The mirror image of the point P(-1, 2, -4) in | vedclass

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(B) Let the point be $P(x_1, y_1, z_1) = (-1, 2, -4)$ and the plane be $ax + by + cz + d = 0$,where $a = 1, b = -1, c = -2, d = 1$. Let the mirror ima

Vedclass · Accepted Answer

$(-3, 4, 0)$

Vedclass · Answer

(B) Let the point be $P(x_1, y_1, z_1) = (-1, 2, -4)$ and the plane be $ax + by + cz + d = 0$,where $a = 1, b = -1, c = -2, d = 1$. Let the mirror image be $P'(x', y', z')$. The formula for the mirror image of a point $(x_1, y_1, z_1)$ in the plane $ax + by + cz + d = 0$ is given by: $\frac{x' - x_1}{a} = \frac{y' - y_1}{b} = \frac{z' - z_1}{c} = -2 \frac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2}$. First,calculate the value of $ax_1 + by_1 + cz_1 + d$: $1(-1) - 1(2) - 2(-4) + 1 = -1 - 2 + 8 + 1 = 6$. Next,calculate $a^2 + b^2 + c^2$: $1^2 + (-1)^2 + (-2)^2 = 1 + 1 + 4 = 6$. Now,substitute these values into the formula: $\frac{x' - (-1)}{1} = \frac{y' - 2}{-1} = \frac{z' - (-4)}{-2} = -2 \frac{6}{6} = -2$. Solving for $x', y', z'$: $x' + 1 = -2 \implies x' = -3$. $y' - 2 = 2 \implies y' = 4$. $z' + 4 = 4 \implies z' = 0$. Thus,the mirror image is $(-3, 4, 0)$.

The mirror image of the point $P(-1, 2, -4)$ in the plane $x - y - 2z + 1 = 0$ is

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