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

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(D) The plane passes through the midpoint $M$ of the line segment $PP'$.$M = \left( \frac{1 - \frac{7}{3}}{2}, \frac{2 - \frac{4}{3}}{2}, \frac{3 - \f

Vedclass · Accepted Answer

$(1, -1, 1)$

Vedclass · Answer

(D) The plane passes through the midpoint $M$ of the line segment $PP'$.$M = \left( \frac{1 - \frac{7}{3}}{2}, \frac{2 - \frac{4}{3}}{2}, \frac{3 - \frac{1}{3}}{2} ight) = \left( \frac{-\frac{4}{3}}{2}, \frac{\frac{2}{3}}{2}, \frac{\frac{8}{3}}{2} ight) = \left( -\frac{2}{3}, \frac{1}{3}, \frac{4}{3} ight)$.The normal vector $\vec{n}$ to the plane is given by $\vec{PP'} = \left( -\frac{7}{3} - 1, -\frac{4}{3} - 2, -\frac{1}{3} - 3 ight) = \left( -\frac{10}{3}, -\frac{10}{3}, -\frac{10}{3} ight)$.We can take the normal vector as $\vec{n} = (1, 1, 1)$.The equation of the plane is $1(x - (-\frac{2}{3})) + 1(y - \frac{1}{3}) + 1(z - \frac{4}{3}) = 0$.$x + \frac{2}{3} + y - \frac{1}{3} + z - \frac{4}{3} = 0 \implies x + y + z - 1 = 0$.Checking the options:For $(-1, -1, -1): -1 - 1 - 1 - 1 = -4 
eq 0$.For $(-1, -1, 1): -1 - 1 + 1 - 1 = -2 
eq 0$.For $(1, 1, 1): 1 + 1 + 1 - 1 = 2 
eq 0$.For $(1, -1, 1): 1 - 1 + 1 - 1 = 0$.Thus,the point $(1, -1, 1)$ lies on the plane.

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

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