1.1+(-2)(-2))=2`
`[because((alpha+2)/(2))-1((beta+1)/(2))-2((gamma+6)/(2))=3]`
`impliesalpha=6,beta=5,gamma=-2impliesQ(6,5,-2
)`