Point symmetrical

**Apprentice123** · June 24th 2008, 05:34 PM

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)

**earboth** · June 24th 2008, 10:37 PM

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

Answer:
(-6,7,1)

1. Calculate the equation of a line l through P perpendicular to the given plane $E: (4,-3,1) \cdot (x,y,z)+18=0$

$l: (x,y,z)=(2,1,3)+t \cdot (4,-3,1)$

2. Let M denote the intersection point between l and the given plane:

$E \cap l = \{M\}~\implies~ (4,-3,1) \cdot \left((2,1,3)+t \cdot (4,-3,1)\right)+18=0$ $~\implies~8+26t+18=0~\implies~ t = -1$

and therefore $M(-2,4,2)$

3. M is the midpoint of $\overline{PQ}$ . Let $Q(q_1 ,q_2 ,q_3 )$ denote the reflection point of P with respect to E. Then you know by the formula calculating the midpoint between 2 points:

$\frac{2+q_1}{2} =-2 \ \wedge\ \frac{1+q_2}{2} =4 \ \wedge\ \frac{3+q_3}{2} =2$

Solve these equations for $q_i$ . You'll get $Q(-6,7,1)$

**Apprentice123** · June 25th 2008, 05:35 AM

Thank you

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