I will ignore the hint and use a very straight-forward method
the shortest distance will of course be the normal to the plane going through (1,0,0)
plane:
6x + 9y - z + 1 = 0
a normal to the plane is (6,9,-1)
and passing through (1,0,0) would be the line
x = 6t + 1
y = 9t
z = -t
intersecting this with the plane:
6(6t+1) + 9(9t) -(-t) + 1 = 0
36t + 6 + 81t + t = -1
118t = -7
t = -7/118
point of contact of the normal with plane is
(-42/118 + 1, -63/118, 7/118) = (38/59 , -63/118, 7/118)
expected nicer numbers, but ....
distance between (1,0,0) and the above weird point
= √((-42/118+1 - 1)^2 + (-63/118)^2 + (7/118)^2 )
= √(5782)/118
= appr .6444
Find the point on the plane z = 6x + 9y + 1 closest to the point P = (1, 0, 0). Hint: Minimize the square of the distance. (x, y, z) =
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