Question
Sun February 06, 2011 By:

# FIND THE COORDINATES OF THE FOOT OF THE PERPENDICULAR AND THE PERPENDICULAR DISTANCE OF THE POINT P(3,2,1) FROM THE PLANE 2X-Y+Z+1=0. FIND ALSO, THE IMAGE OF THE POINT IN THE PLANE

Sat March 05, 2011
dear student,
we know that the distance of a point  from the plane Ax+By+Cz+d=0
is
modulus of
here the point is (3,2,1) and plane is 2X-Y+Z+1=0
now put these values in the formula and get the answer.
By the reflection of a point in a plane we mean that,
a point which is at the same distance from the plane but on the other side of the plane and the line joining a point and its reflection is perpendicular to the plane.
normal to the given plane is along the direction (2,-1,1)
so the reflection lies along the line
so
x=2k+3
y=-k+2
z=k+1
let for some k, it represents the foot of the perpendicular from the given point to the given plane.
so
2(2k+3)-(-k+2)+(k+1)+1=0
solving we get the value of k
put this value in x,y and z found above to get the reflection.
z=
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