Question

Can someone help me with this as fast as possible...... find the distance of the point ( 2,3,4) from the plane 3x+2y+2z+5=0 measured parallel to the line x+3/3 , y+2/2 , z/2

Answer 1

Point Q = (2,3,4)
Plane P :  3 x + 2y + 2z + 5 = 0
Line L:  (x+3)/3 = (y+2)/2 = z/2 = t
             x = -3 + 3t,    y = -2 + 2t,    z = 2t

Equation of line L' parallel to L and passing through Q(2,3,4):
          x = 2+3t,   y = 3+2t,    z = 4+2t

Intersection point S of plane P and L':
            3 (2+3t) + 2 (3+2t) + 2 (4+2t)+5 = 0              =>       t = -25/17
            S = (2-75/17,  3-50/17, 4-50/17) = (-41/17, 1/17, 18/17)

Distance QS between Q(2,3,4) and S :
$=\sqrt{(2+\frac{41}{17})^2+(3-\frac{1}{17})^2+(4-\frac{18}{17})^2}\\\\=\frac{1}{17}\sqrt{75^2+50^2+50^2}=\frac{25\sqrt{17}}{17}=\frac{25}{\sqrt{17}}$

Answer = 25/√17
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equation of a plane P' parallel to plane P:  3 x + 2y + 2 z + k = 0
Let P' pass through point Q(2,3,4).   Then
       3 *2+2*3+2*4 + k = 0       => k = - 20
equation of P' plane = 3x+2y+2z -20 =0

Given Line L is perpendicular to the given plane.  So we need perpendicular distance between P and P' planes = |5 - (-20)|/√(3²+2²+2²) = 25/√17. 

Perpendicular distance from point Q(2,3,4) onto plane P = 25/√17.