Answer :

Given :


Equation of plane : 2x - y + z + 3 = 0


P = (1, 3, 4)


To Find : Image of point P in the plane.


Note :


If two vectors with direction ratios (a1, a2, a3) & (b1, b2, b3) are parallel then



Given equation of the plane is


2x - y + z + 3 = 0


2x - y + z = -3 ………..eq(1)


From eq(1) direction ratios of normal vector of the plane are


(2, -1, 1)


Therefore, equation of normal vector is



Given point is P = (1, 3, 4)


Let, R(a, b, c) be image of point P in the given plane.


Therefore, the power of points P and R in the given plane will be equal and opposite.


2a – b + c + 3 = - (2(1) – 3 + 4 + 3)


2a – b + c + 3 = - 6


2a – b + c = - 9 ………eq(2)


Now,


As are parallel



a = 2k+1, b = -k+3, c = k+4


substituting a, b, c in eq(2)


2(2k+1) - (-k+3) + (k+4) = -9


4k + 2 + k – 3 + k + 4 = -9


6k = -12



,




Therefore, co-ordinates of the image of P are


R(a, b, c) =



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