Question

Perpendicular distance between the planes 2x-y+2z=1 and 4x-2y+4z=1 is.......,Select the correct option from the given options.
(a) 1/3
(b) 3
(c) 1/6
(d) 6

Answer 1

for solution refer attachment and option c is correct

Answer 2

perpendicular distance between two parallel planes ax + by + cz + d = 0 and ax + by + cz + k = 0 is given by $\frac{|k-d|}{\sqrt{a^2+b^2+c^2}}$

here, first plane : 2x - y + 2z = 1 => 2x - y + 2z - 1= 0
2nd plane : 4x - 2y + 4z = 1 => 2x - y + 2z -0.5 = 0
here, d = -1 , k = -0.5 , a = 2, b = -1 and c = 2

now, distance between planes = $\frac{|-0.5-(-1)|}{\sqrt{2^2+(-1)^2+2^2}}$

=$\frac{0.5}{\sqrt{4+1+4}}$

=$\frac{1}{6}$

hence, option (c) is correct.