The image of the point (3, 2, 1) in the plane | vedclass

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(C) The formula for the image $(x, y, z)$ of a point $(x_1, y_1, z_1)$ in the plane $ax + by + cz + d = 0$ is given by: $\frac{x - x_1}{a} = \frac{y -

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$(3, 2, 1)$

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(C) The formula for the image $(x, y, z)$ of a point $(x_1, y_1, z_1)$ in the plane $ax + by + cz + d = 0$ is given by: $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \frac{-2(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2}$ Here, the point is $(3, 2, 1)$ and the plane equation is $2x - y + 3z - 7 = 0$. Substituting the values: $\frac{x - 3}{2} = \frac{y - 2}{-1} = \frac{z - 1}{3} = \frac{-2(2(3) - 1(2) + 3(1) - 7)}{2^2 + (-1)^2 + 3^2}$ $\frac{x - 3}{2} = \frac{y - 2}{-1} = \frac{z - 1}{3} = \frac{-2(6 - 2 + 3 - 7)}{4 + 1 + 9}$ $\frac{x - 3}{2} = \frac{y - 2}{-1} = \frac{z - 1}{3} = \frac{-2(0)}{14} = 0$ Setting each part to $0$: $x - 3 = 0 \Rightarrow x = 3$ $y - 2 = 0 \Rightarrow y = 2$ $z - 1 = 0 \Rightarrow z = 1$ Thus, the image of the point is $(3, 2, 1)$, which means the point lies on the plane.

The image of the point $(3, 2, 1)$ in the plane $2x - y + 3z = 7$ is

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