Find the reflection of the point P(2, -1, 3) | vedclass

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(B) Let the point be $P(2, -1, 3)$ and the plane be $3x - 2y - z = 9$.The normal vector to the plane is $\vec{n} = (3, -2, -1)$.The line passing throu

Vedclass · Accepted Answer

$\left( \frac{26}{7}, \frac{-15}{7}, \frac{17}{7} \right)$

Vedclass · Answer

(B) Let the point be $P(2, -1, 3)$ and the plane be $3x - 2y - z = 9$.The normal vector to the plane is $\vec{n} = (3, -2, -1)$.The line passing through $P$ and perpendicular to the plane is given by $\frac{x - 2}{3} = \frac{y + 1}{-2} = \frac{z - 3}{-1} = \lambda$.Any point on this line is $M(3\lambda + 2, -2\lambda - 1, -\lambda + 3)$.Since $M$ lies on the plane,we have $3(3\lambda + 2) - 2(-2\lambda - 1) - (-\lambda + 3) = 9$.$9\lambda + 6 + 4\lambda + 2 + \lambda - 3 = 9$.$14\lambda + 5 = 9 \implies 14\lambda = 4 \implies \lambda = \frac{2}{7}$.The coordinates of $M$ are $M\left(3(\frac{2}{7}) + 2, -2(\frac{2}{7}) - 1, -(\frac{2}{7}) + 3\right) = M\left(\frac{20}{7}, -\frac{11}{7}, \frac{19}{7}\right)$.Let the reflection of $P$ be $P'(x', y', z')$. Since $M$ is the midpoint of $PP'$,we have $\frac{x' + 2}{2} = \frac{20}{7} \implies x' = \frac{40}{7} - 2 = \frac{26}{7}$.$\frac{y' - 1}{2} = -\frac{11}{7} \implies y' = -\frac{22}{7} + 1 = -\frac{15}{7}$.$\frac{z' + 3}{2} = \frac{19}{7} \implies z' = \frac{38}{7} - 3 = \frac{17}{7}$.Thus,the reflection is $\left( \frac{26}{7}, -\frac{15}{7}, \frac{17}{7} \right)$.

Find the reflection of the point $P(2, -1, 3)$ in the plane $3x - 2y - z = 9$.

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