Symmetry by plane

Determine the coordinates of a image of point A (3, -4, -6) at a symmetry that is determined by the plane x-y-4z-13 = 0

Correct result:

x =  1
y =  -2
z =  2

Solution:

r:xy4z13=0 A(3,4,6)  n=s n=(1,1,4) p:x=t+a;y=t+b;z=4t+c;tR  3=t+a 4=t+b 6=4 t+c t=1  a+t=3 bt=4 c4t=6 t=1  a=2 b=3 c=2  p:x=t+2;y=t3;z=4t2;tR  p(t=1)=A  x=1+a=1+2=1
y=(1)+b=(1)+(3)=2y=-(-1)+b=-(-1)+(-3)=-2
z=4 (1)+c=4 (1)+(2)=2z=-4 \cdot \ (-1)+c=-4 \cdot \ (-1)+(-2)=2



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