Question

Find the coordinates of the foot of the perpendicular Q  drawn from P(3, 2, 1) to the plane 2x − y + z + 1 = 0. Also, find the distance PQ and the image of the point P treating this plane as a mirror

Answer 1

The given equation of the plane is 2x - y + z + 1 = 0

A foot of the perpendicular is given by

`("x"-"x"_1)/"a"=("y"-"y"_1)/"b"=("z"-"z"_1)/"c"=-(("ax"_1+"by"_1+"cz"_1-"d"))/("a"^1+"b"^2+"c"^2)`

Therefore, foot of perpendicular Q from point P(3,2,1) to the given plane is given by

`("x"-3)/2=("y"-2)/-1=("z"-1)/1=-((2xx3+(-1)xx2+1xx1+1))/(2^2+(-1)+1^2)=-1`

⇒ x = 1, y = 3, z  = 0

Hence, coordinates of foot of perpendicular Q are (1, 3, 0).

Distance PQ`=|((2xx3+(-1)xx1+1xx1+1))/(sqrt(2^2+(-1) ^2 +1^2))|=sqrt6`

Image of the point P is given by

`("x"-"x"_1)/"a"=("y"-"y"_1)/"b"=("z"-"z"_1)/"c"=-(2("ax"_1+"by"_1+"cz"_1-"d"))/("a"^2+"b"^2+"c"^2)`

Therefore, coordinates of image of the point P are given by

`("x"-3)/2=("y"-2)/-1=("z"-1)/1=-(2[(2xx3+(-1)xx2+1xx1+1)])/(2^2+(-1)^2+1^2)=-2`

⇒ x = - 1, y = 4, z = - 1

Hence, coordinates of image of P are (-1, 4, -1)