The image of the point (5, 2, 6) with respect | vedclass

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(C) Let the given point be $P(x_1, y_1, z_1) = (5, 2, 6)$ and the plane be $ax + by + cz + d = 0$, where $x + y + z - 9 = 0$. Here, $a = 1, b = 1, c =

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$(\frac{7}{3}, -\frac{2}{3}, \frac{10}{3})$

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(C) Let the given point be $P(x_1, y_1, z_1) = (5, 2, 6)$ and the plane be $ax + by + cz + d = 0$, where $x + y + z - 9 = 0$. Here, $a = 1, b = 1, c = 1$ and $d = -9$. The formula for the image $(x, y, z)$ of a point $(x_1, y_1, z_1)$ with respect to the plane $ax + by + cz + d = 0$ is given by: $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = -2 \frac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2}$ Substituting the values: $\frac{x - 5}{1} = \frac{y - 2}{1} = \frac{z - 6}{1} = -2 \frac{(1)(5) + (1)(2) + (1)(6) - 9}{1^2 + 1^2 + 1^2}$ $\frac{x - 5}{1} = \frac{y - 2}{1} = \frac{z - 6}{1} = -2 \frac{5 + 2 + 6 - 9}{3} = -2 \frac{4}{3} = -\frac{8}{3}$ Now, solving for $x, y, z$: $x - 5 = -\frac{8}{3} \Rightarrow x = 5 - \frac{8}{3} = \frac{15 - 8}{3} = \frac{7}{3}$ $y - 2 = -\frac{8}{3} \Rightarrow y = 2 - \frac{8}{3} = \frac{6 - 8}{3} = -\frac{2}{3}$ $z - 6 = -\frac{8}{3} \Rightarrow z = 6 - \frac{8}{3} = \frac{18 - 8}{3} = \frac{10}{3}$ Thus, the image of the point $(5, 2, 6)$ is $(\frac{7}{3}, -\frac{2}{3}, \frac{10}{3})$.

The image of the point $(5, 2, 6)$ with respect to the plane $x + y + z = 9$ is

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