Let P be a point in the first octant,whose im | vedclass

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(A) Let $P \equiv (x_0, y_0, z_0)$.The line passing through $P$ perpendicular to the plane $x+y=3$ is given by $\frac{x-x_0}{1} = \frac{y-y_0}{1} = \f

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$8$

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(A) Let $P \equiv (x_0, y_0, z_0)$.The line passing through $P$ perpendicular to the plane $x+y=3$ is given by $\frac{x-x_0}{1} = \frac{y-y_0}{1} = \frac{z-z_0}{0} = k$.The image $Q$ is given by $\frac{x-x_0}{1} = \frac{y-y_0}{1} = \frac{z-z_0}{0} = -2 \frac{x_0+y_0-3}{1^2+1^2} = -(x_0+y_0-3)$.Thus,$x_Q = x_0 - (x_0+y_0-3) = 3-y_0$ and $y_Q = y_0 - (x_0+y_0-3) = 3-x_0$.Since $Q$ lies on the $z$-axis,$x_Q = 0$ and $y_Q = 0$,which implies $x_0 = 3$ and $y_0 = 3$.The distance of $P(3, 3, z_0)$ from the $x$-axis is $\sqrt{y_0^2 + z_0^2} = 5$.Substituting $y_0 = 3$,we get $\sqrt{3^2 + z_0^2} = 5$,so $9 + z_0^2 = 25$,which gives $z_0^2 = 16$,so $z_0 = 4$ (since $P$ is in the first octant).$P$ is $(3, 3, 4)$. The image $R$ of $P$ in the $xy$-plane is $(3, 3, -4)$.The length of $PR$ is the distance between $(3, 3, 4)$ and $(3, 3, -4)$,which is $|4 - (-4)| = 8$.

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