The reflection of the point (-1, 3, 4) with r | vedclass

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(D) The formula for the reflection $(x, y, z)$ of a point $(x', y', z')$ with respect to the plane $ax + by + cz + d = 0$ is given by:$\frac{x - x'}{a

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$\left( \frac{9}{5}, \frac{-13}{5}, 4 \right)$

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(D) The formula for the reflection $(x, y, z)$ of a point $(x', y', z')$ with respect to the plane $ax + by + cz + d = 0$ is given by:$\frac{x - x'}{a} = \frac{y - y'}{b} = \frac{z - z'}{c} = \frac{-2(ax' + by' + cz' + d)}{a^2 + b^2 + c^2}$Given the plane $x - 2y + 0z + 0 = 0$ and the point $(-1, 3, 4)$:$\frac{x + 1}{1} = \frac{y - 3}{-2} = \frac{z - 4}{0} = \frac{-2((-1) - 2(3) + 0(4) + 0)}{1^2 + (-2)^2 + 0^2}$$\frac{x + 1}{1} = \frac{y - 3}{-2} = \frac{z - 4}{0} = \frac{-2(-1 - 6)}{1 + 4} = \frac{-2(-7)}{5} = \frac{14}{5}$Equating each part:$x + 1 = \frac{14}{5} \implies x = \frac{14}{5} - 1 = \frac{9}{5}$$\frac{y - 3}{-2} = \frac{14}{5} \implies y - 3 = \frac{-28}{5} \implies y = 3 - \frac{28}{5} = \frac{15 - 28}{5} = \frac{-13}{5}$$z - 4 = 0 \implies z = 4$Thus,the reflection is $\left( \frac{9}{5}, \frac{-13}{5}, 4 \right)$.

The reflection of the point $(-1, 3, 4)$ with respect to the plane $x - 2y = 0$ is .....

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