Let P be a plane passing through the points ( | vedclass

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(A) First,find the equation of the plane $P$ passing through $A(2,1,0), B(4,1,1)$ and $C(5,0,1)$.Vectors in the plane are $\vec{AB} = (4-2, 1-1, 1-0)

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$(6,5,-2)$

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(A) First,find the equation of the plane $P$ passing through $A(2,1,0), B(4,1,1)$ and $C(5,0,1)$.Vectors in the plane are $\vec{AB} = (4-2, 1-1, 1-0) = (2,0,1)$ and $\vec{AC} = (5-2, 0-1, 1-0) = (3,-1,1)$.The normal vector $\vec{n} = \vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 0 & 1 \ 3 & -1 & 1 \end{vmatrix} = \hat{i}(0+1) - \hat{j}(2-3) + \hat{k}(-2-0) = (1, 1, -2)$.The equation of the plane is $1(x-2) + 1(y-1) - 2(z-0) = 0$,which simplifies to $x + y - 2z - 3 = 0$.Let $R(2,1,6)$ be the point and $I(x,y,z)$ be its image in the plane. The line passing through $R$ and $I$ is perpendicular to the plane,so its direction ratios are $(1, 1, -2)$.The equation of the line $RI$ is $\frac{x-2}{1} = \frac{y-1}{1} = \frac{z-6}{-2} = k$.Any point on this line is $(k+2, k+1, -2k+6)$.The foot of the perpendicular $F$ is the midpoint of $RI$,where $k$ is such that $F$ lies on the plane. For the image $I$,the point $I$ corresponds to $2k$,so $I = (2k+2, 2k+1, -4k+6)$.Substituting $F(k+2, k+1, -2k+6)$ into the plane equation: $(k+2) + (k+1) - 2(-2k+6) - 3 = 0$.$k + 2 + k + 1 + 4k - 12 - 3 = 0 \Rightarrow 6k - 12 = 0 \Rightarrow k = 2$.The image $I$ is at $2k$,so $I = (2(2)+2, 2(2)+1, -4(2)+6) = (6, 5, -2)$.

Let $P$ be a plane passing through the points $(2,1,0), (4,1,1)$ and $(5,0,1)$ and $R$ be any point $(2,1,6)$. Then the image of $R$ in the plane $P$ is

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