Symmetric point of a point wrt a line

Question

I am trying to solve the following question.

Let $d$ be the $$ \frac{x-1}{2}=\frac{y-2}{4}=\frac{z-3}{5}, $$ and let $P$ be the point of coordinates $(4,3,10)$. Find the coordinates of the symmetric point $P^{\prime}$ of $P$ with respect to $d$.

Book has the following solution.

Let $M\left(x_0, y_0, z_0\right)$ be the midpoint of $P P^{\prime}$. Then $M \in d, P M \perp$ d, i.e., $$ \begin{gathered} \frac{x_0-1}{2}=\frac{y_0-2}{4}=\frac{z_0-3}{5} \\ 2\left(x_0-4\right)+3\left(y_0-3\right)+5\left(z_0-10\right)=0 . \end{gathered} $$ This yields $x_0=129 / 41, y_0=258 / 41, z_0=343 / 41$. Now, the coordinates of $P^{\prime}$ are given by $\left(P, M ; P^{\prime}\right)=-2$ (which is the simple ratio), and we have $P^{\prime}(-94 / 41$, $-393 / 41,-276 / 41)$

However, I don't understand why he takes $2\left(x_0-4\right)+3\left(y_0-3\right)+5\left(z_0-10\right)=0$ instead of $2\left(x_0-4\right)+4\left(y_0-3\right)+5\left(z_0-10\right)=0$. Since $PM$ is orthogonal to $d$, I am in $PM\cdot (2,4,5)$ must be equal to zero since $(2,4,5)$ is direction vector of the line. If we use the dot product that I wrote, I calculated $x_0=135/13, y_0=-270/13$.

Is my solution wrong?

I appreciate any help in advance.

Answer 1

The equations are

$ (x_0, y_0, z_0) = (1,2,3) + t (2, 4, 5) $

$PM = M - P = (1, 2, 3) - (4, 3, 10) + t (2, 4, 5) = (-3, -1, -7) + t (2, 4, 5)$

$ PM \perp d $ , so

$ \bigg( (-3, -1, -7) + t(2, 4, 5) \bigg) \cdot (2, 4, 5) = 0 $

So that

$ -45 + 45 t = 0 $

$\Longrightarrow t = 1 $

Thus $M = (1,2,3) + (2,4,5) = (3, 6, 8) $

$M = \dfrac{1}{2} (P + P') $

So

$P' = 2 M - P = 2 (3, 6, 8) - (4,3,10) = (2, 9, 6) $

Answer 2

Why don't they write lines like this?: $$x=2t+1,y=4t+2,z=5t+3$$ Its direction vector is $<2,4,5>.$ The plane passing through $(4,3,10)$ and perpendicular to this vector is $$2(x-4)+4(y-3)+5(z-10)=0.$$

Common solution of the plane and line: $2(2t+1-4)+4(4t+2-3)+5(5t+3-10)=0\implies t=1.$

So, the mid point $M=(2\times1+1,4\times1+2,5\times1+3)=(3,6,8)$ and $P'=2M-P=(2,9,6)$.