[Math] Mirror a line over a plane

analytic geometrylinear algebrasolution-verification

I am trying to mirror a line over a plane, but I am not sure if I am doing it right, so please tell me if something that I do is wrong.

I have 2 points $A(1, 2, 1);B(-1,0,2)$ and I have to mirror the line that goes through these 2 points over the plane $x – 2y + 2z = -5$

To get the line I need a direction vector, in this case $\vec{e}=\vec{AB} = \vec{r_b} – \vec{r_a} = \begin{bmatrix}-2\\-2\\1 \end{bmatrix} $ and a point – $A$ so I get:

$$
\vec{r} = \begin{bmatrix}1 \\ 2 \\ 1 \end{bmatrix} + \lambda\begin{bmatrix}-2\\-2\\1\end{bmatrix}
$$
So, next I need the projection of $\vec{e}$ onto $\vec{n}$ and I get it using this formula:
$$
\vec{s} = \frac{\vec{n}^T\vec{e}}{\vec{n}^T\vec{n}}\vec{n} = \frac{4}{9}\begin{bmatrix}1 \\ 2 \\ 1 \end{bmatrix}
$$
Then I simply substract twice the result ($\vec{s}$) from the directional vector ($\vec{e}$)

Is this the right way to do it?

Best Answer

It's one way to do it, but you haven't finished the solution. After finding the new direction vector you still need a constant term in the parametric equation; it will not be the same.

You also have a typo in the last formula: should have $-2$ instead of $2$ as the second coordinate.

It is more efficient to reflect the points $A,B$ about the plane, and then find the line through them. To reflect point $A$:

  1. Take any point $C$ on the plane, say $(-5,0,0)$.
  2. find the vector $\vec{CA} = (6, 2,1)$.
  3. Take its normal component $$ \vec v = \frac{\vec{CA}\cdot \vec{n}}{|\vec n|^2}\vec n$$
  4. Subtract $2\vec v$ from $A$.
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