Find the reflection of the point (-1, 3, 4) i | vedclass

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(D) Let the point be $P(-1, 3, 4)$ and the plane be $x - 2y = 0$. The normal vector to the plane is $\vec{n} = (1, -2, 0)$.Let $Q(x_1, y_1, z_1)$ be t

Vedclass · Accepted Answer

None of these

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(D) Let the point be $P(-1, 3, 4)$ and the plane be $x - 2y = 0$. The normal vector to the plane is $\vec{n} = (1, -2, 0)$.Let $Q(x_1, y_1, z_1)$ be the reflection of $P$ in the plane. The line $PQ$ passes through $P$ and is parallel to the normal vector $\vec{n}$.The equation of line $PQ$ is $\frac{x + 1}{1} = \frac{y - 3}{-2} = \frac{z - 4}{0} = \lambda$.Any point on this line is $Q(\lambda - 1, -2\lambda + 3, 4)$.The midpoint $R$ of $PQ$ is $\left( \frac{\lambda - 1 - 1}{2}, \frac{-2\lambda + 3 + 3}{2}, \frac{4 + 4}{2} \right) = \left( \frac{\lambda - 2}{2}, 3 - \lambda, 4 \right)$.Since $R$ lies on the plane $x - 2y = 0$,we have $\left( \frac{\lambda - 2}{2} \right) - 2(3 - \lambda) = 0$.$\frac{\lambda - 2}{2} - 6 + 2\lambda = 0 \Rightarrow \lambda - 2 - 12 + 4\lambda = 0 \Rightarrow 5\lambda = 14 \Rightarrow \lambda = \frac{14}{5}$.The reflection $Q$ is given by $x_1 = \lambda - 1 = \frac{14}{5} - 1 = \frac{9}{5}$,$y_1 = -2\lambda + 3 = -2(\frac{14}{5}) + 3 = \frac{-28 + 15}{5} = -\frac{13}{5}$,and $z_1 = 4$.Thus,the reflection is $\left( \frac{9}{5}, -\frac{13}{5}, 4 \right)$. Since this is not among the options,the correct answer is $D$.

Find the reflection of the point $(-1, 3, 4)$ in the plane $x - 2y = 0$.

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