Finding the coordinates of the point Q, symmetrical from the point P (2,1,3) in relation to the plane 4x -3y + z +18 = 0
Answer:
(-6,7,1)
1. Calculate the equation of a line l through P perpendicular to the given plane $\displaystyle E: (4,-3,1) \cdot (x,y,z)+18=0$
$\displaystyle l: (x,y,z)=(2,1,3)+t \cdot (4,-3,1)$
2. Let M denote the intersection point between l and the given plane:
$\displaystyle E \cap l = \{M\}~\implies~ (4,-3,1) \cdot \left((2,1,3)+t \cdot (4,-3,1)\right)+18=0 $ $\displaystyle ~\implies~8+26t+18=0~\implies~ t = -1$
and therefore $\displaystyle M(-2,4,2)$
3. M is the midpoint of $\displaystyle \overline{PQ}$ . Let $\displaystyle Q(q_1 ,q_2 ,q_3 )$ denote the reflection point of P with respect to E. Then you know by the formula calculating the midpoint between 2 points:
$\displaystyle \frac{2+q_1}{2} =-2 \ \wedge\ \frac{1+q_2}{2} =4 \ \wedge\ \frac{3+q_3}{2} =2$
Solve these equations for $\displaystyle q_i$. You'll get $\displaystyle Q(-6,7,1)$