A construction company has a job to knock down the old mall

A construction company has a job to knock down the old mall. They havea crane to help them do this. The mall roof is given by the equation .5ry+5z-1 The crane is located at the point (0,0,0).

Solution

The equation of the roof is that of a plane 0.5x +y +5z = 1 or. 0.5 x +y +5z – 1 = 0.

The present distance of the crane from the roof is that of the poinr (0,0,0) from this plane. We know that the formula for the distance of a point (x,y,z) from the plane ax+by+cz+d= 0 is ( |ax+by +cz + d|)/(a² + b² + c² ). Here, a = 0.5, b = 1, c = 5 and d = -1. Also x = y = z = 0. Therefore, the distance of the point (0,0,0) from the above plane is |-1 |/( 0.5²+ 1² + 5² ) = 1/ ( 0.25 + 1 + 25) = 1/(26.25) ft. Since a,b,c, d are fixed, we have to shift the crane from the point (0,0,0) to a point (x, y,z) (say) so that the distance of the point (x,y,z) from the plane is 10 ft. Then, (|0.5x+y+5z -1|)/(26.25) =10 or, (|0.5x+y+5z -1|) = 10(26.25) or (0.5x + y +5z-1) = ± 10(26.25) or, x + 2y +5z -2 = ± 20(26.25). Shifting the crane from the point (0,0,0) to any of the points (x,y,z) which are the solutions to the equation x + 2y +5z -2 = ± 20(26.25) will enable the crane to be 10 feet from the roof.