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

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 t=1  p:x=t+2;y=t3;z=4t2;tR  p(t=1)=A  x=1+a=1+2=1r:x-y-4z-13=0 \ \\ A(3,-4,-6) \ \\ \ \\ n=s \ \\ n=(1,-1,-4) \ \\ p: x=t+a; y=-t+b; z=-4t + c; t \in R \ \\ \ \\ 3=t + a \ \\ -4=-t + b \ \\ -6=-4 \cdot \ t + c \ \\ t=1 \ \\ \ \\ a+t=3 \ \\ b-t=-4 \ \\ c-4t=-6 \ \\ t=1 \ \\ \ \\ a=2 \ \\ b=-3 \ \\ c=-2 \ \\ t=1 \ \\ \ \\ p: x=t+2; y=-t-3; z=-4t-2; t \in R \ \\ \ \\ 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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For Basic calculations in analytic geometry is helpful line slope calculator. From coordinates of two points in the plane it calculate slope, normal and parametric line equation(s), slope, directional angle, direction vector, the length of segment, intersections the coordinate axes etc.

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