# Point symmetrical

• Jun 24th 2008, 05:34 PM
Apprentice123
Point symmetrical
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

(-6,7,1)
• Jun 24th 2008, 10:37 PM
earboth
Quote:

Originally Posted by Apprentice123
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

(-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 $\displaystyle Q(-6,7,1)$
• Jun 25th 2008, 05:35 AM
Apprentice123
Thank you