The mirror image of the point A(1, 2, 3) in a | vedclass

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(A) Let $A = (1, 2, 3)$ and $B = \left(-\frac{7}{3}, -\frac{4}{3}, -\frac{1}{3}\right)$. The plane passes through the midpoint $M$ of $AB$ and is perp

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$(1, -1, 1)$

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(A) Let $A = (1, 2, 3)$ and $B = \left(-\frac{7}{3}, -\frac{4}{3}, -\frac{1}{3}\right)$. The plane passes through the midpoint $M$ of $AB$ and is perpendicular to the line segment $AB$.The midpoint $M$ is given by:$M = \left( \frac{1 - \frac{7}{3}}{2}, \frac{2 - \frac{4}{3}}{2}, \frac{3 - \frac{1}{3}}{2} \right) = \left( \frac{-\frac{4}{3}}{2}, \frac{\frac{2}{3}}{2}, \frac{\frac{8}{3}}{2} \right) = \left( -\frac{2}{3}, \frac{1}{3}, \frac{4}{3} \right)$.The direction ratios of the normal to the plane are the same as the direction ratios of the line $AB$:$D.r.s = \left( 1 - (-\frac{7}{3}), 2 - (-\frac{4}{3}), 3 - (-\frac{1}{3}) \right) = \left( \frac{10}{3}, \frac{10}{3}, \frac{10}{3} \right)$.We can take the normal vector as $\vec{n} = (1, 1, 1)$.The equation of the plane is $1(x - x_0) + 1(y - y_0) + 1(z - z_0) = 0$ where $(x_0, y_0, z_0) = M = \left( -\frac{2}{3}, \frac{1}{3}, \frac{4}{3} \right)$:$1(x + \frac{2}{3}) + 1(y - \frac{1}{3}) + 1(z - \frac{4}{3}) = 0$$x + y + z + \frac{2}{3} - \frac{1}{3} - \frac{4}{3} = 0$$x + y + z - 1 = 0 \Rightarrow x + y + z = 1$.Checking the options:$(A)$ $(1, -1, 1) \Rightarrow 1 - 1 + 1 = 1$. This point lies on the plane.

The mirror image of the point $A(1, 2, 3)$ in a plane is $B\left(-\frac{7}{3}, -\frac{4}{3}, -\frac{1}{3}\right)$. Which of the following points lies on this plane?

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