Symmetry by plane

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

Correct answer:

x =  1
y =  -2
z =  2

Step-by-step explanation:

r: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 ∈ R  3=t+a −4=−t+b −6=−4⋅ t+c t=1  a+t=3 b−t=−4 c−4t=−6 t=1  t=11​=1 c=−6+4t=−6+4⋅ 1=−2 b=−4+t=−4+1=−3 a=3−t=3−1=2  a=2 b=−3 c=−2 t=1  p: x = t+2; y=−t−3; z = −4t−2; t ∈ R  p(t=−1) = A′  x=−1+a=−1+2=1
y=−(−1)+b=−(−1)+(−3)=−2
z=−4⋅ (−1)+c=−4⋅ (−1)+(−2)=2

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