Given plane is 2x – 2y + 4z + 5 = 0 and point (1, 3/2, 2)
The direction ratios of the normal to the plane are 2, -2, 4
So, the equation of the line passing through (1, 3/2, 2) and direction ratios are equal to the direction ratios of the normal to the plane i.e. 2, -2, 4 is
Now, any point in the plane is 2λ + 1, -2λ + 3/2, 4λ + 2
Since, the point lies in the plane, then
2(2λ + 1) – 2(-2λ + 3/2) + 4(4λ + 2) + 5 = 0
4λ + 2 + 4λ – 3 + 16λ + 8 + 5 = 0
24λ + 12 = 0 λ = ½
So, the coordinates of the point in the plane are
2(-1/2) + 1, -2(-1/2) + 3/2, 4(-1/2) + 2 = 0, 5/2, 0
Thus, the foot of the perpendicular is (0, 5/2, 0) and the required length