Let T be the plane 3x2y2z 13 Find the shortest distance d f

Let T be the plane 3x2y2z = 13. Find the shortest distance d from the point P0=(5, 3, 4) to T, and the point Q in T that is closest to P0. Use the square root symbol \'\' where needed to give an exact value for your answer.

d = ? Q = (?, ?, ?)

Solution

Distance of a point from a plane : d = |ax+by+cz+d |/sqrt(a^2 +b^2+c^2)

= | 3*-5 - 2*-3 -2*-4 |/sqrt(3^2 + 2^2+2^2)

= | -15 +6 +8|/sqrt(9+4+4) = 1/sqrt17

d = 1/sqrt17

Form the line that passes through point Q with a direction vector that is parallel to the normal vector of the plane.

L(x, y, z) = (-5, -3, -4) + t(3, -2, -2)

x = -5 +3t ; y = -3 -2t ; z= -4 -2t

Now substitute this point in the equation plane :

3(-5+3t) - 2(-3 -2t) -2(-4 -2t) = 13

-15 +9t + 6 + 4t +8 +4t =13

17t = 14 ; t= 14/17

Point Q : ( -5 +42/17 , -3 -28/17 , -4 -28/17 )

( -43/17 , -79/17 , -96/17 )