If the mirror image of the point (2, 4, 7) in | vedclass

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(C) The formula for the mirror image $(a, b, c)$ of a point $(x_1, y_1, z_1)$ in the plane $Ax + By + Cz + D = 0$ is given by $\frac{a - x_1}{A} = \fr

Vedclass · Accepted Answer

$-6$

Vedclass · Answer

(C) The formula for the mirror image $(a, b, c)$ of a point $(x_1, y_1, z_1)$ in the plane $Ax + By + Cz + D = 0$ is given by $\frac{a - x_1}{A} = \frac{b - y_1}{B} = \frac{c - z_1}{C} = \frac{-2(Ax_1 + By_1 + Cz_1 + D)}{A^2 + B^2 + C^2}$.Given the point $(2, 4, 7)$ and the plane $3x - y + 4z - 2 = 0$,we have $A=3, B=-1, C=4, D=-2$.Substituting these values into the formula:$\frac{a - 2}{3} = \frac{b - 4}{-1} = \frac{c - 7}{4} = \frac{-2(3(2) - 1(4) + 4(7) - 2)}{3^2 + (-1)^2 + 4^2}$$\frac{a - 2}{3} = \frac{b - 4}{-1} = \frac{c - 7}{4} = \frac{-2(6 - 4 + 28 - 2)}{9 + 1 + 16} = \frac{-2(28)}{26} = \frac{-56}{26} = \frac{-28}{13}$.Now,solving for $a, b, c$:$a = 2 + 3(\frac{-28}{13}) = 2 - \frac{84}{13} = \frac{26 - 84}{13} = \frac{-58}{13}$.$b = 4 - 1(\frac{-28}{13}) = 4 + \frac{28}{13} = \frac{52 + 28}{13} = \frac{80}{13}$.$c = 7 + 4(\frac{-28}{13}) = 7 - \frac{112}{13} = \frac{91 - 112}{13} = \frac{-21}{13}$.Finally,calculating $2a + b + 2c$:$2(\frac{-58}{13}) + \frac{80}{13} + 2(\frac{-21}{13}) = \frac{-116 + 80 - 42}{13} = \frac{-78}{13} = -6$.

If the mirror image of the point $(2, 4, 7)$ in the plane $3x - y + 4z = 2$ is $(a, b, c)$,then $2a + b + 2c$ is equal to

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