Question

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-2/6=z/2.

Answer 1

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-2/6=z/2 would be √42 ≈ 6.48 units

• Firstly, the distance is to be measured parallel to the line

                             (x + 3)/3 = (y-2)/6 = z/2                           . . .  (1)

• So, the line (say, L) is parallel to the above line and it is connecting the point (2,3,4) to the plane; hence,all it's points are following the relation of the line. So, the point of intersection( say (a,b,c) ) of the plane and line L also follows the relation (1) .

• Lets assume for the point of intersection (a,b,c) the relation is:
•          (a + 3)/3 = (b-2)/6 = c/2  = s                      (s is a real parameter)  
• ∴ a = (3*s - 3) ;   b = (6*s +2);  c = 2*s
• Now, since the point (a,b,c) is also on the plane, it follows the equation of the plane, so:

                3*a + 2*b + 2*c +5 =0

or,            3*( 3*s - 3) + 2*(6*s + 2) + 2*(2*s) + 5 = 0

or,           25*s - 9 + 4+ 5 = 0

or,            s = 0

• Hence, the point of intersection is (a,b,c) = (-3,2,0)
• Now the distance between (2,3,4) and (a,b,c) is

                   √(2+3)² + (3-2)² + (4-0)² = √42 ≈ 6.48 units